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Find a solution of the Dirichlet problem in the disk having boundary values \( u\left( {1,\theta }\right) = \cos {4\theta } - 1 \) . Express the solution in rectangular coordinates! | It is immediately seen that in polar coordinates the solution must be\n\n\[ u\left( {r,\theta }\right) = - 1 + {r}^{4}\cos {4\theta }.\]\n\nWe rewrite the cosine to introduce cos and sin of the single value \( \theta \) :\n\n\[ u = - 1 + {r}^{4}\left( {{\cos }^{2}{2\theta } - {\sin }^{2}{2\theta }}\right)\]\n\n\[ = - 1... | Yes |
Example 6.7. Let \( V = {L}^{2}\left( \mathbf{T}\right) ,{\mathcal{D}}_{A} = V \cap {C}^{2}\left( \mathbf{T}\right) \) and let \( A \) be the operator \( - {D}^{2} \), so that \( {Au} = - {u}^{\prime \prime } \) . Since \( u \in {C}^{2}\left( \mathbf{T}\right) \), the image \( {Au} \) is a continuous function and thus ... | \[ \langle {Au}, v\rangle = - {\int }_{\mathbf{T}}{u}^{\prime \prime }\left( x\right) \overline{v\left( x\right) }{dx} = - {\left\lbrack {u}^{\prime }\left( x\right) \overline{v\left( x\right) }\right\rbrack }_{-\pi }^{\pi } + {\int }_{\mathbf{T}}{u}^{\prime }\left( x\right) \overline{{v}^{\prime }\left( x\right) }{dx}... | Yes |
Lemma 6.1 A symmetric operator has only real eigenvalues, and eigenvectors corresponding to different eigenvalues are orthogonal. | Proof. Suppose that \( {Au} = {\lambda u} \) and \( {Av} = {\mu v} \), where \( u \neq 0 \) and \( v \neq 0 \) . Then we can write\n\n\[ \n\lambda \langle u, v\rangle = \langle {\lambda u}, v\rangle = \langle {Au}, v\rangle = \langle u,{Av}\rangle = \langle u,{\mu v}\rangle = \bar{\mu }\langle u, v\rangle .\n\]\n\n\( \... | Yes |
Theorem 6.2 The problem (E)+(B) has solutions for an infinite number of values of the parameter \( \lambda \), which can be arranged in an increasing sequence:\n\n\[ \n{\lambda }_{1} < {\lambda }_{2} < {\lambda }_{3} < \cdots ,\;\text{where}\;{\lambda }_{n} \rightarrow \infty \;\text{as}\;n \rightarrow \infty .\n\]\n\n... | Proofs can be found in texts on ordinary differential equations. | No |
If \( f\left( t\right) = {e}^{-\left| t\right| }, t \in \mathbf{R} \), then \( f \in {L}^{1}\left( \mathbf{R}\right) \), and | \[\n\widehat{f}\left( \omega \right) = {\int }_{\mathbf{R}}{e}^{-\left| t\right| }{e}^{-{i\omega t}}{dt} = {\int }_{0}^{\infty }{e}^{-\left( {1 + {i\omega }}\right) t}{dt} + {\int }_{-\infty }^{0}{e}^{\left( {1 - {i\omega }}\right) t}{dt}\n\]\n\n\[= \frac{1}{1 + {i\omega }} + \frac{1}{1 - {i\omega }} = \frac{2}{1 + {\o... | Yes |
Example 7.2. Let \( f\left( t\right) = 1 \) for \( \left| t\right| < 1, = 0 \) for \( \left| t\right| > 1 \) (i.e., \( f\left( t\right) = \) \( H\left( {t + 1}\right) - H\left( {t - 1}\right) \), where \( H \) is the Heaviside function as in Sec. 2.6). Then clearly \( f \in {L}^{1}\left( \mathbf{R}\right) \), and | \[ \widehat{f}\left( \omega \right) = {\int }_{-1}^{1}{e}^{-{i\omega t}}{dt} = {\left\lbrack \frac{{e}^{-{i\omega t}}}{-{i\omega }}\right\rbrack }_{-1}^{1} = \frac{2}{\omega } \cdot \frac{{e}^{i\omega } - {e}^{-{i\omega }}}{2i} = \frac{2\sin \omega }{\omega },\;\omega \neq 0. \] For \( \omega = 0 \) one has \( {e}^{-{i... | Yes |
Theorem 7.1 If \( f \in {L}^{1}\left( \mathbf{R}\right) \), the following holds for the Fourier transform \( \widehat{f} \) :\n\n(a) \( \widehat{f} \) is bounded; more precisely, \( \left| {\widehat{f}\left( \omega \right) }\right| \leq {\int }_{\mathbf{R}}\left| {f\left( t\right) }\right| {dt} \) . | \( \textit{Proof.}\;\left( a\right) \; \) follows immediately from the estimate \( \;\left| {{\int }_{I}\varphi \left( t\right) \;{dt}}\right| \leq {\int }_{I}\left| {\varphi \left( t\right) }\right| {dt}, \) which holds for any interval \( I \) and any Riemann-integrable function \( \varphi \) (even if it is complex-v... | Yes |
Theorem 7.2 The mapping \( \mathcal{F} : f \mapsto \widehat{f} \) is a linear map from the space \( {L}^{1}\left( \mathbf{R}\right) \) to the space \( {C}_{0}\left( \widehat{\mathbf{R}}\right) \) of those continuous functions defined on \( \widehat{\mathbf{R}} \) that tend to 0 at \( \pm \infty \) . | Proof. The fact that \( \widehat{f} \in {C}_{0}\left( \widehat{\mathbf{R}}\right) \) is the content of (b) and (c) in Theorem 7.1. The linearity of \( \mathcal{F} \) means just that\n\n\[ \mathcal{F}\left\lbrack {f + g}\right\rbrack = \mathcal{F}\left\lbrack f\right\rbrack + \mathcal{F}\left\lbrack g\right\rbrack ,\;\m... | Yes |
Theorem 7.3 Suppose that \( f \in {L}^{1}\left( \mathbf{R}\right) \) and let a be a real number. Then the translated function \( {f}_{a}\left( t\right) = f\left( {t - a}\right) \) and the function \( {e}^{iat}f\left( t\right) \) also belong to \( {L}^{1}\left( \mathbf{R}\right) \), and\n\n\[{\widehat{f}}_{a}\left( \ome... | Proof. For the first formula, start with \( {\widehat{f}}_{a}\left( \omega \right) = {\int }_{\mathbf{R}}f\left( {t - a}\right) {e}^{-{i\omega t}}{dt} \) . The change of variable \( t - a = y \) gives the result. The proof of the second formula is maybe even simpler. | Yes |
Theorem 7.4 Suppose that \( f \) is differentiable and that both \( f \) and \( {f}^{\prime } \) belong to \( {L}^{1}\left( \mathbf{R}\right) \) . Then\n\n\[ \left( \widehat{Df}\right) \left( \omega \right) = \widehat{{f}^{\prime }}\left( \omega \right) = \mathcal{F}\left\lbrack {f}^{\prime }\right\rbrack \left( \omega... | The proof of (7.13) relies, in principle, on integration by parts:\n\n\[ {\widehat{f}}^{\prime }\left( \omega \right) = {\int }_{-\infty }^{\infty }{f}^{\prime }\left( t\right) {e}^{-{i\omega t}}{dt} = {\left\lbrack f\left( t\right) {e}^{-{i\omega t}}\right\rbrack }_{-\infty }^{\infty } - {\int }_{-\infty }^{\infty }f\... | Yes |
Lemma 7.1\n\n\\[ \n{\\int }_{0}^{\\infty }\\frac{\\sin {Au}}{u}{du} = \\frac{\\pi }{2}\\;\\text{ for }A > 0.\n\\] | It is easy to check, by the change of variable \\( {Au} = t \\), that the integral is independent of \\( A \\) (if \\( A > 0 \\) ), so one can just as well assume that \\( A = 1 \\) . The integral is not absolutely convergent, and \\( {\\int }_{0}^{\\infty } \\) in this case stands for the limit of \\( {\\int }_{0}^{X}... | Yes |
For \( f\left( t\right) = {e}^{-\left| t\right| } \) we have \( \widehat{f}\left( \omega \right) = \frac{2}{1 + {\omega }^{2}} \) . Since \( f \) is piecewise smooth it follows that\n\n\[ \n{e}^{-\left| t\right| } = \frac{1}{\pi }\mathop{\lim }\limits_{{A \rightarrow \infty }}{\int }_{-A}^{A}\frac{{e}^{i\omega t}}{1 + ... | We can switch letters in this formula \( - t \) and \( \omega \) are exchanged for each other\n\n- and then we also change the sign of \( \omega \), and we get (after multiplication by \( \pi \) ) the formula\n\n\[ \n\pi {e}^{-\left| \omega \right| } = {\int }_{\mathbf{R}}\frac{{e}^{-{i\omega t}}}{1 + {t}^{2}}{dt}\n\]\... | Yes |
Suppose that a table contains an item like this:\n\n\[ \n\\begin{array}{ll} f\\left( t\\right) & \\widehat{f}\\left( \\omega \\right) \\\\ t{e}^{-\\left| t\\right| } & \\frac{-{4i\\omega }}{{\\left( 1 + {\\omega }^{2}\\right) }^{2}} \\end{array} \n\]\n\nFrom this one can find the transform of the function\n\n\[ \ng\\le... | by performing two steps:\n\n1. Switch sides in the table, switching variables at the same time:\n\n\[ \n\\frac{-{4it}}{{\\left( 1 + {t}^{2}\\right) }^{2}}\\;\\omega {e}^{-\\left| \\omega \\right| } \n\]\n\n2. Multiply the right-hand side by \\( {2\\pi } \\) and change the sign of the variable there:\n\n\[ \n\\frac{-{4i... | Yes |
Theorem 7.6 (Convolution theorem)\n\n\[ \mathcal{F}\left\lbrack {f * g}\right\rbrack = \mathcal{F}\left\lbrack f\right\rbrack \mathcal{F}\left\lbrack g\right\rbrack \] | Formally, the proof runs like this:\n\n\[ \mathcal{F}\left\lbrack {f * g}\right\rbrack \left( \omega \right) = {\int }_{\mathbf{R}}{e}^{-{i\omega t}}\left( {{\int }_{\mathbf{R}}f\left( {t - y}\right) g\left( y\right) {dy}}\right) {dt} \]\n\n\[ = {\iint }_{{\mathbf{R}}^{2}}{e}^{-{i\omega }\left( {t - y + y}\right) }f\le... | Yes |
What function \( f \) has the Fourier transform \n\n\[ \n\widehat{f}\left( \omega \right) = \frac{1}{{\left( 1 + {\omega }^{2}\right) }^{2}}? \n\] | Solution. We start from the formula \n\n\[ \n\widehat{g}\left( \omega \right) = \frac{2}{1 + {\omega }^{2}}\;\text{ if }\;g\left( t\right) = {e}^{-\left| t\right| }. \n\] \n\nBy the convolution theorem we get \n\n\[ \n\mathcal{F}\left\lbrack {g * g}\right\rbrack \left( \omega \right) = {\left( \widehat{g}\left( \omega ... | Yes |
If \( f\left( t\right) = 1/\left( {1 + {t}^{2}}\right) \), find \( f * f \) . | Put \( g = f * f \) . Computing the convolution directly is toilsome. Instead, we make use of Theorem 7.6. Let us start from the fact that \( \widehat{f}\left( \omega \right) = \pi {e}^{-\left| \omega \right| } \) (Example 7.3, p. 174), which means that\n\n\[ \n{\int }_{\mathbf{R}}\frac{{e}^{-{i\omega t}}}{1 + {t}^{2}}... | Yes |
Prove the formula\n\n\\[ \n{\\int }_{-1}^{1}\\frac{\\sin \\left( {t - y}\\right) }{t - y}{dy} = {\\int }_{-1}^{1}\\frac{\\sin y}{y}{e}^{iyt}{dy},\\;t \\in \\mathbf{R}. \n\\]\n\n(7.19) | Solution. Let \\( f\\left( t\\right) = H\\left( {t + 1}\\right) - H\\left( {t - 1}\\right) \\), and \\( g\\left( t\\right) = \\left( {\\sin t}\\right) /t \\) . From the table of Fourier transforms we recognize that \\( \\widehat{f}\\left( \\omega \\right) = g\\left( \\omega \\right) /\\pi \\), and by the inversion form... | Yes |
Because of the formal symmetry between the Fourier transformation and the inversion formula, one can expect that there exists a formula involvning the convolution of transforms. Indeed, if \( \widehat{f} \) and \( \widehat{g} \) are sufficiently nice, to ensure that the necessary integrals converge, it is true that the... | \[ \widehat{fg}\left( \omega \right) = \frac{1}{2\pi }{\int }_{\widehat{\mathbf{R}}}\widehat{f}\left( {\omega - \alpha }\right) g\left( \alpha \right) {d\alpha } = \frac{1}{2\pi }\widehat{f} * \widehat{g}\left( \omega \right) .\] | Yes |
The Plancherel formula enables us to compute certain integrals. If \( f\left( t\right) = 1 \) for \( \left| t\right| < 1 \) and \( = 0 \) otherwise, then (see Example 7.2 p. 167) \( \widehat{f}\left( \omega \right) = \frac{2\sin \omega }{\omega } \) . Plancherel now gives | \[ {\int }_{-1}^{1}{1dt} = \frac{1}{2\pi }{\int }_{-\infty }^{\infty }\frac{4{\sin }^{2}\omega }{{\omega }^{2}}{d\omega } \] or, after rewriting, \[ {\int }_{-\infty }^{\infty }\frac{{\sin }^{2}t}{{t}^{2}}{dt} = \pi \] | Yes |
\[ {\int }_{-1}^{1}\frac{\sin \left( {t - y}\right) }{t - y}{dy} = {\int }_{-1}^{1}\frac{\sin y}{y}{e}^{iyt}{dy},\;t \in \mathbf{R}. \] | Let \( f\left( y\right) = H\left( {y + 1}\right) - H\left( {y - 1}\right) \) and \( g\left( \omega \right) = \) \( {e}^{it\omega }f\left( \omega \right) \) . Then\n\n\[ \widehat{f}\left( \omega \right) = \frac{2\sin \omega }{\omega }\;\text{ and }\;\widehat{g}\left( y\right) = \frac{2\sin \left( {y - t}\right) }{y - t}... | Yes |
Theorem 7.7 (Shannon’s sampling theorem) Suppose that \( f \) is continuous on \( \mathbf{R} \), that \( f \in {L}^{1}\left( \mathbf{R}\right) \) and that \( \widehat{f}\left( \omega \right) = 0 \) for \( \left| \omega \right| > c \) . Then\n\n\[ f\left( t\right) = \mathop{\sum }\limits_{{n \in \mathbf{Z}}}f\left( \fra... | Proof. By the Fourier inversion formula, we have\n\n\[ f\left( t\right) = \frac{1}{2\pi }{\int }_{-c}^{c}\widehat{f}\left( \omega \right) {e}^{it\omega }{d\omega } \]\n\n(7.26)\n\nWe shall rewrite this integral. We introduce a function \( g \) as follows:\n\n\[ g\left( \omega \right) = \frac{c}{\pi }\widehat{f}\left( \... | Yes |
Find the function whose Laplace transform is \( \widetilde{f}\left( s\right) = 1/\left( {{s}^{2} + 1}\right) \). | Solution. We want to compute\n\n\[ \frac{1}{2\pi i}\mathop{\lim }\limits_{{A \rightarrow \infty }}{\int }_{\sigma - {iA}}^{\sigma + {iA}}\frac{{e}^{ts}}{{s}^{2} + 1}{ds} \]\n\nThe integrand has simple poles at \( s = \pm i \) and is analytic in the rest of the \( s \) -plane. We can choose \( \sigma = 1 \), say, and ma... | Yes |
Find a solution of the equation \( {y}^{\prime \prime }\left( t\right) - y\left( t\right) = \delta \left( t\right) \) . | Solution. Fourier transformation, and using the rule for the transform of a derivative, gives\n\n\[ \n{\left( i\omega \right) }^{2}\widehat{y}\left( \omega \right) - \widehat{y}\left( \omega \right) = 1 \n\]\n\nIf we solve for \( \widehat{y} \), we find \( \widehat{y}\left( \omega \right) = - 1/\left( {1 + {\omega }^{2... | Yes |
Example 8.5. There is also a sort of inverted problem, compared with the ones in Examples 8.2-3 above. Suppose that we want to measure a physical quantity \( f\left( t\right) \), that depends on time \( t \) . Is it really possible to determine \( f\left( t\right) \) at a particular point in time? Every measurement tak... | HEISENBERG's undecidedness principle actually tells us that certain types of measurement cannot be exact at all; the best we can hope for is to get some mean value. | No |
In Sec. 7.7 and 7.8 we solved a couple of problems for partial differential equations using Fourier transformation. In order to be able to use this method we had to impose rather restrictive conditions on the solutions - they had to be integrable in a certain way, and differentiability past an integral sign had to be e... | The problems in Examples 8.2 and 8.3 above have been treated by many physicists ever since the later years of the nineteenth century by using the following trick. Let us assume that the independent variable is \( t \) . Introduce a \ | No |
If \( \varphi \) is defined by \( \varphi \left( x\right) = \sin x \) for \( \left| x\right| < \pi \) and 0 elsewhere, then \( \operatorname{supp}\varphi = \left\lbrack {-\pi ,\pi }\right\rbrack \). | Although \( \varphi \left( 0\right) = 0 \), the point \( x = 0 \) belongs to the support, because every neighborhood of this point contains points where \( \varphi \left( x\right) \neq 0 \). | Yes |
Lemma 8.1 Suppose that \( f \) is continuous in \( \left\lbrack {a, b}\right\rbrack \) and differentiable in \( \rbrack a, b\lbrack \) , and suppose that the derivative \( {f}^{\prime }\left( x\right) \) has a limit \( A \), as \( x \searrow a \) . Then \( f \) has a right-hand derivative for \( x = a \), and this deri... | Proof. The mean-value theorem of Lagrange can be used on the subinterval \( \lbrack a, a + \) \( h\rbrack \), where \( h > 0 \) :\n\n\[ f\left( {a + h}\right) - f\left( a\right) = {f}^{\prime }\left( \xi \right) \cdot h\; \Leftrightarrow \;\frac{f\left( {a + h}\right) - f\left( a\right) }{h} = {f}^{\prime }\left( \xi \... | Yes |
First define a function \( \varphi \) by putting\n\n\[ \varphi \left( x\right) = \left\{ \begin{array}{ll} 0, & x \leq 0 \\ {e}^{-1/x}, & x > 0 \end{array}\right. \] | The substitution \( 1/x = t \) and letting \( t \rightarrow + \infty \), corresponding to \( x \searrow 0 \), shows that \( \varphi \) is continuous also at \( x = 0 \) . For \( x > 0 \), a few differentiations give\n\n\[ {\varphi }^{\prime }\left( x\right) = {e}^{-1/x} \cdot \frac{1}{{x}^{2}},{\varphi }^{\prime \prime... | Yes |
With \( \psi \) and \( B \) as before, let\n\n\[ \Psi \left( x\right) = \frac{1}{B}{\int }_{-\infty }^{x}\psi \left( y\right) {dy} \]\n\nThis gives a function in \( \mathcal{E} \), having the value 0 for \( x \leq 0 \) and 1 for \( x \geq 1 \) . Then put \( \Omega \left( x\right) = \Psi \left( x\right) - \Psi \left( {x... | This can be shown in the following way (compare Figure 8.2). For any fixed \( x \) , at most two terms in the sum are different from zero, and so the series is very much convergent. Furthermore, it is easy to see that \( \Phi \left( {x + 1}\right) = \Phi \left( x\right) \), so that \( \Phi \) has period 1 . Thus we can... | Yes |
Let \( f \) be a continuous function on \( \mathbf{R} \) such that there are constants \( M \) and \( m \) such that \( \left| {f\left( x\right) }\right| \leq M{\left( 1 + \left| x\right| \right) }^{m} \) for all \( x \) . Then we can define a tempered distribution \( {T}_{f} \) by letting\n\n\[ \n{T}_{f}\left\lbrack \... | It is clear that \( {T}_{f} \) is a linear mapping; the fact that it is continuous follows from the fact that convergence in \( \mathcal{S} \) implies uniform convergence even after multiplication by expressions such as \( {\left( 1 + \left| x\right| \right) }^{m + 2} \) :\n\n\[ \n\left| {{T}_{f}\left\lbrack {\varphi }... | Yes |
Define \( \delta \in {\mathcal{S}}^{\prime } \) by \( \delta \left\lbrack \varphi \right\rbrack = \varphi \left( 0\right) \). (The reader is asked to check linearity and continuity.) This distribution is called the Dirac distribution (Dirac function, Dirac measure), and this is the object announced in Examples 8.2 and ... | If \( f \) is an arbitrary distribution, as a rule there exists no \ | No |
The function \( 1/x \) cannot play the role of \( f \) in Example 8.11, since the integral will be divergent at the origin, However, we can define \( f\left\lbrack \varphi \right\rbrack \) using a symmetric limit: | \[ f\left\lbrack \varphi \right\rbrack = \mathop{\lim }\limits_{{\varepsilon \searrow 0}}{\int }_{\left| x\right| > \varepsilon }\frac{\varphi \left( x\right) }{x}{dx} = \mathop{\lim }\limits_{{\varepsilon \searrow 0}}\left( {{\int }_{-\infty }^{-\varepsilon } + {\int }_{\varepsilon }^{\infty }}\right) \frac{\varphi \l... | Yes |
If \( f\left( x\right) = {e}^{{x}^{2}} \), then \( f \) does not describe a tempered distribution according to Example 8.11. | Indeed, if we choose the test function \( \varphi \left( x\right) = {e}^{-{x}^{2}} \), which clearly belongs to \( \mathcal{S} \), the integral will diverge. The function \( f \) increases too fast. | Yes |
If \( \chi \) is a multiplicator function and \( f \in {\mathcal{S}}^{\prime } \), we can define the product \( {\chi f} \in {\mathcal{S}}^{\prime } \) by putting \( \left( {\chi f}\right) \left\lbrack \varphi \right\rbrack = f\left\lbrack {\chi \varphi }\right\rbrack \) for all \( \varphi \in \mathcal{S} \) . | Check that this is reasonable if \( f \) is an ordinary function, and that \( {\chi f} \) actually turns out to be a tempered distribution! | No |
What is the product \( {\chi \delta } \) ? | According to the definition we have\n\n\[ \left( {\chi \delta }\right) \left\lbrack \varphi \right\rbrack = \delta \left\lbrack {\chi \varphi }\right\rbrack = \chi \left( 0\right) \varphi \left( 0\right) = \chi \left( 0\right) \cdot \delta \left\lbrack \varphi \right\rbrack .\n\nThis result is often written in the form... | Yes |
Let \( f \) be P.V. \( 1/x \), and \( \chi \left( x\right) = x \) . What is \( {\chi f} \) ? | Indeed, \[ {\chi f}\left\lbrack \varphi \right\rbrack = f\left\lbrack {\chi \varphi }\right\rbrack = \mathop{\lim }\limits_{{\varepsilon \searrow 0}}{\int }_{\left| x\right| > \varepsilon }x \cdot \frac{\varphi \left( x\right) }{x}{dx} = \mathop{\lim }\limits_{{\varepsilon \searrow 0}}{\int }_{\left| x\right| > \vareps... | Yes |
Find the derivative of the Heaviside function \( H \) ! | By definition, it should emerge from the following calculation:\n\n\[ \n{H}^{\prime }\left\lbrack \varphi \right\rbrack = - H\left\lbrack {\varphi }^{\prime }\right\rbrack = - {\int }_{0}^{\infty }{\varphi }^{\prime }\left( x\right) {dx} = - {\left\lbrack \varphi \left( x\right) \right\rbrack }_{x = 0}^{\infty } = - \l... | Yes |
Find the derivatives of \( \delta \) : | \[ {\delta }^{\prime }\left\lbrack \varphi \right\rbrack = - \delta \left\lbrack {\varphi }^{\prime }\right\rbrack = - {\varphi }^{\prime }\left( 0\right) ,\;{\delta }^{\prime \prime }\left\lbrack \varphi \right\rbrack = - {\delta }^{\prime }\left\lbrack {\varphi }^{\prime }\right\rbrack = \delta \left\lbrack {\varphi ... | Yes |
What is the derivative of \( {\chi f} \), where \( \chi \) belongs to \( \mathcal{M} \) and \( f \in {\mathcal{S}}^{\prime } \)? | On the one hand,\n\n\[ \n{\left( \chi f\right) }^{\prime }\left\lbrack \varphi \right\rbrack = - \left( {\chi f}\right) \left\lbrack {\varphi }^{\prime }\right\rbrack = - f\left\lbrack {\chi {\varphi }^{\prime }}\right\rbrack \n\] \n\nand on the other,\n\n\[ \n\left( {{\chi }^{\prime }f + \chi {f}^{\prime }}\right) \le... | Yes |
Consider the function \( f : \mathbf{R} \rightarrow \mathbf{R} \) that is given by\n\n\[ f\left( t\right) = \left\{ \begin{array}{ll} 1 - {t}^{2} & \text{ for }t < - 2 \\ t + 2 & \text{ for } - 2 < t < 1 \\ 1 - t & \text{ for }t > 1 \end{array}\right. \]\n | This can now be compressed into one formula:\n\n\[ f\left( t\right) = \left( {1 - {t}^{2}}\right) \left( {1 - H\left( {t + 2}\right) }\right) + \left( {t + 2}\right) \left( {H\left( {t + 2}\right) - H\left( {t - 1}\right) }\right) \]\n\n\[ + \left( {1 - t}\right) H\left( {t - 1}\right) \]\n\n\[ = \left( {1 - {t}^{2}}\r... | Yes |
The function \( f\left( x\right) = \left| {{x}^{2} - 1}\right| \) can be rewritten as | \[ f\left( x\right) = \left( {{x}^{2} - 1}\right) H\left( {x - 1}\right) + \left( {1 - {x}^{2}}\right) \left( {H\left( {x + 1}\right) - H\left( {x - 1}\right) }\right) + \left( {{x}^{2} - 1}\right) \left( {1 - H\left( {x + 1}\right) }\right) = \left( {{x}^{2} - 1}\right) \left( {{2H}\left( {x - 1}\right) - {2H}\left( {... | Yes |
What connection holds between the derivatives of \( f \) and \( g \) ? | By the definition of \( {g}^{\prime } \) we should have\n\n\[ \n{g}^{\prime }\left\lbrack \varphi \right\rbrack = - g\left\lbrack {\varphi }^{\prime }\right\rbrack = - \int f\left( {{ax} + b}\right) {\varphi }^{\prime }\left( x\right) {dx}. \n\] | No |
Theorem 8.1 If \( f \in {\mathcal{S}}^{\prime } \), then \( {f}^{\prime } = 0 \) if and only if \( f \) is a constant function. | To prove the theorem we need a lemma. | No |
Lemma 8.2 A test function \( \varphi \in \mathcal{S} \) is the derivative of another function in \( \mathcal{S} \) if and only if \( {\int }_{\mathbf{R}}\varphi \left( x\right) {dx} = 0 \) . | Clearly, if \( \varphi = {\Phi }^{\prime } \), then \( \int \varphi = \Phi \left( \infty \right) - \Phi \left( {-\infty }\right) = 0 \), so the hard part is to prove the converse statement. If \( \int \varphi = 0 \), we can define a primitive function \( \Phi \) by\n\n\[ \Phi \left( x\right) = {\int }_{-\infty }^{x}\va... | Yes |
Let \( f\left( x\right) = 1 \) for all \( x \) . What is the Fourier transform \( \widehat{f} \) ? | We should have\n\n\[ \widehat{f}\left\lbrack \varphi \right\rbrack = f\left\lbrack \widehat{\varphi }\right\rbrack = {\int }_{\mathbf{R}}f\left( x\right) \widehat{\varphi }\left( x\right) {dx} = {\int }_{\mathbf{R}}\widehat{\varphi }\left( x\right) {dx} \]\n\n\[ = {2\pi } \cdot \frac{1}{2\pi }{\int }_{\mathbf{R}}\wideh... | Yes |
Take \( f\left( x\right) = {x}^{n}\left( {n\text{integer} \geq 0}\right) \) . This function defines a tempered distribution, and its transform satisfies\n\n\[ \widehat{f}\left\lbrack \varphi \right\rbrack = f\left\lbrack \widehat{\varphi }\right\rbrack = {\int }_{\mathbf{R}}{x}^{n}\widehat{\varphi }\left( x\right) {dx}... | By the ordinary rules for Fourier transforms we have that \( {\left( ix\right) }^{n}\widehat{\varphi }\left( x\right) \) is the transform of the function \( {\varphi }^{\left( n\right) } \), which means that \( {x}^{n}\widehat{\varphi }\left( x\right) \) is the transform of \( {\left( -i\right) }^{n}{\varphi }^{\left( ... | Yes |
Theorem 8.3 If \( f, g \in {\mathcal{S}}^{\prime } \), then\n\n(a) \( f \) is even/odd \( \Leftrightarrow \widehat{f} \) is even/odd.\n\n(b) \( \widehat{\widehat{f}} = {2\pi }\check{f} \).\n\n(c) \( {\widehat{f}}_{a} = {e}^{-{ia\omega }}\widehat{f} \).\n\n(d) \( \mathcal{F}\left( {{e}^{iax}f}\right) = {\left( \widehat{... | (d): The effect of the left-hand member on a test function \( \varphi \) is rewritten:\n\n\[ \mathcal{F}\left( {{e}^{iax}f}\right) \left\lbrack \varphi \right\rbrack = \left( {{e}^{iax}f}\right) \left\lbrack \widehat{\varphi }\right\rbrack = f\left\lbrack {{e}^{iax}\widehat{\varphi }}\right\rbrack = f\left\lbrack {\mat... | Yes |
Transform \( f = \) P.V. \( 1/x \) (Example 8.15, page 205): we have seen that \( {xf} = 1 \) . Transformation gives \( {iD}\widehat{f} = {2\pi \delta } = {2\pi }{H}^{\prime } \), which can be rewritten as \( {iD}\left( {\widehat{f} + {2\pi iH}}\right) = 0 \) . | By Theorem 8.1 it follows that \( \widehat{f} + {2\pi iH} = c = \) constant, whence \( \widehat{f} = c - {2\pi iH} \) . To determine the constant we notice that \( f \) is odd, and thus \( \widehat{f} \) must also be odd. This gives \( c = {\pi i} \) and \( \widehat{f} = {\pi i}\left( {1 - {2H}}\right) \) . If we intro... | Yes |
Example 8.30. \( H = \) the Heaviside function. In Example 8.29 we saw that\n\n\[ \n\mathcal{F}\left( {\text{ P.V. }\frac{1}{x}}\right) \left( \omega \right) = {\pi i}\left( {1 - {2H}\left( \omega \right) }\right) .\n\] | Since both sides are odd distributions, rule (b) gives\n\n\[ \n{\pi i}\mathcal{F}\left( {1 - {2H}}\right) \left( \omega \right) = {2\pi }{\left( \text{ P.V. }\frac{1}{\omega }\right) }^{ \vee } = - {2\pi }\text{ P.V. }\frac{1}{\omega }.\n\]\n\nOn the other hand, \( \mathcal{F}\left( {1 - {2H}}\right) = \widehat{1} - 2\... | Yes |
Theorem 8.4 If \( f \in {\mathcal{S}}^{\prime } \), then \( {xf}\left( x\right) \) is the zero distribution if and only if \( f = {A\delta } \) for some constant \( A \) . | Proof. Transformation of the equation \( {xf}\left( x\right) = 0 \) gives \( i{\widehat{f}}^{\prime } = 0 \), and by Theorem 8.1 this means that \( \widehat{f} = C \), where \( C \) is a constant. Transform back again: since \( \widehat{1} = {2\pi \delta } \), we find that \( f \) must be a constant times \( \delta \),... | Yes |
A simple example of a distribution with period \( {2\pi } \) is the so-called pulse train (see Figure 8.5): \n\n\[ \n\mathrm{{III}} = \mathop{\sum }\limits_{{n \in \mathbf{Z}}}{\delta }_{2\pi n}\;\text{ or }\;\mathrm{{III}}\left( x\right) = \mathop{\sum }\limits_{{n \in \mathbf{Z}}}\delta \left( {x - n \cdot {2\pi }}\r... | That this is actually a tempered distribution hinges essentially on the fact that the sum \n\n\[ \n\mathrm{{III}}\left\lbrack \varphi \right\rbrack = \mathop{\sum }\limits_{{n \in \mathbf{Z}}}\varphi \left( {x + n \cdot {2\pi }}\right) \n\] \n\nis convergent, which follows from the estimate \( \left| {\varphi \left( {x... | Yes |
Let \( a > 0 \) . Let us find a fundamental solution of the familiar operator \( P\left( D\right) = {D}^{2} + {a}^{2} \), i.e., we want to find a distribution \( E \) such that \( {E}^{\prime \prime } + {a}^{2}E = \delta \) . | Fourier transformation gives\n\n\[ \n{\left( i\omega \right) }^{2}\widehat{E} + {a}^{2}\widehat{E} = \widehat{\delta } = 1 \n\]\n\nand at least one solution of this ought to be found by solving for \( \widehat{E} \) like this:\n\n\[ \n\widehat{E} = \frac{1}{-{\omega }^{2} + {a}^{2}} = \frac{1}{2a}\left( {\frac{1}{\omeg... | Yes |
Theorem 9.1 An absolutely convergent series remains absolutely convergent and its sum does not change as a result of any rearrangement of its terms. | Proof. Put \( s = \mathop{\sum }\limits_{{k = 1}}^{\infty }{a}_{k} \) . Let \( \varphi \) be a rearrangement bijection and put \( {t}_{n} = \mathop{\sum }\limits_{{k = 1}}^{n}{a}_{\varphi \left( k\right) } \) . The starting point is that \( \sum {a}_{k} \) is absolutely convergent, i.e., that\n\n\[ \mathop{\sum }\limit... | Yes |
Example 9.2. It is well known that \( \frac{1}{1 - x} = \mathop{\sum }\limits_{{k = 0}}^{\infty }{x}^{k} \), where the series is absolutely convergent for \( \left| x\right| < 1 \) . Then, by Theorem 9.3, we also have | \[ \frac{1}{{\left( 1 - x\right) }^{2}} = \frac{1}{1 - x} \cdot \frac{1}{1 - x} = \mathop{\sum }\limits_{{j = 0}}^{\infty }{x}^{j}\mathop{\sum }\limits_{{k = 0}}^{\infty }{x}^{k} = \mathop{\sum }\limits_{{j, k = 0}}^{\infty }{x}^{j + k}. \] We choose to sum this series \ | No |
Example 9.3. Convolutions have been encountered a few times in different situations earlier in the book. (A unified discussion of this notion is found in Appendix A.) Here we take a look at the case of convolutions of sequences. If \( a = {\left\{ {a}_{i}\right\} }_{i = - \infty }^{\infty } \) and \( b = {\left\{ {b}_{... | If, say, \( a \) and \( b \) are absolutely summable, which means that the series \( \mathop{\sum }\limits_{{-\infty }}^{\infty }\left| {a}_{i}\right| = s \) and \( \mathop{\sum }\limits_{{-\infty }}^{\infty }\left| {b}_{i}\right| = t \) are convergent, this is true, as is seen by the following computations:\n\n\[ \n\m... | Yes |
Let \( f \) be defined by \( f\left( {x, y}\right) = {xy} \) for \( \left| x\right| < \pi ,\left| y\right| < \pi \) . We compute its Fourier coefficients \( {c}_{mn} \) . | We shall need the value of the integral\n\n\[ \n{\alpha }_{k} \mathrel{\text{:=}} {\int }_{-\pi }^{\pi }t{e}^{-{ikt}}{dt} = {\left\lbrack t\frac{{e}^{-{ikt}}}{-{ik}}\right\rbrack }_{-\pi }^{\pi } - \frac{1}{-{ik}}{\int }_{-\pi }^{\pi }{e}^{-{ikt}}{dt} \n\]\n\n\[ \n= {2\pi i}\frac{{\left( -1\right) }^{k}}{k}\;\text{ for... | Yes |
Define \( f \) by \( f\left( {x, y}\right) = \) \( {xy} - {x}^{2} \) for \( - 1 < x < 1 \) and \( 0 < y < {2\pi } \), and assume the period to be 2 in the \( x \) variable and \( {2\pi } \) in the \( y \) variable. The Fourier coefficients are given by the formula | \[ {c}_{mn} = \frac{1}{2 \cdot {2\pi }}{\int }_{-1}^{1}{dx}{\int }_{0}^{2\pi }\left( {{xy} - {x}^{2}}\right) {e}^{-i\left( {{m\pi x} + {ny}}\right) }{dy}. \] | Yes |
An important application of the transform is found in the theory of probability, where the multi-dimensional normal distribution is used. In dimension \( d \), a normalized version of this is described by the density function\n\n\[ f\left( \mathbf{x}\right) = \frac{1}{{\left( 2\pi \right) }^{d/2}}\exp \left( {-\frac{1}... | It holds that \( \parallel f{\parallel }_{1} = 1 \), and it is easy to compute the Fourier transform or characteristic function\n\n\[ \widehat{f}\left( \mathbf{\omega }\right) = \frac{1}{{\left( 2\pi \right) }^{d/2}}{\int }_{{\mathbf{R}}^{d}}{e}^{-{\left| \mathbf{x}\right| }^{2}/2}{e}^{-i\mathbf{\omega } \cdot \mathbf{... | Yes |
Proposition 3.1. Let \( \left( {U, x}\right) \) be a chart around \( p \) . Then any tangent vector \( v \in {M}_{p} \) can be uniquely written as a linear combination \( v = \mathop{\sum }\limits_{i}{\alpha }_{i}\partial /\partial {x}^{i}\left( p\right) \) . In fact, \( {\alpha }_{i} = v\left( {x}^{i}\right) \) . | Proof. We may assume without loss of generality that \( x\left( p\right) = 0 \), and that \( x\left( U\right) \) is star-shaped. By Lemma 3.1, any \( f \in \mathcal{F}M \) satisfies \( f \circ {x}^{-1} = \) \( f\left( p\right) + \sum {u}^{i}{\psi }_{i} \), with \( {\psi }_{i}\left( 0\right) = \partial /\partial {x}^{i}... | Yes |
Proposition 4.1. With notation as in Definition 4.1, let \( x \) be a coordinate map around \( p \in U, y \) a coordinate map around \( f\left( p\right) \in N \) . Then the matrix of \( {f}_{*p} \) with respect to the bases \( \left\{ {\partial /\partial {x}^{i}\left( p\right) }\right\} \) and \( \left\{ {\partial /\pa... | Proof.\n\n\[ \n{f}_{*p}\frac{\partial }{\partial {x}^{j}}\left( p\right) = \mathop{\sum }\limits_{i}{f}_{*p}\frac{\partial }{\partial {x}^{j}}\left( p\right) \left( {y}^{i}\right) \frac{\partial }{\partial {y}^{i}}\left( {f\left( p\right) }\right) = \mathop{\sum }\limits_{i}\frac{\partial }{\partial {x}^{j}}\left( p\ri... | Yes |
Proposition 6.1. If \( f : {M}^{n} \rightarrow {N}^{k} \) is an immersion, then for any \( p \in M \), there exists a neighborhood \( U \) of \( p \), and a coordinate map \( y \) defined on some neighborhood \( V \) of \( f\left( p\right) \) such that (1) A point \( q \) belongs to \( f\left( U\right) \cap V \) iff \(... | Proof. Consider the inclusion \( \imath : {\mathbb{R}}^{n} \rightarrow {\mathbb{R}}^{k} \), and let \( x \) be a coordinate map around \( p \) with \( x\left( p\right) = 0,\widetilde{y} \) a coordinate map around \( f\left( p\right) \) with \( \left( {\widetilde{y} \circ f}\right) \left( p\right) = 0 \) . Since \( \wid... | Yes |
Proposition 6.2. Let \( q \) be a regular value of \( f : {M}^{n} \rightarrow {N}^{k} \), where \( n \geq k \), and suppose that \( A \mathrel{\\text{:=}} {f}^{-1}\left( q\right) \neq \varnothing \). Then for \( p \in A,{\imath }_{*p}{A}_{p} = \ker {f}_{*p} \). | Proof. Since both subspaces have common dimension \( n - k \), it suffices to check that \( {\imath }_{*p}{A}_{p} \subset \ker {f}_{*p} \). Let \( v \in {A}_{p} \). For \( \phi \in \mathcal{F}N \), we have\n\n\[ \n\\left( {{f}_{*p}{\imath }_{*p}v}\\right) \\left( \\phi \\right) = {\\left( f \\circ \\imath \\right) }_{*... | Yes |
Proposition 7.1. Let \( X : U \rightarrow {TM} \) be a map such that \( \pi \circ X = {1}_{U} \) . The following statements are equivalent:\n\n(1) \( X \) is a vector field on \( U \) (i.e., \( X \), as a map, is differentiable).\n\n(2) If \( \left( {V, x}\right) \) is a chart with \( V \subset U \), then \( X{x}^{i} \... | Proof. (1) \( \Rightarrow \) (2): Recall that \( \left( {V, x}\right) \) induces a coordinate map \( \bar{x} \) on \( {\pi }^{-1}\left( V\right) \) , where \( \bar{x}\left( v\right) = \left( {x \circ \pi \left( v\right), v\left( {x}^{1}\right) ,\ldots, v\left( {x}^{n}\right) }\right) \) . Since \( X \) is smooth, \( {\... | Yes |
Proposition 8.1. Let \( \\left( {U, x}\\right) \) denote a chart of \( {M}^{n} \) . Then \( \\left\\lbrack {\\partial /\\partial {x}^{i},\\partial /\\partial {x}^{j}}\\right\\rbrack \\equiv \) \( 0 \) for \( 1 \\leq i, j \\leq n \) . | Proof. For \( \\phi \\in \\mathcal{F}U \) ,\n\n\[ \n\\left\\lbrack {\\frac{\\partial }{\\partial {x}^{i}},\\frac{\\partial }{\\partial {x}^{j}}}\\right\\rbrack \\phi = \\frac{\\partial }{\\partial {x}^{i}}\\frac{\\partial }{\\partial {x}^{j}}\\phi - \\frac{\\partial }{\\partial {x}^{j}}\\frac{\\partial }{\\partial {x}^... | Yes |
Proposition 8.2. Let \( f : M \rightarrow N \) be differentiable, \( {X}_{i} \in \mathfrak{X}M,{Y}_{i} \in \mathfrak{X}N \) , \( i = 1,2 \) . If \( {X}_{i} \) and \( {Y}_{i} \) are \( f \) -related, then \( \left\lbrack {{X}_{1},{X}_{2}}\right\rbrack \) and \( \left\lbrack {{Y}_{1},{Y}_{2}}\right\rbrack \) are \( f \) ... | Proof. If \( \phi \in \mathcal{F}N \), then for \( p \in M \) ,\n\n\[{\left\lbrack {Y}_{1},{Y}_{2}\right\rbrack }_{f\left( p\right) }\phi = {Y}_{1 \mid f\left( p\right) }\left( {{Y}_{2}\phi }\right) - {Y}_{2 \mid f\left( p\right) }\left( {{Y}_{1}\phi }\right) = {f}_{ * }{X}_{1 \mid p}\left( {{Y}_{2}\phi }\right) - {f}_... | Yes |
Proposition 8.3. Let \( {\Phi }_{t} \) and \( {\Psi }_{s} \) denote local flows of \( X \) and \( Y \in \mathfrak{X}M \) . Then \( \left\lbrack {X, Y}\right\rbrack \equiv 0 \) iff \( {\Phi }_{t} \circ {\Psi }_{s} = {\Psi }_{s} \circ {\Phi }_{t} \) for all \( s, t \) . | Proof. Suppose that \( {\Phi }_{t} \circ {\Psi }_{s} = {\Psi }_{s} \circ {\Phi }_{t} \) . By Exercise 21 below, \( Y \) is \( {\Phi }_{t} \) - related to itself; i.e., \( {\Phi }_{t * }Y = Y \circ {\Phi }_{t} \), so that \( {\Phi }_{-t * }Y \circ {\Phi }_{t} = Y.{L}_{X}Y \) then vanishes by definition.\n\nConversely, s... | No |
Proposition 9.1. If for every \( p \in M \) there exists an integral manifold \( N\left( p\right) \) of \( \Delta \) with \( p \in N\left( p\right) \), then \( \Delta \) is integrable. | Proof. Let \( X, Y \in \Delta, p \in M \) . We must show that \( {\left\lbrack X, Y\right\rbrack }_{p} \in {\Delta }_{p} \) . Since \( {i}_{*q} : N{\left( p\right) }_{q} \rightarrow {\Delta }_{q} \) is an isomorphism for every \( q \in N\left( p\right) \), there exist vector fields \( \widetilde{X} \) and \( \widetilde... | Yes |
Proposition 10.1. \( {V}^{ * } \otimes W \) is canonically isomorphic to \( \operatorname{Hom}\left( {V, W}\right) \) . In particular, \( \dim \left( {V \otimes W}\right) = \dim V \cdot \dim W \) . In fact if \( \left\{ {e}_{i}\right\} \) and \( \left\{ {f}_{j}\right\} \) are bases of \( V \) and \( W \) respectively, ... | Proof. The map\n\n\[ \n{V}^{ * } \times W \rightarrow \operatorname{Hom}\left( {V, W}\right) \n\]\n\n\[ \n\left( {\alpha, w}\right) \mapsto \left( {v \mapsto \left( {\alpha v}\right) \cdot w}\right) \n\]\n\nis bilinear, and by Lemma 10.1 induces a unique linear map \( L : {V}^{ * } \otimes W \rightarrow \) \( \operator... | Yes |
Proposition 10.2. \( {T}_{r, s}\left( V\right) \) is canonically isomorphic to \( {M}_{s, r}\left( V\right) \) . | Proof. Define a nonsingular pairing \( b \) of \( {T}_{r, s}\left( V\right) \) with \( {T}_{r, s}\left( {V}^{ * }\right) \) as follows: for \( u = {u}_{1} \otimes \cdots \otimes {u}_{r} \otimes {v}_{r + 1}^{ * } \otimes \cdots \otimes {v}_{r + s}^{ * } \in {T}_{r, s}\left( V\right) \) and \( {v}^{ * } = {v}_{1}^{ * } \... | Yes |
Proposition 10.3. If \( \left\{ {{e}_{1},\ldots ,{e}_{n}}\right\} \) is a basis of \( V \), then \( \left\{ {{e}_{{i}_{1}} \land \cdots \land {e}_{{i}_{k}} \mid }\right. \) \( \left. {1 \leq {i}_{1} < \cdots < {i}_{k} \leq n}\right\} \) is a basis of \( {\Lambda }_{k}\left( V\right) \) . Thus, \( \dim {\Lambda }_{k}\le... | Proof. Since \( \left( {{e}_{i} + {e}_{j}}\right) \land \left( {{e}_{i} + {e}_{j}}\right) = 0,{e}_{i} \land {e}_{j} = - {e}_{j} \land {e}_{i} \), so the set in the statement spans \( {\Lambda }_{k}\left( V\right) \) . To see that it is linearly independent, suppose that \( \mathop{\sum }\limits_{j}{\alpha }_{j}{e}_{{i}... | Yes |
Proposition 10.4. There are canonical isomorphisms \( {\Lambda }_{k}\left( {V}^{ * }\right) \cong {\Lambda }_{k}{\left( V\right) }^{ * } \cong \) \( {A}_{k}\left( V\right) \) . | Proof. The second isomorphism is the one induced from Exercise 26. For the first one, there is a unique bilinear map \( b : {\Lambda }_{k}\left( {V}^{ * }\right) \times {\Lambda }_{k}\left( V\right) \rightarrow \mathbb{R} \) which is given on decomposable elements by\n\n\[ b\left( {{v}_{1}^{ * } \land \cdots \land {v}_... | Yes |
Proposition 10.5. For \( \alpha \in {A}_{k}\left( V\right) \) and \( \beta \in {A}_{l}\left( V\right) \) ,\n\n\[ \alpha \land \beta \left( {{v}_{1},\ldots ,{v}_{k + l}}\right) = \mathop{\sum }\limits_{{\sigma \in {\bar{P}}_{k + l}}}\left( {\operatorname{sgn}\sigma }\right) \alpha \left( {{v}_{\sigma \left( 1\right) },\... | Proof. It suffices to establish the result for decomposable elements \( \alpha = \) \( {u}_{1}^{ * } \land \cdots \land {u}_{k}^{ * } \) and \( \beta = {w}_{1}^{ * } \land \cdots \land {w}_{l}^{ * } \) . Notice that\n\n\[ \alpha \left( {{v}_{1},\ldots ,{v}_{k}}\right) = \det \left( {{u}_{i}^{ * }{v}_{j}}\right) = \math... | Yes |
Proposition 11.1. The differentiable structure on \( M \) induces differentiable structures on \( {T}_{r, s}\left( M\right) ,{\Lambda }_{k}^{ * }\left( M\right) \), and \( {\Lambda }^{ * }\left( M\right) \) for which the projection onto \( M \) are submersions. | Proof. Given a chart \( \left( {U, x}\right) \) of \( M \), consider bases \( \left\{ {\partial /\partial {x}_{\mid p}^{i}}\right\} \) and \( \left\{ {d{x}_{\mid p}^{i}}\right\} \) of \( {M}_{p} \) and \( {M}_{p}^{ * } \) respectively. These in turn induce bases for the vector spaces \( {T}_{r, s}\left( {M}_{p}\right) ... | Yes |
Proposition 11.2. \( T \) is a tensor field of type \( \left( {r, s}\right) \) on \( M \) iff for any chart \( \left( {U, x}\right) \) of \( M \) , \n\n\[ \n{T}_{\mid U} = \sum {T}_{{i}_{1},\ldots ,{i}_{r};{j}_{1},\ldots ,{j}_{s}}\frac{\partial }{\partial {x}^{{i}_{1}}} \otimes \cdots \otimes \frac{\partial }{\partial ... | Observe that the functions in Proposition 11.2 satisfy \n\n\[ \n{T}_{{i}_{1},\ldots ,{i}_{r};{j}_{1},\ldots ,{j}_{s}} = T\left( {d{x}^{{i}_{1}},\ldots, d{x}^{{i}_{r}},\frac{\partial }{\partial {x}^{{j}_{1}}},\ldots ,\frac{\partial }{\partial {x}^{{j}_{s}}}}\right) , \n\] \n\n\[ \n{\alpha }_{{i}_{1},\ldots ,{i}_{k}} = \... | Yes |
Proposition 11.3. A multilinear map \( T : \mathfrak{X}\left( M\right) \times \cdots \times \mathfrak{X}\left( M\right) \rightarrow \mathbb{R} \) is a tensor field iff it is linear over \( \mathcal{F}\left( M\right) \) . | Proof. The condition is necessary by the above remark. Conversely, suppose \( T \) is multilinear and linear over \( \mathcal{F}\left( M\right) \) . We claim that \( T \) \ | No |
Proposition 12.1. \( {\partial }^{2} = 0 \) . | Proof. Since \( \partial \) is linear, it suffices to show that \( \partial \partial c = 0 \) for a singular \( n \) -cube \( c \) . Now,\n\n\[ \partial \left( {\partial c}\right) = \partial \left( {\mathop{\sum }\limits_{{i = 1}}^{n}\mathop{\sum }\limits_{{k = 0,1}}{\left( -1\right) }^{i + k}{c}_{i, k}}\right) = \math... | No |
Proposition 14.1. For \( f : G \rightarrow \mathbb{R}, g \in G,{\int }_{G}f = {\int }_{G}f \circ {L}_{g} = {\int }_{G}f \circ {R}_{g} \) . | Proof. For the first identity, notice that \( {\int }_{G}{L}_{g}^{ * }\left( {f\omega }\right) = {\int }_{G}{f\omega } \) by Exercise 39 below, since \( {L}_{g} \) is an orientation-preserving diffeomorphism of \( G \) . But \( \omega \) is left-invariant, so that\n\n\[ \n{\int }_{G}f = {\int }_{G}{f\omega } = {\int }_... | No |
Proposition 15.1. Let \( {M}^{n} \) be an oriented, compact manifold without boundary, \( n > 0 \) . Then \( {H}^{n}\left( M\right) \neq 0 \) . | Proof. Let \( \omega \) be a volume form on \( M \) ; i.e., \( \omega \) is a nowhere-zero \( n \) -form with \( \omega \left( {{v}_{1},\ldots ,{v}_{n}}\right) > 0 \) for any positively oriented basis \( {v}_{1},\ldots ,{v}_{n} \) of \( {M}_{p}, p \in M \) . If \( c : {\left\lbrack 0,1\right\rbrack }^{n} \rightarrow M ... | Yes |
Proposition 15.3. Let \( {M}^{n} \) be connected and orientable (without boundary). Given \( \omega \in {Z}_{c}^{n}\left( M\right) ,\omega \) belongs to \( {B}_{c}^{n}\left( M\right) \) iff \( {\int }_{M}\omega = 0 \) . | Proof. If \( \omega \) belongs to \( {B}_{c}^{n}\left( M\right) \), then \( \omega = {d\alpha } \), where \( \alpha \) has compact support. By Stokes' theorem,\n\n\[ \n{\int }_{M}\omega = {\int }_{M}{d\alpha } = {\int }_{\partial M}\alpha = 0.\n\]\n\nWe merely illustrate the proof of the converse in the case \( M = \ma... | No |
Proposition 2.1. Let \( {\left\{ {U}_{\alpha }\right\} }_{\alpha \in A} \) be an open cover of a manifold \( B \), and \( G \) a Lie group acting effectively on a manifold \( F \) . Suppose there is a collection of maps \( {f}_{\alpha ,\beta } : {U}_{\alpha } \cap {U}_{\beta } \rightarrow G \) such that\n\n(2.1)\n\n\[ ... | Proof. Consider the disjoint union \( { \cup }_{\alpha \in A}\left( {{U}_{\alpha } \times F}\right) \), and the quotient space \( M \) under the equivalence relation:\n\n\[ \n\left( {p,{q}_{1}}\right) \sim \left( {p,{q}_{2}}\right) \text{iff}{q}_{2} = {f}_{\alpha ,\beta }\left( p\right) {q}_{1}\text{for some}\alpha ,\b... | Yes |
Proposition 3.1. If \( G \) acts effectively on the homogeneous space \( M = \) \( G/H \), then the tangent bundle of \( M \) is equivalent to the bundle \( G{ \times }_{H}{M}_{p} \rightarrow M \) , where \( H \) acts on \( {M}_{p} \) via the linear isotropy representation at \( p \) . | Proof. Consider the map \( f : G{ \times }_{H}{M}_{p} \rightarrow {TM} \) defined by \( f\left\lbrack {g, u}\right\rbrack = {g}_{ * }u \) , which is clearly smooth, and linear on each fiber. Its inverse is given as follows: if \( v \in {M}_{q} \), then by transitivity of the action of \( G \), there exists some \( g \i... | Yes |
Proposition 4.1. Let \( \xi = \pi : E \rightarrow B \) be a rank \( n \) vector bundle, and define \( {E}^{ * } = { \cup }_{b \in B}{E}_{b}^{ * } \) . For \( \alpha \in {E}_{b}^{ * } \), let \( {\pi }^{ * }\left( \alpha \right) = b \) . There exists a natural rank \( n \) vector bundle structure on \( {\xi }^{ * } = {\... | Proof. Let \( \left( {\pi ,\phi }\right) \) be a bundle chart of \( \xi \) over \( U \subset B \) . Since \( {\phi }_{\mid {E}_{b}} : {E}_{b} \rightarrow {\mathbb{R}}^{n} \) is an isomorphism for each \( b \in U \), so is \( {\bar{\phi }}_{\mid {E}_{b}^{ * }} : {E}_{b}^{ * } \rightarrow {\mathbb{R}}^{n * } \cong {\math... | Yes |
Proposition 5.1. \( {f}^{ * }\xi = {\pi }_{1} : {f}^{ * }M \rightarrow \bar{B} \) is a fiber bundle with fiber \( F \) and group \( G \), called the pullback bundle of \( \xi \) via \( f \), and \( {\pi }_{2} : {f}^{ * }M \rightarrow M \) is a bundle map covering \( f \) . Furthermore, \( {f}^{ * }\xi \) is uniquely ch... | Proof. A bundle chart \( \left( {\pi ,\phi }\right) \) of \( \xi \) over \( U \subset B \) induces a chart \( \left( {{\pi }_{1},\bar{\phi }}\right) \) of \( {f}^{ * }\xi \) over \( {f}^{-1}\left( U\right) \), where \( \bar{\phi } = \phi \circ {\pi }_{2} \) . It is easily checked that the transition functions satisfy \... | Yes |
Proposition 5.2. Let \( {\xi }_{i} = {\pi }_{i} : {E}_{i} \rightarrow {B}_{i} \) be vector bundles over \( {B}_{i}, i = 1,2 \) . If \( h : {E}_{1} \rightarrow {E}_{2} \) maps each fiber \( {\pi }_{1}^{-1}\left( {b}_{1}\right) \) linearly into a fiber \( {\pi }_{2}^{-1}\left( {b}_{2}\right) \), then \( h = f \circ g \),... | Proof. Consider the pullback bundle \( {\bar{h}}^{ * }{\xi }_{2} \), where \( \bar{h} : {B}_{1} \rightarrow {B}_{2} \) is the map induced by \( h \) . If \( p{r}_{2} : {\bar{h}}^{ * }{E}_{2} \rightarrow {E}_{2} \) is the bundle map given by projection onto the second factor, then \( h = p{r}_{2} \circ g \), where \( g ... | Yes |
Proposition 5.3. If \( 0 \rightarrow {\xi }_{1}\overset{h}{ \rightarrow }{\xi }_{2}\overset{f}{ \rightarrow }{\xi }_{3} \rightarrow 0 \) is an exact sequence of homomorphisms, then there exists an equivalence \( g : {\xi }_{2} \rightarrow {\xi }_{1} \oplus {\xi }_{3} \) with \( g \circ h : {\xi }_{1} \rightarrow {\xi }... | Proof. Consider a Euclidean metric \( \langle \) , \( \rangle {on}{\xi }_{2} \) (cf. Theorem 4.1). Since the metric is smooth as a section, the orthogonal projection \( \pi : {\xi }_{2} \rightarrow h\left( {\xi }_{1}\right) \) is a bundle homomorphism. Being onto, its kernel \( h{\left( {\xi }_{1}\right) }^{ \bot } \) ... | Yes |
Corollary 6.3 (The Homotopy Covering Property). Let \( \xi \) denote a fiber bundle over \( B \) . If \( f, g : \bar{B} \rightarrow B \) are homotopic, then \( {f}^{ * }\xi \cong {g}^{ * }\xi \) . | Proof. Let \( H : \bar{B} \times I \rightarrow B \) be a homotopy with \( H \circ {\imath }_{0} = f \) and \( H \circ {\imath }_{1} = g \) . By Corollary 6.2,\n\n\[ \n{f}^{ * }\xi = {\imath }_{0}^{ * }\left( {{H}^{ * }\xi }\right) \cong {\imath }_{1}^{ * }\left( {{H}^{ * }\xi }\right) = {g}^{ * }\xi .\n\] | Yes |
Proposition 2.1. The fundamental group \( {\pi }_{1}\left( {X, p}\right) \) acts on \( {\pi }_{n}\left( {X, p}\right) \) by means of \( \left\lbrack c\right\rbrack \left( \left\lbrack f\right\rbrack \right) = {c}_{\# }\left( \left\lbrack f\right\rbrack \right) \) . | Although we used a particular homotopy \( H \) of \( f \) to construct \( c\left( f\right) \), the only condition required of \( H \) is that \( H\left( {\partial {I}^{n}, t}\right) = c\left( t\right) \) :\n\nLEMMA 2.1. Let \( c \) be a curve in \( X \) from \( {p}_{0} \) to \( {p}_{1} \) . If \( f \in {C}^{n}\left( {X... | Yes |
Proposition 2.2. If \( X \) is path-connected, then there is a bijection between the set of orbits \( {\pi }_{n}\left( {X, p}\right) /{\pi }_{1}\left( {X, p}\right) \) and the set \( \left\lbrack {{S}^{n}, X}\right\rbrack \) of (free) homotopy classes of maps \( {S}^{n} \rightarrow X \) . | Proof. We have already seen that any \( f : \left( {{I}^{n},\partial {I}^{n}}\right) \rightarrow \left( {X, p}\right) \) can be viewed as \( f : \left( {{S}^{n}, q}\right) \rightarrow \left( {X, p}\right) \), yielding a map \( h : {\pi }_{n}\left( {X, p}\right) \rightarrow \left\lbrack {{S}^{n}, X}\right\rbrack \) . Fu... | Yes |
Proposition 5.1. If \( {\xi }^{n + k} \) is a rank \( n + k \) bundle over \( {S}^{n} \), then there exists a rank \( n \) bundle \( {\eta }^{n} \) over \( {S}^{n} \) such that \( {\xi }^{n + k} \cong {\eta }^{n} \oplus {\epsilon }^{k} \), where \( {\epsilon }^{k} \) denotes the trivial bundle of rank \( k \) . | Proof. The exact homotopy sequence of the bundle \( {SO}\left( {n + 1}\right) \rightarrow {S}^{n} \) implies that \( {\pi }_{n - 1}\left( {{SO}\left( n\right) }\right) \rightarrow {\pi }_{n - 1}\left( {{SO}\left( {n + 1}\right) }\right) \) is surjective. By Examples and Remarks 3.1(iii), the inclusion homomorphism \( {... | Yes |
Proposition 1.1. Let \( \mathcal{H} \) be a connection on a vector bundle \( \xi = \pi : E \rightarrow \) \( M, c : \left\lbrack {a, b}\right\rbrack \rightarrow M \) a curve in \( M \) . For any \( u \in {E}_{c\left( a\right) } \), there exists a unique parallel section \( {X}_{u} \) of \( \xi \) along \( c \) such tha... | Proof. Let \( D \) denote the standard coordinate vector field on \( \left\lbrack {a, b}\right\rbrack ,\bar{D} \) its \( {c}^{ * }\mathcal{H} \) -horizontal lift to \( {c}^{ * }E \) . If \( {c}_{u} \) denotes the integral curve of \( \bar{D} \) with \( {c}_{u}\left( a\right) = \) \( \left( {a, u}\right) \), then\n\n\[ ... | Yes |
Proposition 1.2. If \( \mathcal{H} \) is a connection on a bundle over a connected manifold, then \( \mathcal{H} \) is trivial iff its holonomy group is trivial. | Proof. The only if part of the statement is clear. Conversely, if the holonomy group is trivial, then the parallel translate of any vector \( u \in {E}_{p} \) to a point \( q \) is independent of the chosen curve, and therefore defines a parallel section of the bundle. | Yes |
Proposition 1.3. Let \( \mathcal{H} \) be a connection on a vector bundle \( \xi = \pi : E \rightarrow \) M. The following statements are equivalent:\n\n(1) \( \mathcal{H} \) is flat.\n\n(2) For any open, simply connected subset \( U \) of \( M \), the restriction of \( \mathcal{H} \) to \( {\pi }^{-1}\left( U\right) \... | Proof. We have already remarked that a locally trivial connection is flat. Conversely, if \( \mathcal{H} \) is flat, then given \( u \in {\pi }^{-1}\left( U\right) \), there exists by Frobenius’ theorem (Theorem 9.2 of Chapter 1), a maximal connected integral manifold \( \widetilde{U} \) of \( {\mathcal{H}}_{\mid {\pi ... | Yes |
Proposition 2.1. \( \left( {{\pi }_{ * },\kappa }\right) : {\tau E} \rightarrow {\tau M} \oplus \xi \) is a bundle map covering \( \pi \) : \( E \rightarrow M \) . | Proof. Both \( {\tau E} \) and \( {\tau M} \oplus \xi \) have the same fiber dimension, and since \( \left( {{\pi }_{ * },\kappa }\right) \) is linear on each fiber, it suffices to show that it has trivial kernel. But \( \ker \left( {{\pi }_{ * },\kappa }\right) = \ker {\pi }_{ * } \cap \ker \kappa = E\left( \mathcal{V... | Yes |
Lemma 2.1. \( \kappa \circ {\mu }_{a * } = {\mu }_{a} \circ \kappa ,\;a \in \mathbb{R} \). | Proof of Lemma 2.1. It suffices to consider vertical vectors, since both sides vanish when applied to horizontal ones. So let \( w \in {\left( {E}_{p}\right) }_{u}, v \mathrel{\text{:=}} {\mathcal{J}}_{u}^{-1}w \in \) \( {E}_{p} \). If \( \imath : {E}_{p} \hookrightarrow E \) denotes inclusion, then \( {\mu }_{a} \circ... | Yes |
Let \( \xi \) be a vector bundle with connection over \( M, f \) : \( N \rightarrow M, u \in {N}_{p} \), and \( X \) a section of \( \xi \) along \( f \) . Given a curve \( c : I \rightarrow N \) with \( \dot{c}\left( 0\right) = u \), denote by \( {X}_{t} \) the parallel section along \( f \circ c \) with \( {X}_{t}\le... | Proof. Choose linearly independent parallel sections \( {X}_{1},\ldots ,{X}_{k} \) of \( \xi \) along \( f \circ c \), so that \( X \circ c = \sum {h}^{i}{X}_{i} \), and \( {X}_{t}\left( s\right) = \sum {h}^{i}\left( t\right) {X}_{i}\left( s\right) \) . Then\n\n\[ \n\mathop{\lim }\limits_{{t \rightarrow 0}}\frac{{X}_{t... | Yes |
Proposition 3.1. The operator \( R \) is a 2-form on \( M \) with values in \( \operatorname{End}\xi \) ; i.e., \( R \in {A}_{2}\left( {M,\operatorname{End}\xi }\right) \), cf. Examples and Remarks 2.1(ii). In particular, \( R \) is tensorial in all three arguments. | Proof. It is clear that \( R\left( {{U}_{1} + {U}_{2}, V}\right) X = R\left( {{U}_{1}, V}\right) X + R\left( {{U}_{2}, V}\right) X \) . Since \( R \) is skew-symmetric in the first two arguments, a similar identity holds for the second argument. We apply Proposition 11.3 of Chapter 1 to show tensoriality: Given \( f \i... | Yes |
Proposition 3.2. Let \( u, v \in {M}_{p}, x \in {E}_{p} \). Consider a map \( f \) from a neighborhood of the origin in \( {\mathbb{R}}^{2} \) into \( \dot{M} \) with \( f\left( 0\right) = p,{f}_{ * }{D}_{1}\left( 0\right) = u,{f}_{ * }{D}_{2}\left( 0\right) = v \). For \( t, s \) small enough, let \( {x}_{t, s} \) den... | Proof. Define a section \( X \) along \( f \) by letting \( X\left( {t, s}\right) \) be the parallel translate of \( x \) along the curves in (1) and (2) from \( p \) to \( f\left( {t, s}\right) \). By Lemma 3.1,\n\n\[ R\left( {u, v}\right) x = R\left( {{f}_{ * }{D}_{1}\left( 0\right) ,{f}_{ * }{D}_{2}\left( 0\right) }... | Yes |
Proposition 3.3. Let \( \nabla \) and \( \widetilde{\nabla } \) denote two connections on \( \xi ,\omega \mathrel{\text{:=}} \widetilde{\nabla } - \nabla \in \) \( {A}_{1}\left( {M,\operatorname{End}\xi }\right) \) . Then\n\n\[ \widetilde{R} = R + {d}^{\nabla }\omega + \left\lbrack {\omega ,\omega }\right\rbrack \]\n\n... | Proof. Given \( U, V \in \mathfrak{X}M \) ,\n\n\[ {\widetilde{\nabla }}_{U}{\widetilde{\nabla }}_{V}X = {\nabla }_{U}\left( {{\nabla }_{V}X + \omega \left( V\right) X}\right) + \omega \left( U\right) \left( {{\nabla }_{V}X + \omega \left( V\right) X}\right) \]\n\n\[ = {\nabla }_{U}{\nabla }_{V}X + {\nabla }_{U}\left( {... | Yes |
Proposition 5.2. The connection form \( \omega \) of a connection \( \mathcal{H} \) satisfies\n\n(1) \( {\omega }_{\mid \mathcal{H}} \equiv 0,\;{l}_{b * } \circ {\omega }_{\mid \ker {\pi }_{ * }} = {1}_{\ker {\pi }_{ * }} \), \n\n(2) \( {R}_{g}^{ * }\omega = {\operatorname{Ad}}_{{g}^{-1}} \circ \omega ,\;g \in G \). \n... | Proof. Part (1) is immediate from Definition 5.2. It suffices to verify (2) for a vertical vector \( u \in \ker {\pi }_{*b} \), since both sides vanish when applied to a horizontal one. Now, by (1), \n\n\[ \n\left( {{R}_{g}^{ * }\omega }\right) \left( b\right) \left( u\right) = \left( {\omega \circ {R}_{g}}\right) \lef... | Yes |
Proposition 5.4. Given \( b \in P \), and \( x, y \in {T}_{b}P \), the matrix of \( R\left( {{\pi }_{ * }x,{\pi }_{ * }y}\right) \in \) \( \mathfrak{o}\left( {E}_{\pi \left( b\right) }\right) \) with respect to the basis \( b \) is \( \Omega \left( b\right) \left( {x, y}\right) \) . | Proof. It must be shown that \( R\left( {{\pi }_{ * }x,{\pi }_{ * }y}\right) \left\lbrack {b, u}\right\rbrack = \left\lbrack {b,\Omega \left( b\right) \left( {x, y}\right) u}\right\rbrack \) for \( u \in \) \( {\mathbb{R}}^{n} \) . Both sides of the equation vanish if \( x \) or \( y \) are vertical, so we may assume t... | Yes |
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