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Lemma 2.1 Suppose that \( {\mu }_{n - 2} \in \left( {n - 3, n - 2}\right) \) and \( x \in {C}^{n - 2}\left\lbrack {0,1}\right\rbrack, x\left( 0\right) = {x}^{\prime }\left( 0\right) = \cdots = {x}^{n - 3}\left( 0\right) = 0 \) . Then \( {D}_{0 + }^{{\mu }_{n - 2}}x \in C\left\lbrack {0,1}\right\rbrack \) and | \[ {D}_{0 + }^{{\mu }_{n - 2}}x\left( t\right) = \frac{1}{\Gamma \left( {n - 2 - {\mu }_{n - 2}}\right) }{\int }_{0}^{t}{\left( t - s\right) }^{n - 3 - {\mu }_{n - 2}}{x}^{\left( n - 2\right) }\left( s\right) \mathrm{d}s. \] | Yes |
Lemma 2.2 Suppose that \( {\mu }_{i} \in (i - 1, i\rbrack \) for \( i = 1,2,\cdots, n - 3 \) and \( x \in {C}^{n - 2}\left\lbrack {0,1}\right\rbrack \) , \( x\left( 0\right) = {x}^{\prime }\left( 0\right) = \cdots = {x}^{\left( n - 3\right) }\left( 0\right) = 0 \) . Then \( {D}_{0 + }^{{\mu }_{i}}x \in C\left\lbrack {0... | \[ {D}_{0 + }^{{\mu }_{i}}x\left( t\right) = \frac{1}{\Gamma \left( {n - 2 - {\mu }_{i}}\right) }{\int }_{0}^{t}{\left( t - s\right) }^{n - 3 - {\mu }_{i}}{x}^{\left( n - 2\right) }\left( s\right) \mathrm{d}s. \] | Yes |
Lemma 2.3 Given \( h \in {L}^{1}\left\lbrack {0,1}\right\rbrack \), then for \( t \in \left\lbrack {0,1}\right\rbrack \) ,\n\n\[ u\left( t\right) = {\int }_{0}^{1}G\left( {t, s}\right) h\left( s\right) \mathrm{d}s \]\n\n\( \left( {2.3}\right) \)\n\nis the unique solution in \( {C}^{n - 2}\left\lbrack {0,1}\right\rbrack... | Proof Because\n\n\[ u\left( t\right) = \left\{ \begin{array}{l} - {I}_{0 + }^{\alpha }h\left( t\right) + {c}_{1}{t}^{\alpha - 1} + {c}_{2}{t}^{\alpha - 2} + \cdots + {c}_{n}{t}^{\alpha - n},\;\text{ for }n - 1 < \alpha < n; \\ - {I}_{0 + }^{\alpha }h\left( t\right) + {c}_{1}{t}^{\alpha - 1} + {c}_{2}{t}^{\alpha - 2} + ... | Yes |
Lemma 3.1 \( {Q}_{l} : P \rightarrow P \) is a completely continuous operator. | Proof Suppose that \( x \in P \) and \( \rho \left( t\right) = {f}_{l}\left( {t, x\left( t\right) ,{D}_{0 + }^{{\mu }_{1}}x\left( t\right) ,{D}_{0 + }^{{\mu }_{2}}x\left( t\right) ,\cdots ,{D}_{0 + }^{{\mu }_{n - 2}}x\left( t\right) }\right) \) . Then, by (1.4) and (3.1), we have that \( \rho \in {L}^{1}\left\lbrack {0... | Yes |
Theorem 4.1 Let \( \left( {\mathrm{H}}_{1}\right) \) and \( \left( {\mathrm{H}}_{2}\right) \) (or \( \left( {\mathrm{H}}_{3}\right) \) ) hold. Then the problem (1.1),(1.2) has a positive solution \( u \) and, for \( t \in \left\lbrack {0,1}\right\rbrack \) , \[ \left\{ \begin{array}{l} u\left( t\right) \geq M{t}^{\alph... | Proof Theorem 3.1 shows us that problem (1.7),(1.2) has a solution \( {u}_{l} \in P \) . In addition, Lemma 3.2 provides us that \( \left\{ {u}_{l}\right\} \) is relatively compact in \( X \) and satisfies inequality (3.19) for \( t \in \left\lbrack {0,1}\right\rbrack, l \in \mathbb{N} \) . Now the Arzelà - Ascoli theo... | Yes |
Example 4.1 Suppose that \( \rho \in {L}^{1}\left\lbrack {0,1}\right\rbrack \) and \( m \) is a positive constant. Let \( a \in \) \( \left( {0,\frac{1}{\alpha - 1}}\right) ,{b}_{1} \in \left( {0,\frac{1}{n - 1 - {\mu }_{1}}}\right) ,{b}_{2} \in \left( {0,\frac{1}{n - 1 - {\mu }_{2}}}\right) ,\cdots ,{b}_{n - 2} \in \l... | Theorem 4.1 guarantees that for each \( \alpha \in (n - 1, n\rbrack \), and \( {\mu }_{i} \in (i - 1, i\rbrack \), for \( i = 1,2,\cdots, n - 2 \), the problem has a positive solution \( u \) satisfying the boundary condition (1.2) and inequality (4.1) for \( t \in \left\lbrack {0,1}\right\rbrack \) , where \( M = \fra... | Yes |
Corollary 1.1 For any integer \( b \geq 2,\mu \left( {\xi }_{{\varepsilon }^{1}, b}\right) = 1 + \frac{1 + \sqrt{5}}{2} \) . | In fact, Corollary 1.1 gives a new proof of Adamczewski and Allouche's result. To see this, take \( \alpha = \frac{1 + \sqrt{5}}{2} = \left\lbrack {1,1,\cdots }\right\rbrack \), define two binary sequences\n\n\[ \n{g}_{\alpha }\left( n\right) = \left\{ \begin{array}{ll} 1, & \text{ if }n = \lfloor {k\alpha }\rfloor \te... | Yes |
Theorem 1.2 \( {}^{\left\lbrack 1\right\rbrack } \) If \( \alpha = \frac{1 + \sqrt{5}}{2} = \left\lbrack {1,1,\cdots }\right\rbrack \), then for any integer \( b \geq 2 \) , | \[ \mu \left( {{S}_{b}\left( \alpha \right) }\right) = 1 + \frac{1 + \sqrt{5}}{2}. \] | No |
Lemma 2.1 For any integer \( n \geq 2 \), we have the following statements:\n\n1) \( {U}_{n - 1}{U}_{n} = {U}_{n}{U}_{n - 1}^{ * } \) | Proof 1) By induction on \( n \), the case \( n = 2 \) is trivial. Assume that the result is true for all \( n \leq m \) ; we need to prove it for \( n = m + 1 \) .\n\nHence, we have\n\n\[ \n{U}_{m}{U}_{m + 1} = {U}_{m}{U}_{m}^{k}{U}_{m - 1} = {U}_{m}^{k}{U}_{m}{U}_{m - 1} \n\]\n\n\[ \n= {U}_{m}^{k}{U}_{m - 1}^{k}{U}_{... | Yes |
Lemma 2.2 For any \( n \in \mathbb{N} \), we have\n\n1) \( {\varepsilon }_{n} = 0 \) if and only if \( {\tau }_{0}\left( n\right) \in \{ 0,1,\cdots, k - 1\} \),\n\n2) \( {\varepsilon }_{n} = 1 \) if and only if \( {\tau }_{0}\left( n\right) = k \) . | Proof It’s only need to prove that \( {\varepsilon }_{n} = 0 \) if and only if \( {\tau }_{0}\left( n\right) \in \{ 0,1,\cdots, k - 1\} \) by induction on \( n \) . The other statement has the same proof.\n\nIt is true for \( n = 0,1,\cdots, k \), since \( \varepsilon = {\varepsilon }_{0}{\varepsilon }_{1}{\varepsilon ... | No |
Lemma 2.3 Let \( n = {\sum }_{i \geq 0}{n}_{i}{f}_{i} \) with \( {n}_{i} \in \{ 0,1,\cdots, k\} \), Assume that for \( 0 \leq i \leq t \) if \( {n}_{i + 1} = k \), then \( {n}_{i} = 0 \) . Then we have \( {n}_{i} = {\tau }_{i}\left( n\right) \) for \( 0 \leq i \leq t \) . | Proof If there exists \( i \in \mathbb{N} \) such that \( {n}_{i + 1} = k \), but \( {n}_{i} \neq 0 \) . Let \( {i}_{0} = \max \{ i \geq 0 \) : \( \left. {{n}_{i + 1} = k,{n}_{i} \neq 0}\right\} \) . Let \( {j}_{0} = \max \left\{ {j \geq 0 : {n}_{{i}_{0} + 1} = {n}_{{i}_{0} + 3} = \cdots = {n}_{j} = k}\right\} \) . Cle... | Yes |
For any \( i \in \mathbb{N}, n \geq 0,{\varepsilon }_{i + {f}_{n}} \neq {\varepsilon }_{i} \) if and only if \( i{ \equiv }_{n + 1}{f}_{n + 1} - 2 \) or \( i{ \equiv }_{n + 1}{f}_{n + 1} - 1 \) . Moreover, | Proof If \( n = 0 \), it is easy to check that it is true. Now we assume that \( n \geq 1 \) . Hence there are two cases for discussion.\n\nCase 1: \( n \geq 1 \), and \( {\tau }_{n}\left( i\right) \neq k \) . Then, we have\n\n\[ i + {f}_{n} = \mathop{\sum }\limits_{{j = 0}}^{{n - 1}}{\tau }_{j}\left( i\right) {f}_{j} ... | Yes |
Theorem 1.1 Assume \( \left( {\mathrm{f}}_{1}\right) - \left( {\mathrm{f}}_{4}\right) \) hold. Then for each \( 0 < \mu < {\mu }_{N, p} \), problem (1.1) has a sign-changing solution. If in addition \( f\left( {x, t}\right) \) is odd in \( t \), then problem (1.1) has a sequence of sign-changing solutions. | Motivated by these works, we will prove Theorem 1.1 by using the method of invariant sets of gradient flows (see [16])and employ the general idea from [9]. However we need to deal with some difficulties such as verification the compactness of the functional. What's more, given \( u \in X \) in general, \( {u}^{ + } \) ... | No |
Lemma 2.1 I satisfies the (PS) condition. | Proof Step \( 1\left\{ {u}_{n}\right\} \) is bounded in \( X \) .\n\nLet \( \left\{ {u}_{n}\right\} \subset X \) be a (PS) sequence, then\n\n\[ c + o\left( 1\right) \begin{Vmatrix}{u}_{n}\end{Vmatrix} \geq I\left( {u}_{n}\right) - \frac{1}{q}\left\langle {{I}^{\prime }\left( {u}_{n}\right) ,{u}_{n}}\right\rangle \]\n\n... | Yes |
Lemma 3.1 The operator \( A \) is well-defined and continuous. | Proof Given \( u \in X \), define\n\n\[ J\left( v\right) = \frac{1 - \delta }{p}{\int }_{\Omega }{\left| \Delta v\right| }^{p}\mathrm{\;d}x + \frac{\delta }{p}{\int }_{\Omega }{\left| \Delta \left( u - v\right) \right| }^{p}\mathrm{\;d}x + \delta {\int }_{\Omega }{\left| \Delta u\right| }^{p - 2}{\Delta u\Delta v}\math... | Yes |
Lemma 3.2 It holds that for some constant \( c \)\n\n\[ \left\langle {{I}^{\prime }\left( u\right), u - {Au}}\right\rangle \geq \delta \parallel u - {Au}{\parallel }^{p} \]\n\n\[ \begin{Vmatrix}{{I}^{\prime }\left( u\right) }\end{Vmatrix} \leq c\parallel u - {Au}{\parallel }^{p - 1},\;\text{ if }1 < p \leq 2, \]\n\n\[ ... | The proof of this lemma is almost the same as that of [9, Lemma 3.2], we omit it. | No |
Lemma 3.3 There exists \( {\varepsilon }_{0} > 0 \) such that if \( 0 < \varepsilon < {\varepsilon }_{0} \) then \[ A\left( {\partial {P}_{ + }}\right) \subset {P}_{ + },\;A\left( {\partial {P}_{ - }}\right) \subset {P}_{ - }.\] | Proof Given \( u \in X \), define \( w \in X \) by \[ \left( {1 - \delta }\right) {\int }_{\Omega }{\left| \Delta w\right| }^{p - 2}{\Delta w\Delta \varphi }\mathrm{d}x + \delta {\int }_{\Omega }\left( {{\left| \Delta u\right| }^{p - 2}{\Delta u} - {\left| \Delta \left( u - w\right) \right| }^{p - 2}\Delta \left( {u - ... | Yes |
Lemma 4.1 Let \( K \) be the critical point set of \( I \) :\n\n\[ K = \\left\\{ {u \\in X \\mid {I}^{\\prime }\\left( u\\right) = 0}\\right\\} .\n\]\n\nThen there exists a locally Lipschitz continuous operator \( B \) defined on \( {X}_{0} = X \\smallsetminus K \) which inherits the properties of \( A \) . More precis... | For the proof of Lemma 4.1, one can see [12, Lemma 2.1] for example. By (4.2) (4.3)and (3.4) (3.5) in Lemma 3.2, we have\n\n\[ \\begin{Vmatrix}{{I}^{\\prime }\\left( u\\right) }\\end{Vmatrix} \\leq c\\parallel u - {Bu}{\\parallel }^{p - 1},\\text{ if }1 < p \\leq 2,\n\]\n\n\[ \\begin{Vmatrix}{{I}^{\\prime }\\left( u\\r... | Yes |
1) Assume \( {\varepsilon }_{0} \) sufficiently small. Then for \( 0 < \varepsilon < {\varepsilon }_{0} \)\n\n\[ I\left( u\right) \geq {c\varepsilon }\text{for}u \in \partial {P}_{ + } \cap \partial {P}_{ - }\text{.} \] | 1) By the conditions \( \left( {f}_{2}\right) \) and \( \left( {f}_{3}\right) \), there exist some constants such that\n\n\[ I\left( u\right) = \frac{1}{p}{\int }_{\Omega }\left( {{\left| \Delta u\right| }^{p} - \frac{\mu }{{\left| x\right| }^{2p}}{\left| u\right| }^{p}}\right) \mathrm{d}x - {\int }_{\partial \Omega }F... | Yes |
Lemma 4.3 Let \( N \) be an open neighborhood of \( {\widetilde{K}}_{c} \) . Then there exist \( {\varepsilon }_{0},{\varepsilon }_{1},{\varepsilon }_{2} > 0 \) such that for \( 0 < {\varepsilon }_{1} < {\varepsilon }_{2} < {\varepsilon }_{0} \) there exists a continuous map \( \sigma : \left\lbrack {0,1}\right\rbrack ... | For the proof of Lemma 4.3, one can see [9, Lemma 4.3] for example. | No |
Lemma 2.1 Assume \( d\left( \delta \right) \) is defined in (2.7), then the following properties hold.\n\n1) \( J\left( u\right) \geq a\left( \delta \right) {r}^{2}\left( \delta \right) \), for \( a\left( \delta \right) = \frac{1}{2} - \frac{\delta }{p + 1},0 < \delta < \frac{p + 1}{2}, r\left( \delta \right) = {\left(... | Proof By the definition of \( d\left( \delta \right) \) and fundamental knowledge of calculus, the lemma can be easily proved, and so we omit the proof here. | No |
Lemma 3.1 Let nonlinearity \( f\left( u\right) \) satisfies \( \left( *\right) ,{u}_{0} \in {H}_{0}^{1}\left( \Omega \right) ,{u}_{1} \in H\left( \Omega \right) \), then \( F\left( {u}_{0}\right) \in \) \( {L}^{1}\left( \Omega \right) \), and \( {E}_{m}\left( 0\right) \rightarrow E\left( 0\right) \) as \( m \rightarrow... | Proof Using the definition of \( F\left( u\right) \), we have\n\n\[ {\int }_{\Omega }\left| {F\left( u\right) }\right| \mathrm{d}x = {\int }_{\Omega }{\left| \frac{1}{p + 1}{\left| u\right| }^{p + 1}\right| }^{\mathrm{d}x} = \frac{1}{p + 1}\parallel u{\parallel }_{p + 1}^{p + 1} \leq C\parallel u{\parallel }_{{H}_{0}^{... | Yes |
Lemma 3.2 Let \( {u}_{0} \in {H}_{0}^{1}\left( \Omega \right) ,{u}_{1} \in H\left( \Omega \right), E\left( 0\right) < d \), and let\n\n\[ \n{W}^{\prime } = \left\{ {{u}_{0} \in {H}_{0}^{1}\left( \Omega \right) : I\left( u\right) > 0}\right\} \cup \{ 0\} .\n\]\n\nIf \( {u}_{0} \in {W}^{\prime } \), then \( {u}_{m} \in {... | Proof If there exists a \( {t}^{ * } > 0 \) such that \( {u}_{m}\left( {t}^{ * }\right) \notin {W}^{\prime } \) for sufficiently large \( m \) . From the continuity of \( I\left( {u}_{m}\right) \) corresponding to time variable \( t \), we must have a \( {t}_{0} > 0 \) such that \( {u}_{m}\left( {t}_{0}\right) \in \par... | Yes |
Theorem 3.1 Let \( {u}_{0} \in {H}_{0}^{1}\left( \Omega \right) ,{u}_{1} \in H\left( \Omega \right) \), Assume that \( E\left( 0\right) < d,{u}_{0} \in {W}^{\prime },\forall T > 0 \) , problem (1.1)-(1.3) admits a weak solution \( u \in {L}^{\infty }\left( {\left\lbrack {0, T}\right\rbrack ;{H}_{0}^{1}\left( \Omega \ri... | Proof Let \( {u}_{m}\left( {x, t}\right) \) be the approximate solution of problem (1.1)-(1.3) constructed by (3.3)-(3.6). From (3.20), we find that \n\n\[ {\begin{Vmatrix}{u}_{m}\end{Vmatrix}}_{{H}_{0}^{1}}^{2} \leq \frac{2\left( {p + 1}\right) }{p - 1}d;{\begin{Vmatrix}{u}_{m}\end{Vmatrix}}_{p + 1}^{2} \leq {C}^{2}{\... | Yes |
Lemma 4.1 If \( u \in {H}_{0}^{1}\left( \Omega \right) \), and \( {I}_{\delta }\left( u\right) < 0 \), then \( \parallel u{\parallel }_{{H}_{0}^{1}} > r\left( \delta \right) \), where \( r\left( \delta \right) = {\left( \frac{\delta }{{C}^{p + 1}}\right) }^{\frac{1}{p - 1}} \) . In particular, if \( I\left( u\right) < ... | Proof It is easy to see that \( \parallel u{\parallel }_{{H}_{0}^{1}} \neq 0 \) from \( {I}_{\delta }\left( u\right) < 0 \) . Thus, from\n\n\[ \delta \parallel u{\parallel }_{{H}_{0}^{1}}^{2} < \parallel u{\parallel }_{p + 1}^{p + 1} \leq {C}^{p + 1}\parallel u{\parallel }_{{H}_{0}^{1}}^{p - 1}\parallel u{\parallel }_{... | Yes |
Lemma 4.2 Assume that \( u \in {H}_{0}^{1}\left( \Omega \right) ,0 < E\left( 0\right) < d \), and \( {\delta }_{1} < {\delta }_{2} \) are the two roots of equation \( E\left( 0\right) = d\left( \delta \right) \) . Then the sign of \( {I}_{\delta }\left( u\right) \) is unchangeable for \( \delta \in \left( {{\delta }_{1... | Proof First, \( E\left( 0\right) > 0 \) implies that \( \parallel u{\parallel }_{{H}_{0}^{1}} \neq 0 \) . If the sign of \( {I}_{\delta }\left( u\right) \) is changeable for \( {\delta }_{1} < \delta < {\delta }_{2} \), there exists a \( {\delta }^{ * } \in \left( {{\delta }_{1},{\delta }_{2}}\right) \) such that \( {I... | Yes |
Lemma 4.3 Let \( {u}_{0} \in {H}_{0}^{1}\left( \Omega \right) ,{u}_{1} \in H\left( \Omega \right) ,0 < e < d,{\delta }_{1} < {\delta }_{2} \) are two roots of equation \( d\left( \delta \right) = E\left( 0\right) ,\delta \in \left( {{\delta }_{1},{\delta }_{2}}\right) \) . Then, we have\n\n1) All solutions of problem (... | Proof 1) Let \( u\left( t\right) \) be any weak solution of problem (1.1)-(1.3) with \( 0 < E\left( 0\right) \leq e \) and \( I\left( {u}_{0}\right) > 0 \) or \( {\begin{Vmatrix}{u}_{0}\end{Vmatrix}}_{{H}_{0}^{1}} \neq 0 \) . \( T \) be the maximal existence time of \( u\left( t\right) \) . First, from \( 0 < E\left( 0... | Yes |
Theorem 1.1 If \( 0 \leq {f}_{0}\left( {x, v}\right) \in {L}^{1} \cap {L}^{\infty }\left( {\mathbb{R}}^{6}\right) ,\left( {{\xi }_{0},{\eta }_{0}}\right) \in {\mathbb{R}}^{3} \times {\mathbb{R}}^{3} \) and \( {f}_{0}\left( {x, v}\right) \) satisfies\n\n\[ \n{\int }_{{\mathbb{R}}^{6}}{f}_{0}\left( {x, v}\right) \mathrm{... | ## 2. Proof of Theorem 1.1\n\nIn order to prove Theorem 1.1, we first introduce some notations and definitions and make some estimates of the electric field \( E\left( {t, x}\right) \) .\n\nUnder the same framework of [2], let \( \left( {f\left( {t, x, v}\right) ,\xi \left( t\right) }\right) \) be the classical solutio... | Yes |
Lemma 2.1 Let \( \left( {f\left( {t, x, v}\right) ,\xi \left( t\right) }\right) \) be a classical solution of system (1.1). Then \( \parallel f\left( t\right) {\parallel }_{p} \) for \( p \in \left\lbrack {1,\infty }\right\rbrack \) and the energy \( \varepsilon \left( {f\left( t\right) }\right) \) are conserved, i.e. ... | \[ \parallel f\left( t\right) {\parallel }_{p} = {\begin{Vmatrix}{f}_{0}\end{Vmatrix}}_{p},\;\varepsilon \left( {f\left( t\right) }\right) = \varepsilon \left( {f}_{0}\right) . \] | Yes |
Theorem 2.1 If the \( d \) -band biorthogonal wavelets determined by the filters \( h,{g}^{1},{g}^{2},\cdots \) , \( {g}^{d - 1},\widetilde{h},{\widetilde{g}}^{1},{\widetilde{g}}^{2},\cdots ,{\widetilde{g}}^{d - 1} \) have the perfect reconstruction, then the sub-band operator \( T \) is a reversible bounded linear ope... | The proof of Theorem 2.1 are trivial. | No |
Lemma 2.1 Retaining the definitions and notations as above, we have\n\n1) \( \parallel Q\parallel = \parallel T{\parallel }^{2} \) ;\n\n2) \( {T}^{-1} = {\widetilde{T}}^{ * } = {\overline{\widetilde{A}}}^{\mathrm{T}},{\widetilde{T}}^{-1} = {T}^{ * } = {\bar{A}}^{\mathrm{T}} \) ;\n\n3) \( \parallel Q\parallel = \mathop{... | Proof 1) and 2) are trivial according to the operator theory \( {}^{\left\lbrack {12}\right\rbrack } \) . 3) follows the fact that \( Q \) is a self-adjoint operator due to \( {Q}^{ * } = Q \) . | No |
Example 2.1 Let the lengths of scaling filters be \( \left( {{15},9}\right) \) . Assume that the scaling symbols \( \left( {{H}_{0}\left( z\right) ,{\widetilde{H}}_{0}\left( z\right) }\right) \) have the following form\n\n\[ \n{H}_{0}\left( z\right) = {\left( \frac{1 + z + {z}^{2}}{3}\right) }^{5}Q\left( z\right) ,{\wi... | We can obtain the associated scaling filters as follows \( {}^{\left\lbrack {16}\right\rbrack } \):\n\n\[ \n{h}_{0} \approx \lbrack {0.0302708750},{0.0197271260},{0.0109853080}, - {0.12261759},{0.011382944},{0.24928687}\text{,}\n\]\n\n\[ \n{0.83207454},{0.93777986}\rbrack \text{,}\n\]\n\n\[ \n{\widetilde{h}}_{0} \appro... | Yes |
Theorem 1.1 If \( \mathop{\prod }\limits_{{i = 1}}^{k}\frac{{p}_{i}{q}_{i}}{{p}_{i} - 1} \leq {2}^{k} \) (i.e. \( \left| A\right| \geq 0 \) ), then every solution of (1.1) is global. | Due to the coupled relationship of the boundary flux in (1.1), we say that any blowup of the solutions must be simultaneous blow-up, which means that \( {\begin{Vmatrix}{u}_{i}\left( \cdot, t\right) \end{Vmatrix}}_{\infty } \rightarrow + \infty \) \( \left( {i = 1,2,\ldots, k}\right) \) as \( t \rightarrow T \) is equi... | No |
Theorem 1.2 Assume \( \mathop{\prod }\limits_{{i = 1}}^{k}\frac{{p}_{i}{q}_{i}}{{p}_{i} - 1} > {2}^{k} \) (i.e. \( \left| A\right| < 0 \) ).\n\n(i) If \( {\alpha }_{i}\left( {{k}_{i} - 1}\right) + 1 > 0, i = 1,2,\ldots, k \), then there exist both global solutions and blow-up solutions;\n\n(ii) If \( {\alpha }_{i}\left... | ## 2. Proof of Theorem 1.2\n\nIn order to prove Theorem 1.2, we introduce several lemmas based on the construction of self-similar super-solutions and supsolutions by the comparison principle.\n\nLemma 2.1 The solutions | No |
Lemma 2.2 The set \( {\mathbb{R}}_{ + }^{3} = \{ \left( {x, y, z}\right) \mid x > 0, y > 0, z > 0\} \) is the positively invariant set of system (1.2). | Proof It follows from the initial value condition \( x\left( 0\right) = {\phi }_{1}(0 > 0, y\left( 0\right) = {\phi }_{2}\left( 0\right) > \) \( 0, z\left( 0\right) = {\phi }_{3}\left( 0\right) > 0 \) that\n\n\[ \left\{ \begin{array}{l} x\left( t\right) = x\left( 0\right) \exp \left\{ {{\int }_{0}^{t}\left\lbrack {r\le... | Yes |
Theorem 2.1 Let \( {M}_{1},{M}_{2},{M}_{3},{m}_{1},{m}_{2} \) and \( {m}_{3} \) be defined by \( \left( {2.2}\right) \left( {2.4}\right) \left( {2.6}\right) \left( {2.9}\right) \left( {2.13}\right) \) and (2.16), respectively. In addition to the condition (H1), suppose that the following condition\n\n(H2) \( {\beta }_{... | Proof Obviously, system (1.2) with the initial value condition \( \left( {x\left( 0\right), y\left( 0\right), z\left( 0\right) }\right) \) has positive solution \( \left( {x\left( t\right), y\left( t\right), z\left( t\right) }\right) \) passing through \( \left( {x\left( 0\right), y\left( 0\right), z\left( 0\right) }\r... | Yes |
Theorem 1.1 For all \( k \geq 1 \), we have\n\n\[ g\left( k\right) \leq k + 2 \]\n\nFurthermore, the equality in the above occurs if and only if \( k = {2}^{{2r} + 1} - 1 \) . | This paper mainly proved that for any integer \( k \geq 1 \) there is an integer \( n \leq k + 2 \) such that \( {t}_{nk} = 0 \) . In order to prove our main result, we use different ideas. On the one hand we show for a large set of integers \( k \) that there exists an integer \( n \leq k + 2 \) such that \( {t}_{kn} ... | No |
Lemma 2.1 If \( k = {2}^{{2r} + 1} - 1 \) for some \( r \geq 0 \), we have \( g\left( k\right) = k + 2 \) . | Proof If \( k = {2}^{{2r} + 1} - 1 \), set \( n = k + 2 = {2}^{{2r} + 1} + 1 \), we have\n\n\[ \n{kn} = \left( {{2}^{{2r} + 1} - 1}\right) \left( {{2}^{{2r} + 1} + 1}\right) = {2}^{{4r} + 2} - 1.\n\]\n\nHence\n\n\[ \n{t}_{kn} \equiv {s}_{2}\left( {kn}\right) \equiv {4r} + 2 \equiv 0\left( {\;\operatorname{mod}\;2}\righ... | Yes |
Lemma 2.2 Let \( k \in \mathbb{N} \) . If there exists an even integer \( u \geq 2 \) with \( {L}_{u + 1}\left( k\right) = {01}^{u} \), then we have \( g\left( k\right) \leq k \) . | Proof Let \( \ell = \ell \left( k\right) \) and set \( n = {2}^{\ell - 1} + 1 < k \) . In the following we are going to show \( {t}_{kn} = 0 \) . We have\n\n\[ \n{\left( kn\right) }_{2} = {U}_{\ell - \left( {u + 1}\right) }\left( k\right) {10}^{u}{L}_{\ell - 1}\left( k\right) .\n\]\n\nThe following figure explains the ... | No |
Lemma 2.3 Let \( k \in \mathbb{N} \) . If \( {L}_{2}\left( k\right) = {01} \) . Then we have \( g\left( k\right) < k \) . | Proof The proof will be divided into four cases.\n\nCase 1 If \( {U}_{2}\left( k\right) = {10} \), set \( n = {2}^{\ell - 2} + 1 < k \), we have\n\n\[{\left( nk\right) }_{2} = {U}_{\ell - 2}\left( k\right) {11}{L}_{\ell - 2}\left( k\right) .\]\n\nThe following figure explains this fact:\n\n\[ \begin{matrix} \cdots {01}... | Yes |
Lemma 2.4 Let \( k \in \mathbb{N} \) . If there exists an odd integer \( u \geq 3 \) with \( {U}_{2}\left( k\right) = {10} \) and \( {L}_{u + 1}\left( k\right) = {01}^{u} \) . Then we have \( g\left( k\right) < k \) . | Proof Set \( n = {2}^{\ell - 2} + 1 < k \), and we have\n\n\[ \n{\left( kn\right) }_{2} = {U}_{\ell \left( k\right) - \left( {u + 1}\right) }\left( k\right) {10}^{u - 1}1{L}_{\ell - 2}\left( k\right) , \n\]\n\nas illustrated below:\n\n\[ \n\begin{matrix} \cdots {01}^{u - 2}{11} \\ {10}\cdots \\ \cdots {10}^{u - 2}{01}\... | Yes |
Lemma 2.5 Let \( k \in \mathbb{N} \) . If there exists an odd integer \( u \geq 3 \) with \( {U}_{2}\left( k\right) = {11} \) and \( {L}_{u + 1}\left( k\right) = {01}^{u} \) . Then we have \( g\left( k\right) < k \) . | The proof is similar to Lemma 2.3, here we omit the proof. | No |
Theorem 1.1 (Hölder) Suppose \( 1 < p < \infty \) and \( 1 < q < \infty \) are conjugate exponents. If \( f \in {L}^{p} \) and \( g \in {L}^{q} \), then \( {fg} \in {L}^{1} \) and\n\n\[ \parallel {fg}{\parallel }_{{L}^{1}} \leq \parallel f{\parallel }_{{L}^{p}}\parallel g{\parallel }_{{L}^{q}} \] | The proof of the theorem relies on a simple generalized form of the arithmetic-geometric mean inequality: if \( A, B \geq 0 \), and \( 0 \leq \theta \leq 1 \), then\n\n(2)\n\n\[ {A}^{\theta }{B}^{1 - \theta } \leq {\theta A} + \left( {1 - \theta }\right) B. \]\n\nTo establish (2), we observe first that we may assume \(... | Yes |
Theorem 1.2 (Minkowski) If \( 1 \leq p < \infty \) and \( f, g \in {L}^{p} \), then \( f + g \in \) \( {L}^{p} \) and \( \parallel f + g{\parallel }_{{L}^{p}} \leq \parallel f{\parallel }_{{L}^{p}} + \parallel g{\parallel }_{{L}^{p}} \) . | Proof. The case \( p = 1 \) is obtained by integrating \( \left| {f\left( x\right) + g\left( x\right) }\right| \leq \) \( \left| {f\left( x\right) }\right| + \left| {g\left( x\right) }\right| \) . When \( p > 1 \), we may begin by verifying that \( f + g \in {L}^{p} \) , when both \( f \) and \( g \) belong to \( {L}^{... | Yes |
Proposition 1.4 If \( X \) has finite positive measure, and \( {p}_{0} \leq {p}_{1} \), then \( {L}^{{p}_{1}}\left( X\right) \subset {L}^{{p}_{0}}\left( X\right) \) and\n\n\[ \frac{1}{\mu {\left( X\right) }^{1/{p}_{0}}}\parallel f{\parallel }_{{L}^{{p}_{0}}} \leq \frac{1}{\mu {\left( X\right) }^{1/{p}_{1}}}\parallel f{... | We may assume that \( {p}_{1} > {p}_{0} \) . Suppose \( f \in {L}^{{p}_{1}} \), and set \( F = {\left| f\right| }^{{p}_{0}} \) , \( G = 1, p = {p}_{1}/{p}_{0} > 1 \), and \( 1/p + 1/q = 1 \), in Hölder’s inequality applied to \( F \) and \( G \) . This yields\n\n\[ \parallel f{\parallel }_{{L}^{{p}_{0}}}^{{p}_{0}} \leq... | Yes |
Proposition 1.5 If \( X = \mathbb{Z} \) is equipped with counting measure, then the reverse inclusion holds, namely \( {L}^{{p}_{0}}\left( \mathbb{Z}\right) \subset {L}^{{p}_{1}}\left( \mathbb{Z}\right) \) if \( {p}_{0} \leq {p}_{1} \) . Moreover, \( \parallel f{\parallel }_{{L}^{{p}_{1}}} \leq \parallel f{\parallel }_... | Indeed, if \( f = \{ f\left( n\right) {\} }_{n \in \mathbb{Z}} \), then \( \sum {\left| f\left( n\right) \right| }^{{p}_{0}} = \parallel f{\parallel }_{{L}^{{p}_{0}}}^{{p}_{0}} \), and \( \mathop{\sup }\limits_{n}\left| {f\left( n\right) }\right| \leq \) \( \parallel f{\parallel }_{{L}^{{p}_{0}}} \) . However\n\n\[ \su... | Yes |
Proposition 2.2 Suppose \( f \in {L}^{\infty } \) is supported on a set of finite measure. Then \( f \in {L}^{p} \) for all \( p < \infty \), and\n\n\[ \parallel f{\parallel }_{{L}^{p}} \rightarrow \parallel f{\parallel }_{{L}^{\infty }}\;\text{ as }p \rightarrow \infty . \]\n | Proof. Let \( E \) be a measurable subset of \( X \) with \( \mu \left( E\right) < \infty \), and so that \( f \) vanishes in the complement of \( E \) . If \( \mu \left( E\right) = 0 \), then \( \parallel f{\parallel }_{{L}^{\infty }} = \) \( \parallel f{\parallel }_{{L}^{p}} = 0 \) and there is nothing to prove. Othe... | Yes |
Proposition 3.1 A linear functional on a Banach space is continuous, if and only if it is bounded. | Proof. The key is to observe that \( \ell \) is continuous if and only if \( \ell \) is continuous at the origin.\n\nIndeed, if \( \ell \) is continuous, we choose \( \epsilon = 1 \) and \( g = 0 \) in the above definition so that \( \left| {\ell \left( f\right) }\right| \leq 1 \) whenever \( \parallel f\parallel \leq ... | Yes |
Theorem 3.2 The vector space \( {\mathcal{B}}^{ * } \) is a Banach space. | Proof. It is clear that \( \parallel \cdot \parallel \) defines a norm, so we only check that \( {\mathcal{B}}^{ * } \) is complete. Suppose that \( \left\{ {\ell }_{n}\right\} \) is a Cauchy sequence in \( {\mathcal{B}}^{ * } \) . Then, for each \( f \in \mathcal{B} \), the sequence \( \left\{ {{\ell }_{n}\left( f\rig... | Yes |
Theorem 4.1 Suppose \( 1 \leq p < \infty \), and \( 1/p + 1/q = 1 \) . Then, with \( \mathcal{B} = \) \( {L}^{p} \) we have\n\n\[{\mathcal{B}}^{ * } = {L}^{q}\]\n\nin the following sense: For every bounded linear functional \( \ell \) on \( {L}^{p} \) there is a unique \( g \in {L}^{q} \) so that\n\n\[ \ell \left( f\ri... | The proof of the theorem is based on two ideas. The first, as already seen, is Hölder's inequality; to which a converse is also needed. The second is the fact that a linear functional \( \ell \) on \( {L}^{p},1 \leq p < \infty \), leads naturally to a (signed) measure \( \nu \) . Because of the continuity of \( \ell \)... | Yes |
Lemma 4.2 Suppose \( 1 \leq p, q \leq \infty \), are conjugate exponents.\n\n(i) If \( g \in {L}^{q} \), then \( \parallel g{\parallel }_{{L}^{q}} = \mathop{\sup }\limits_{{\parallel f{\parallel }_{{L}^{p}} \leq 1}}\left| {\int {fg}}\right| \) . | Proof. We start with (i). If \( g = 0 \), there is nothing to prove, so we may assume that \( g \) is not 0 a.e., and hence \( \parallel g{\parallel }_{{L}^{q}} \neq 0 \) . By Hölder’s inequality, we have that\n\n\[ \parallel g{\parallel }_{{L}^{q}} \geq \mathop{\sup }\limits_{{\parallel f{\parallel }_{{L}^{p}} \leq 1}... | Yes |
Theorem 5.2 Suppose \( {V}_{0} \) is a linear subspace of \( V \), and that we are given a linear functional \( {\ell }_{0} \) on \( {V}_{0} \) that satisfies\n\n\[{\ell }_{0}\left( v\right) \leq p\left( v\right) ,\;\text{ for all }v \in {V}_{0}.\n\]\n\nThen \( {\ell }_{0} \) can be extended to a linear functional \( \... | Proof. Suppose \( {V}_{0} \neq V \), and pick \( {v}_{1} \) a vector not in \( {V}_{0} \) . We will first extend \( {\ell }_{0} \) to the subspace \( {V}_{1} \) spanned by \( {V}_{0} \) and \( {v}_{1} \), as we did before. We can do this by defining a putative extension \( {\ell }_{1} \) of \( {\ell }_{0} \), defined o... | No |
Proposition 5.3 Suppose \( {f}_{0} \) is a given element of \( \mathcal{B} \) with \( \begin{Vmatrix}{f}_{0}\end{Vmatrix} = M \) . Then there exists a continuous linear functional \( \ell \) on \( \mathcal{B} \) so that \( \ell \left( {f}_{0}\right) = M \) and \( \parallel \ell {\parallel }_{{\mathcal{B}}^{ * }} = 1 \)... | Proof. Define \( {\ell }_{0} \) on the one-dimensional subspace \( {\left\{ \alpha {f}_{0}\right\} }_{\alpha \in \mathbb{R}} \) by \( {\ell }_{0}\left( {\alpha {f}_{0}}\right) = {\alpha M} \), for each \( \alpha \in \mathbb{R} \) . Note that if we set \( p\left( f\right) = \parallel f\parallel \) for every \( f \in \ma... | Yes |
Proposition 5.4 Let \( {\mathcal{B}}_{1},{\mathcal{B}}_{2} \) be a pair of Banach spaces and \( \mathcal{S} \subset {\mathcal{B}}_{1} \) a dense linear subspace of \( {\mathcal{B}}_{1} \) . Suppose \( {T}_{0} \) is a linear transformation from \( \mathcal{S} \) to \( {\mathcal{B}}_{2} \) that satisfies \( {\begin{Vmatr... | Proof. If \( f \in {\mathcal{B}}_{1} \), let \( \left\{ {f}_{n}\right\} \) be a sequence in \( \mathcal{S} \) which converges to \( f \) . Then since \( {\begin{Vmatrix}{T}_{0}\left( {f}_{n}\right) - {T}_{0}\left( {f}_{m}\right) \end{Vmatrix}}_{{\mathcal{B}}_{2}} \leq M{\begin{Vmatrix}{f}_{n} - {f}_{m}\end{Vmatrix}}_{{... | Yes |
Theorem 5.5 The operator \( {T}^{ * } \) defined by (13) is a bounded linear transformation from \( {\mathcal{B}}_{2}^{ * } \) to \( {\mathcal{B}}_{1}^{ * } \) . Its norm \( \begin{Vmatrix}{T}^{ * }\end{Vmatrix} \) satisfies \( \parallel T\parallel = \begin{Vmatrix}{T}^{ * }\end{Vmatrix} \) . | Proof. First, if \( {\begin{Vmatrix}{f}_{1}\end{Vmatrix}}_{{\mathcal{B}}_{1}} \leq 1 \), we have that\n\n\[ \left| {{\ell }_{1}\left( {f}_{1}\right) }\right| = \left| {{\ell }_{2}\left( {T\left( {f}_{1}\right) }\right) }\right| \leq \begin{Vmatrix}{\ell }_{2}\end{Vmatrix}{\begin{Vmatrix}T\left( {f}_{1}\right) \end{Vmat... | Yes |
Theorem 5.6 There is an extended-valued non-negative function \( \widehat{m} \), defined on all subsets of \( \mathbb{R} \) with the following properties:\n\n(i) \( \widehat{m}\left( {{E}_{1} \cup {E}_{2}}\right) = \widehat{m}\left( {E}_{1}\right) + \widehat{m}\left( {E}_{2}\right) \) whenever \( {E}_{1} \) and \( {E}_... | From (i) we see that \( \widehat{m} \) is finitely additive; however it cannot be countably additive as the proof of the existence of non-measurable sets shows. (See Section 3, Chapter 1 in Book III.) | No |
Corollary 5.8 There is a non-negative function \( \widehat{m} \) defined on all subsets of \( \mathbb{R}/\mathbb{Z} \) so that:\n\n(i) \( \widehat{m}\left( {{E}_{1} \cup {E}_{2}}\right) = \widehat{m}\left( {E}_{1}\right) + \widehat{m}\left( {E}_{2}\right) \) for all disjoint subsets \( {E}_{1} \) and \( {E}_{2} \) .\n\... | We need only take \( \widehat{m}\left( E\right) = I\left( {\chi }_{E}\right) \), with \( I \) as in Theorem 5.7, where \( {\chi }_{E} \) denotes the characteristic function of \( E \) . | Yes |
Theorem 7.3 Let \( X \) be a compact metric space and \( C\\left( X\\right) \) the Banach space of continuous real-valued functions on \( X \) . Then, given any bounded linear functional \( \\ell \) on \( C\\left( X\\right) \), there exists a unique finite signed Borel measure \( \\mu \) on \( X \) so that\n\n\\[ \n\\e... | Proof. By the proposition, there exist two positive linear functionals \( {\\ell }^{ + } \) and \( {\\ell }^{ - } \) so that \( \\ell = {\\ell }^{ + } - {\\ell }^{ - } \) . Applying Theorem 7.1 to each of these positive linear functionals yields two finite Borel measures \( {\\mu }_{1} \) and \( {\\mu }_{2} \) . If we ... | Yes |
Theorem 7.4 Suppose \( X \) is a metric space and \( \ell \) a positive linear functional on \( {C}_{b}\left( X\right) \) . For simplicity assume that \( \ell \) is normalized so that \( \ell \left( 1\right) = 1 \) . Assume also that for each \( \epsilon > 0 \), there is a compact set \( {K}_{\epsilon } \subset X \) so... | The proof of this theorem proceeds as that of Theorem 7.1, save for one key aspect. First we define\n\n\[ \rho \left( \mathcal{O}\right) = \sup \left\{ {\ell \left( f\right) ,\text{ where }f \in {C}_{b}\left( X\right) ,\operatorname{supp}\left( f\right) \subset \mathcal{O}\text{, and }0 \leq f \leq 1}\right\} .\n\]\n\n... | Yes |
Lemma 2.2 (Three-lines lemma) Suppose \( \Phi \left( z\right) \) is a holomorphic function in the strip \( S = \{ z \in \mathbb{C} : 0 < \operatorname{Re}\left( z\right) < 1\} \), that is also continuous and bounded on the closure of \( S \) . If\n\n\[ \n{M}_{0} = \mathop{\sup }\limits_{{y \in \mathbb{R}}}\left| {\Phi ... | Proof. We begin by proving the lemma under the assumption that \( {M}_{0} = {M}_{1} = 1 \) and \( \mathop{\sup }\limits_{{0 \leq x \leq 1}}\left| {\Phi \left( {x + {iy}}\right) }\right| \rightarrow 0 \) as \( \left| y\right| \rightarrow \infty \) . In this case, let \( M = \sup \left| {\Phi \left( z\right) }\right| \) ... | Yes |
Corollary 2.3 With \( T \) as before:\n\n(a) The Riesz diagram of \( T \) is a convex set.\n\n(b) \( \log {M}_{x, y} \) is a convex function on this set. | Conclusion (a) means that if \( \left( {{x}_{0},{y}_{0}}\right) = \left( {1/{p}_{0},1/{q}_{0}}\right) \) and \( \left( {{x}_{1},{y}_{1}}\right) = \) \( \left( {1/{p}_{1},1/{q}_{1}}\right) \) are points in the Riesz diagram of \( T \), then so is the line segment joining them. This is an immediate consequence of Theorem... | Yes |
Corollary 2.4 If \( 1 \leq p \leq 2 \) and \( 1/p + 1/q = 1 \), then\n\n\[ \parallel T\left( f\right) {\parallel }_{{L}^{q}\left( \mathbb{Z}\right) } \leq \parallel f{\parallel }_{{L}^{p}\left( \left\lbrack {0,{2\pi }}\right\rbrack \right) }.\] | Note that since \( {L}^{2}\left( \left\lbrack {0,{2\pi }}\right\rbrack \right) \subset {L}^{1}\left( \left\lbrack {0,{2\pi }}\right\rbrack \right) \) and \( {L}^{2}\left( \mathbb{Z}\right) \subset {L}^{\infty }\left( \mathbb{Z}\right) \) we have \( {L}^{2}\left( \left\lbrack {0,{2\pi }}\right\rbrack \right) + {L}^{1}\l... | Yes |
Corollary 2.5 If \( 1 \leq p \leq 2 \) and \( 1/p + 1/q = 1 \), then\n\n\[ \n{\begin{Vmatrix}{T}^{\prime }\left( \left\{ {a}_{n}\right\} \right) \end{Vmatrix}}_{{L}^{q}\left( \left\lbrack {0,{2\pi }}\right\rbrack \right) } \leq {\begin{Vmatrix}\left\{ {a}_{n}\right\} \end{Vmatrix}}_{{L}^{p}\left( \mathbb{Z}\right) }.\n... | The proof is parallel to that of the previous corollary. The case \( {p}_{0} = \) \( {q}_{0} = 2 \) is, as has already been mentioned, a consequence of Parseval’s identity, while the case \( {p}_{1} = 1 \) and \( {q}_{1} = \infty \) follows directly from the fact that\n\n\[ \n\left| {\mathop{\sum }\limits_{{n = - \inft... | No |
Theorem 3.2 Suppose \( 1 < p < \infty \) . Then the Hilbert transform \( H \), initially defined on \( {L}^{2} \cap {L}^{p} \) by (13) or (14), satisfies the inequality\n\n\[ \parallel H\left( f\right) {\parallel }_{{L}^{p}} \leq {A}_{p}\parallel f{\parallel }_{{L}^{p}},\;\text{ whenever }f \in {L}^{2} \cap {L}^{p}, \]... | ## 3.3 Proof of Theorem 3.2\n\nThe main idea of the proof was already outlined at the end of Section 1 in the context of Fourier series and the corresponding theorem for the conjugate function. While this proof, which depends on complex analysis, is elegant, its approach is essentially limited to this operator and cann... | No |
Theorem 4.1 Suppose \( f \in {L}^{p}\left( {\mathbb{R}}^{d}\right) \) with \( 1 < p \leq \infty \) . Then \( {f}^{ * } \in {L}^{p}\left( {\mathbb{R}}^{d}\right) \) , and (26) holds, namely\n\n\[ \n{\begin{Vmatrix}{f}^{ * }\end{Vmatrix}}_{{L}^{p}} \leq {A}_{p}\parallel f{\parallel }_{{L}^{p}}\n\]\n\nThe bound \( {A}_{p}... | Let us first see why \( {f}^{ * }\left( x\right) < \infty \), for a.e. \( x \), whenever \( f \in {L}^{p} \) . Observe that we can decompose \( f = {f}_{1} + {f}_{\infty } \), where \( {f}_{1}\left( x\right) = f\left( x\right) \) if \( \left| {f\left( x\right) }\right| > 1 \) , and \( {f}_{1}\left( x\right) = 0 \) else... | Yes |
Proposition 5.1 Suppose \( f \in {L}^{p}\left( {\mathbb{R}}^{d}\right), p > 1 \), and \( f \) has bounded support. Then \( f \) belongs to \( {\mathbf{H}}_{r}^{1}\left( {\mathbb{R}}^{d}\right) \) if and only if \( {\int }_{{\mathbb{R}}^{d}}f\left( x\right) {dx} = 0 \) . | Note that \( f \) is automatically in \( {L}^{1} \), by Hölder’s inequality (see Proposition 1.4 in Chapter 1), and the cancelation condition is necessary as has been pointed out.\n\nTo prove the sufficiency we assume that \( f \) is supported in a ball \( {B}_{1} \) of unit radius, and that \( {\int }_{{B}_{1}}\left| ... | Yes |
Lemma 5.2 Suppose \( \Omega \subset {\mathbb{R}}^{d} \) is a non-trivial open set. Then there is a collection \( \left\{ {Q}_{j}\right\} \) of dyadic cubes with disjoint interiors so that \( \Omega = \) \( \mathop{\bigcup }\limits_{{j = 1}}^{\infty }{Q}_{j} \), and\n\n\[ \operatorname{diam}\left( {Q}_{j}\right) \leq d\... | Proof. We claim first that every point \( \bar{x} \in \Omega \) belongs to some dyadic cube \( {Q}_{\bar{x}} \) for which (37) holds (with \( {Q}_{\bar{x}} \) in place of \( {Q}_{j} \) ).\n\nLet \( \delta = d\left( {\bar{x},{\Omega }^{c}}\right) > 0 \) . Now the dyadic cubes containing \( \bar{x} \) have diameters vary... | Yes |
Corollary 5.3 Fix \( p > 1 \) . Then any p-atom \( \mathfrak{a} \) is in \( {\mathbf{H}}_{r}^{1} \) . Moreover there is a bound \( {c}_{p} \), independent of the atom \( \mathfrak{a} \), so that \[ \parallel \mathfrak{a}{\parallel }_{{\mathbf{H}}_{r}^{1}} \leq {c}_{p} \] | Proof. One can rescale a \( p \) -atom \( \mathfrak{a} \), associated to a ball \( B \) of radius \( r \), by replacing \( \mathfrak{a} \) by \( {\mathfrak{a}}_{r} \), with \( {\mathfrak{a}}_{r}\left( x\right) = {r}^{d}\mathfrak{a}\left( {rx}\right) \) . Then clearly \( {\mathfrak{a}}_{r}\left( x\right) \) is supported... | Yes |
Theorem 5.4 If \( f \) belongs to the Hardy space \( {\mathbf{H}}_{r}^{1}\left( \mathbb{R}\right) \), then \( {H}_{\epsilon }\left( f\right) \in \) \( {L}^{1}\left( \mathbb{R}\right) \), for every \( \epsilon > 0 \) . Moreover \( {H}_{\epsilon }\left( f\right) \) (see (14)) converges in the \( {L}^{1} \) norm, as \( \e... | Proof. The argument below illustrates a nice feature of \( {\mathbf{H}}_{r}^{1}\left( \mathbb{R}\right) \) : to show the boundedness of an operator on \( {\mathbf{H}}_{r}^{1} \) it often suffices merely to verify it for atoms, and this is usually a simple task.\n\nLet us first see that for all atoms \( \mathfrak{a} \),... | Yes |
Theorem 6.1 Suppose \( \Phi \) is a \( {C}^{1} \) function with compact support on \( {\mathbb{R}}^{d} \) . With \( M \) defined by (41) we have that \( M\left( f\right) \in {L}^{1}\left( {\mathbb{R}}^{d}\right) \), whenever \( f \in \) \( {\mathbf{H}}_{r}^{1}\left( {\mathbb{R}}^{d}\right) \) . Moreover\n\n\[ \parallel... | Proof. Suppose \( f \) is in \( {\mathbf{H}}_{r}^{1}\left( {\mathbb{R}}^{d}\right) \) and \( f = \sum {\lambda }_{k}{\mathfrak{a}}_{k} \) is an atomic decomposition. Then clearly \( M\left( f\right) \leq \sum \left| {\lambda }_{k}\right| M\left( {\mathfrak{a}}_{k}\right) \), and thus it suffices to prove (42) when \( f... | Yes |
Theorem 6.2 Suppose \( g \in \mathrm{{BMO}} \) . Then the linear functional \( \ell \) defined by (44), initially considered for \( f \in {H}_{0}^{1} \), has a unique extension to \( {\mathbf{H}}_{r}^{1} \) that satisfies\n\n\[ \parallel \ell \parallel \leq c\parallel g{\parallel }_{\mathrm{{BMO}}} \]\n\nConversely, ev... | Proof. Let us first assume that \( g \in \mathrm{{BMO}} \) is bounded. Start with a general \( f \in {\mathbf{H}}_{r}^{1} \), and let \( f = \mathop{\sum }\limits_{{k = 1}}^{\infty }{\lambda }_{k}{\mathfrak{a}}_{k} \) be an atomic decomposition. Then by the convergence of the sum in the \( {L}^{1} \) norm we get \( \el... | Yes |
Proposition 1.1 Suppose \( F \) is a distribution and \( \psi \in \mathcal{D} \). Then\n\n(a) The two definitions of \( F * \psi \) given above coincide.\n\n(b) The distribution \( F * \psi \) is a \( {C}^{\infty } \) function. | Proof. Let us observe first that \( F\left( {\psi }_{x}^{ \sim }\right) \) is continuous in \( x \) and in fact indefinitely differentiable. Note that if \( {x}_{n} \rightarrow {x}_{0} \) as \( n \rightarrow \infty \), then \( {\psi }_{{x}_{n}}^{ \sim }\left( y\right) = \psi \left( {{x}_{n} - y}\right) \rightarrow \psi... | Yes |
Corollary 1.2 Suppose \( F \) is a distribution on \( {\mathbb{R}}^{d} \) . Then there exists a sequence \( \left\{ {f}_{n}\right\} \), with \( {f}_{n} \in {C}^{\infty } \), and \( {f}_{n} \rightarrow F \) in the weak sense. | Proof. Let \( \left\{ {\psi }_{n}\right\} \) be an approximation to the identity constructed as follows. Fix a \( \psi \in \mathcal{D} \) with \( {\int }_{{\mathbb{R}}^{d}}\psi \left( x\right) {dx} = 1 \) and set \( {\psi }_{n}\left( x\right) = {n}^{d}\psi \left( {nx}\right) \) . Form \( {F}_{n} = F * {\psi }_{n} \) . ... | Yes |
Proposition 1.3 Suppose \( F \) is a distribution whose support is \( {C}_{1} \), and \( \psi \) is in \( \mathcal{D} \) and has support \( {C}_{2} \) . Then the support of \( F * \psi \) is contained in \( {C}_{1} + {C}_{2} \) | Indeed for each \( x \) for which \( F\left( {\psi }_{x}^{ \sim }\right) \neq 0 \), we must have that the support of \( F \) intersects the support of \( {\psi }_{x}^{ \sim } \) . Since the support of \( {\psi }_{x}^{ \sim } \) is the set \( x - \) \( {C}_{2} \) this means that the set \( {C}_{1} \) and \( x - {C}_{2} ... | Yes |
Proposition 1.4 Suppose \( F \) is a tempered distribution. Then there is a positive integer \( N \) and a constant \( c > 0 \), so that\n\n\[ \left| {F\left( \varphi \right) }\right| \leq c\parallel \varphi {\parallel }_{N},\;\text{ for all }\varphi \in \mathcal{S}. \]\n | Proof. Assume otherwise. Then the conclusion fails and for each positive integer \( n \) there is a \( {\psi }_{n} \in \mathcal{S} \) with \( {\begin{Vmatrix}{\psi }_{n}\end{Vmatrix}}_{n} = 1 \), while \( \left| {F\left( {\psi }_{n}\right) }\right| \geq n \) . Take \( {\varphi }_{n} = {\psi }_{n}/{n}^{1/2} \) . Then \(... | Yes |
Proposition 1.5 Suppose \( F \) is a tempered distribution and \( \psi \in \mathcal{S} \). Then \( F * \psi \) is a slowly increasing \( {C}^{\infty } \) function, which when considered as a tempered distribution satisfies \( {\left( F * \psi \right) }^{ \land } = {\psi }^{ \land }{F}^{ \land } \). | Proof. The fact that \( F\left( {\psi }_{x}^{ \sim }\right) \) is slowly increasing follows from the proposition in Section 1.4 together with the observation that for any function \( \psi \in \mathcal{D} \) and \( N,{\begin{Vmatrix}{\psi }_{x}^{ \sim }\end{Vmatrix}}_{N} \leq c{\left( 1 + \left| x\right| \right) }^{N}\p... | Yes |
Proposition 1.6 If \( F \) is a distribution of compact support then its Fourier transform \( {F}^{ \land } \) is a slowly increasing \( {C}^{\infty } \) function. In fact, as a function of \( \xi \), one has \( {F}^{ \land }\left( \xi \right) = F\left( {e}_{\xi }\right) \) where \( {e}_{\xi } \) is the element of \( \... | Proof. If we invoke Proposition 1.4, we see immediately that \( \left| {F\left( {e}_{\xi }\right) }\right| \leq \) \( C{\begin{Vmatrix}{e}_{\xi }\end{Vmatrix}}_{N} \leq {c}^{\prime }{\left( 1 + \left| \xi \right| \right) }^{N} \) . By the same estimate, every difference quotient of \( F\left( {e}_{\xi }\right) \) conve... | Yes |
Lemma 1.8 Suppose \( {F}_{1} \) is a distribution supported at the origin that satisfies for some \( N \) the following two conditions:\n\n(a) \( \left| {{F}_{1}\left( \varphi \right) }\right| \leq c\parallel \varphi {\parallel }_{N} \), for all \( \varphi \in \mathcal{D} \).\n\n(b) \( {F}_{1}\left( {x}^{\alpha }\right... | In fact, let \( \eta \in \mathcal{D} \), with \( \eta \left( x\right) = 0 \) for \( \left| x\right| \geq 1 \), and \( \eta \left( x\right) = 1 \) when \( \left| x\right| \leq 1/2 \) , and write \( {\eta }_{\epsilon }\left( x\right) = \eta \left( {x/\epsilon }\right) \). Then since \( {F}_{1} \) is supported at the orig... | Yes |
Theorem 2.1 The distribution \( \operatorname{pv}\left( \frac{1}{x}\right) \) equals:\n\n(a) \( \frac{d}{dx}\left( {\log \left| x\right| }\right) \) .\n\n(b) \( \frac{1}{2}\left( {\frac{1}{x - {i0}} + \frac{1}{x + {i0}}}\right) \) . | Regarding (a), note that \( \log \left| x\right| \) is a locally integrable function. Here \( \frac{d}{dx}\left( {\log \left| x\right| }\right) \) is its derivative taken as a distribution. Now in that sense\n\n\[ \left( {\frac{d}{dx}\log \left| x\right| }\right) \left( \varphi \right) = - {\int }_{-\infty }^{\infty }\... | Yes |
Proposition 2.2 Suppose \( F \) is a tempered distribution on \( {\mathbb{R}}^{d} \) that is homogeneous of degree \( \lambda \) . Then its Fourier transform \( {F}^{ \land } \) is homogeneous of degree \( - d - \lambda \) . | To deal with \( {\left( {F}^{ \land }\right) }_{a} \) we write successively,\n\n\[ \n{\left( {F}^{ \land }\right) }_{a}\left( \varphi \right) = {F}^{ \land }\left( {\varphi }^{a}\right) = F\left( {\left( {\varphi }^{a}\right) }^{ \land }\right) = F\left( {\left( {\varphi }^{ \land }\right) }_{a}\right) \n\]\n\n\[ \n= {... | Yes |
Theorem 2.3 If \( - d < \lambda < 0 \), then\n\n\[ \n{\left( {H}_{\lambda }\right) }^{ \land } = {c}_{\lambda }{H}_{-d - \lambda },\;\text{with}\;{c}_{\lambda } = \frac{\Gamma \left( \frac{d + \lambda }{2}\right) }{\Gamma \left( \frac{-\lambda }{2}\right) }{\pi }^{-d/2 - \lambda }.\n\] | To prove the theorem we start with the fact that \( \psi \left( x\right) = {e}^{-\pi {\left| x\right| }^{2}} \) is its own Fourier transform. Then since \( {\left( {\psi }_{a}\right) }^{ \land } = {\left( {\psi }^{ \land }\right) }^{a} \) we get (with \( a = {t}^{1/2} \) )\n\n\[ \n{\int }_{{\mathbb{R}}^{d}}{e}^{-{\pi t... | Yes |
The Fourier transform of a regular homogeneous distribution \( K \) of degree \( \lambda \) is a regular homogeneous distribution of degree \( - d - \lambda \), and conversely. | Proof. We already know from Proposition 2.2 that \( {K}^{ \land } \) is homogeneous of degree \( - d - \lambda \) . To prove that \( {K}^{ \land } \) agrees with a \( {C}^{\infty } \) function away from the origin, we decompose \( K = {K}_{0} + {K}_{1} \), with \( {K}_{0} \) supported near the origin and \( {K}_{1} \) ... | Yes |
Lemma 2.6 Suppose \( {\lambda }_{1},{\lambda }_{2},\ldots ,{\lambda }_{n} \), are distinct real numbers and that for constants \( {a}_{j} \) and \( {b}_{j},1 \leq j \leq n \), we have\n\n\[ \mathop{\sum }\limits_{{j = 1}}^{n}\left( {{a}_{j}{x}^{{\lambda }_{j}} + {b}_{j}{x}^{{\lambda }_{j}}\log x}\right) = 0\;\text{ for... | To prove the lemma we assume, as one may, that \( {\lambda }_{n} \) is the largest of the \( {\lambda }_{j} \) ’s. Then multiplying the identity by \( {x}^{-{\lambda }_{n}} \) and letting \( x \) tend to infinity we see that \( {b}_{n} \) as well as \( {a}_{n} \) must vanish. Thus we are reduced to the case when \( n \... | No |
Corollary 2.7 There is no distribution \( {K}_{0} \) that is homogeneous of degree \( - d \) and that agrees with the function \( 1/{\left| x\right| }^{d} \) away from the origin. | If such a \( {K}_{0} \) existed, then \( {K}_{0} - \left\lbrack \frac{1}{{\left| x\right| }^{d}}\right\rbrack \) would be supported at the origin, and hence equal to \( \mathop{\sum }\limits_{{\left| \alpha \right| \leq M}}{c}_{\alpha }{\partial }_{x}^{\alpha }\delta \) . Applying this difference to \( {\varphi }^{a} \... | Yes |
Theorem 2.8 For \( d \geq 3 \), the locally integrable function \( F \) defined by \( F\left( x\right) = {C}_{d}{\left| x\right| }^{-d + 2} \) is a fundamental solution for the operator \( \bigtriangleup \), with \( {C}_{d} = - \frac{\Gamma \left( {\frac{d}{2} - 1}\right) }{4{\pi }^{\frac{d}{2}}}. \) | This follows by taking \( \lambda = - d + 2 \) (in Theorem 2.3), then \( \Gamma \left( \frac{d + \lambda }{2}\right) = \) \( \Gamma \left( 1\right) = 1 \), while \( \Gamma \left( {d/2}\right) = \left( {d/2 - 1}\right) \Gamma \left( {d/2 - 1}\right) \) . Therefore \( \widehat{F}\left( \xi \right) \) equals \( 1/\left( {... | Yes |
Theorem 2.10 \( F \) is a fundamental solution of \( L = \frac{\partial }{\partial t} - {\bigtriangleup }_{x} \) . | Proof. Since \( {LF}\left( \varphi \right) = F\left( {{L}^{\prime }\varphi }\right) \) with \( {L}^{\prime } = - \frac{\partial }{\partial t} - {\bigtriangleup }_{x} \), it suffices to see that \( F\left( {{L}^{\prime }\varphi }\right) \), which equals\n\n\[ \mathop{\lim }\limits_{{\epsilon \rightarrow 0}}{\int }_{t \g... | Yes |
Theorem 2.11 Every constant coefficient (linear) partial differential equation \( L \) on \( {\mathbb{R}}^{d} \) has a fundamental solution. | Proof. After a possible change of coordinates consisting of a rotation and multiplication by a constant, we may assume that the characteristic polynomial of \( L \) will be of the form\n\n\[ P\left( \xi \right) = P\left( {{\xi }_{1},{\xi }^{\prime }}\right) = {\xi }_{1}^{m} + \mathop{\sum }\limits_{{j = 0}}^{{m - 1}}{\... | No |
Theorem 2.12 Every elliptic operator has a regular parametrix. | Proof. Observe first by a straightforward inductive argument in \( k \) , that whenever \( \left| \alpha \right| = k \) and \( P \) is any polynomial\n\n\[{\left( \frac{\partial }{\partial \xi }\right) }^{\alpha }\left( \frac{1}{P\left( \xi \right) }\right) = \mathop{\sum }\limits_{{0 \leq \ell \leq k}}\frac{{q}_{\ell ... | Yes |
Corollary 2.13 Given any \( \epsilon > 0 \), the elliptic operator \( L \) has a regular parametrix \( {Q}_{\epsilon } \) that is supported in the ball \( \{ x : \left| x\right| \leq \epsilon \} \) . | In fact, let \( {\eta }_{\epsilon } \) be a cut-off function in \( \mathcal{D} \), that is 1 when \( \left| x\right| \leq \epsilon /2 \) , and that is supported where \( \left| x\right| \leq \epsilon \) . Set \( {Q}_{\epsilon } = {\eta }_{\epsilon }Q \), and observe that \( L\left( {{\eta }_{\epsilon }Q}\right) - {\eta... | Yes |
Theorem 2.14 Suppose the partial differential operator \( L \) has a regular parametrix. Assume \( U \) is a distribution given in an open set \( \Omega \subset {\mathbb{R}}^{d} \) and \( L\left( U\right) = f \), with \( f \) a \( {C}^{\infty } \) function in \( \Omega \) . Then \( U \) is also a \( {C}^{\infty } \) fu... | Proof of the theorem. It suffices to show that \( U \) agrees with a \( {C}^{\infty } \) function on any ball \( B \) with \( \bar{B} \subset \Omega \) . Fix such a ball (say of radius \( \rho \) ), and let \( {B}_{1} \) be the concentric ball having radius \( \rho + \epsilon \), with \( \epsilon > 0 \) so small that \... | Yes |
Theorem 3.2 Let \( T \) be the operator \( T\left( f\right) = f * K \), with \( K \) as in Proposition 3.1. Then \( T \) initially defined for \( f \) in \( \mathcal{S} \) extends to a bounded operator on \( {L}^{p}\left( {\mathbb{R}}^{d}\right) \), for \( 1 < p < \infty \) . | This means that for each \( p,1 < p < \infty \), there is a bound \( {A}_{p} \) so that\n\n(23)\n\n\[ \n\parallel {Tf}{\parallel }_{{L}^{p}\left( {\mathbb{R}}^{d}\right) } \leq {A}_{p}\parallel f{\parallel }_{{L}^{p}\left( {\mathbb{R}}^{d}\right) }\n\]\n\nfor \( f \in \mathcal{S} \) . Thus by Proposition 5.4 in Chapter... | Yes |
Lemma 3.3 For each \( f \) in \( {L}^{1}\left( {\mathbb{R}}^{d}\right) \) and \( \alpha > 0 \), we can find an open set \( {E}_{\alpha } \) and a decomposition \( f = g + b \) so that:\n\n(a) \( m\left( {E}_{\alpha }\right) \leq \frac{c}{\alpha }\parallel f{\parallel }_{{L}^{1}\left( {\mathbb{R}}^{d}\right) } \) .\n\n(... | The proof of the lemma is a simplified version of the argument used to prove Proposition 5.1 in the previous chapter; in particular, here we use the full maximal function \( {f}^{ * } \) instead of the truncated version \( {f}^{ \dagger } \) . The guiding idea is to try to cut the domain of \( f \) into the set when \(... | Yes |
Corollary 1.2 In a complete metric space, a generic set is dense. | Proof of the theorem. We argue by contradiction, and assume that \( X \) is a countable union of nowhere dense sets \( {F}_{n} \) ,\n\n\[ X = \mathop{\bigcup }\limits_{{n = 1}}^{\infty }{F}_{n} \]\n\nBy replacing each \( {F}_{n} \) by its closure, we may assume that each \( {F}_{n} \) is closed. It now suffices to find... | Yes |
Theorem 1.3 Suppose that \( \left\{ {f}_{n}\right\} \) is a sequence of continuous complex-valued functions on a complete metric space \( X \), and \[ \mathop{\lim }\limits_{{n \rightarrow \infty }}{f}_{n}\left( x\right) = f\left( x\right) \] exists for every \( x \in X \) . Then, the set of points where \( f \) is con... | To show that the set \( \mathcal{D} \) of discontinuities of \( f \) is of the first category, we use a characterization of points of continuity of \( f \) in terms of its oscillations. More precisely, we define the oscillation of the function \( f \) at a point \( x \) by \[ \operatorname{osc}\left( f\right) \left( x\... | Yes |
Lemma 1.4 Suppose \( \left\{ {f}_{n}\right\} \) is a sequence of continuous functions on a complete metric space \( X \), and \( {f}_{n}\left( x\right) \rightarrow f\left( x\right) \) for each \( x \) as \( n \rightarrow \infty \) . Then, given an open ball \( B \subset X \) and \( \epsilon > 0 \), there exists an open... | Proof. Let \( Y \) denote a closed ball contained in \( B \) . Note that \( Y \) is itself a complete metric space. Define\n\n\[ \n{E}_{\ell } = \left\{ {x \in Y : \mathop{\sup }\limits_{{j, k \geq \ell }}\left| {{f}_{j}\left( x\right) - {f}_{k}\left( x\right) }\right| \leq \epsilon }\right\} \n\]\n\nThen, since \( {f}... | Yes |
Lemma 1.6 For every \( M > 0 \), the set \( {\mathcal{P}}_{M} \) of zig-zag functions is dense in \( C\left( \left\lbrack {0,1}\right\rbrack \right) \) . | Proof. It is plain that given \( \epsilon > 0 \) and a continuous function \( f \) , there exists a function \( g \in \mathcal{P} \) so that \( \parallel f - g\parallel \leq \epsilon \) . Indeed, since \( f \) is continuous on the compact set \( \left\lbrack {0,1}\right\rbrack \) it must be uniformly continuous, and th... | Yes |
Theorem 2.1 Suppose that \( \mathcal{B} \) is a Banach space, and \( \mathcal{L} \) is a collection of continuous linear functionals on \( \mathcal{B} \). (i) If \( \mathop{\sup }\limits_{{\ell \in \mathcal{L}}}\left| {\ell \left( f\right) }\right| < \infty \) for each \( f \in \mathcal{B} \), then \[ \mathop{\sup }\li... | Proof. It suffices to show (ii) since by Baire’s theorem, \( \mathcal{B} \) is of the second category. So suppose that \( \mathop{\sup }\limits_{{\ell \in \mathcal{L}}}\left| {\ell \left( f\right) }\right| < \infty \) for all \( f \in E \), where \( E \) is of the second category. For each positive integer \( M \), def... | Yes |
Given any point \( {x}_{0} \in \left\lbrack {-\pi ,\pi }\right\rbrack \), there is a continuous function whose Fourier series diverges at \( {x}_{0} \). | We begin with (i), and assume without loss of generality that \( {x}_{0} = 0 \) . Let \( {\ell }_{N} \) denote the linear functional on \( \mathcal{B} \) defined by\n\n\[ \n{\ell }_{N}\left( f\right) = {S}_{N}\left( f\right) \left( 0\right) = \frac{1}{2\pi }{\int }_{-\pi }^{\pi }f\left( {-y}\right) {D}_{N}\left( y\righ... | Yes |
Lemma 2.3 \( \begin{Vmatrix}{\ell }_{N}\end{Vmatrix} = {L}_{N} \) for all \( N \geq 0 \) . | Proof. We already know from the above that \( \begin{Vmatrix}{\ell }_{N}\end{Vmatrix} \leq {L}_{N} \) . To prove the reverse inequality, it suffices to find a sequence of continuous functions \( \left\{ {f}_{k}\right\} \) so that \( \begin{Vmatrix}{f}_{k}\end{Vmatrix} \leq 1 \), and \( {\ell }_{N}\left( {f}_{k}\right) ... | Yes |
Lemma 2.4 There is a constant \( c > 0 \) so that \( {L}_{N} \geq c\log N \) . | Proof. Since \( \left| {\sin y}\right| /\left| y\right| \leq 1 \) for all \( y \), and \( \sin y \) is an odd function, we see that \( {}^{1} \)\n\n\[ \n{L}_{N} \geq c{\int }_{0}^{\pi }\frac{\left| \sin \left( N + 1/2\right) y\right| }{\left| y\right| }{dy} \]\n\n\[ \n\geq c{\int }_{0}^{\left( {N + 1/2}\right) \pi }\fr... | Yes |
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