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degree. We saw that derivations can be thought of as infinitesimal automorphisms. One can similarly consider k[[t]]-linear maps of the form Φ(a) = a + tφ1(a) + t2φ2(a) + · · · and consider whether they define automorphisms of A ⊗ k[[t]]. Working modulo t2, we have already done this problem — we are just considering auto... |
for i ≥ 1. In particular, HH i(A, A) = 0 for i ≥ 2. To show that ker µ is projective, we notice that A ⊗ A = k[X] ⊗k k[X] ∼= k[X, Y ]. So the short exact sequence (∗) becomes 0 (X − Y )k[X, Y ] k[X, Y ] k[X] 0. So (X − Y )k[X, Y ] is a free k[X, Y ] module, and hence projective. We can therefore use our theorems to se... |
��ned by (f g)(a1 ⊗ · · · ⊗ am ⊗ b1 ⊗ · · · ⊗ bn) = f (a1 ⊗ · · · ⊗ am) · g(b1 ⊗ · · · ⊗ bn), where ai, bj ∈ A. Under this product, S·(A, A) becomes an associative graded algebra. Observe that δ(f g) = δf g + (−1)mnf δg. So we say δ is a (left) graded derivation of the graded algebra S·(A, A). In homological (graded) a... |
, It is a grade Lie algebra on but notice that we have a degree shift by 1. S·+1(A, A). Of course, we really should define what a graded Lie algebra is. Definition (Graded Lie algebra). A graded Lie algebra is a vector space L = Li with a bilinear bracket [ ·, · ] : L × L → L such that – [Li, Lj] ⊆ Li+j; – [f, g] − (−1)m... |
k-algebra, and : H 0 × H 1 → H 1 is a module action. Also, H 1 is a Lie algebra, and [ ·, · ] : H 1 × H 0 → H 0 is a Lie module action, i.e. H 0 gives us a Lie algebra representation of H 1. In other words, 64 3 Hochschild homology and cohomology III Algebras the corresponding map [ ·, · ] : H 1 → Endk(H 0) gives us a... |
stated just for the algebra structure, and didn’t think about the Lie algebra. Example. Let A = k[X, Y ], with char k = 0. Then HH 0(A, A) = A and HH 1(A, A) = DerA ∼= p(X, Y ) ∂ ∂y + q(X, Y ) : p, q ∈ A. ∂ ∂Y So we have which is generated as an A-modules by ∂ HH 2(A, A) = DerA ∧A DerA, ∂Y. Then ∂X ∧ ∂ HH i(A, A) = 0 ... |
��cult theorem, and the first proof appeared in 2002. An unnamed lecturer once tried to give a Part III course with this theorem as the punchline, but the course ended up lasting 2 terms, and they never reached the punchline. 3.5 Hochschild homology We don’t really have much to say about Hochschild homology, but we are ... |
A, A) = A [A, A]. 67 4 Coalgebras, bialgebras and Hopf algebras III Algebras 4 Coalgebras, bialgebras and Hopf algebras We are almost at the end of the course. So let’s define an algebra. Definition (Algebra). A k-algebra is a k-vector space A and k-linear maps → xy u : k → A λ → λI called the multiplication/product and ... |
map”. It might be slightly difficult to get one’s head around what a coalgebra actually is. It, of course, helps to look at some examples, and we will shortly do so. It also helps to know that for our purposes, we don’t really care about coalgebras per se, but things that are both algebras and coalgebras, in a compatibl... |
1,...,in xi1 1 · · · xin n, where {xi} is a basis of g (the PBW theorem). Then we define λ + ε λi1,...,in xi1 1 · · · xin n = λ. This time, the specification of ∆ is telling us that if X ∈ g and v, w are elements of a representation of g, then X acts on the tensor product by ∆(X) · (v ⊗ w) = Xv ⊗ w + v ⊗ Xw. Example. Con... |
gebra means we can take tensor products of modules and still get modules. If we want to take duals as well, then it turns out the right notion is that of a Hopf algebra: Definition (Hopf algebra). A bialgebra (H, µ, u, ∆, ε) is a Hopf algebra if there is an antipode S : H → H that is a k-linear map such that µ ◦ (S ⊗ id... |
�nite codimension}. 70 4 Coalgebras, bialgebras and Hopf algebras III Algebras Example. Let G be a finite group. Then (kG)∗ is a commutative non-cocommutative Hopf algebra if G is non-abelian. Let {g} be the canonical basis for kG, and {φg} be the dual basis of (kG)∗. Then ∆(φg) = h1h2=g φh1 ⊗ φh2. There is an easy way ... |
1 = g g (φg ⊗ 1) ⊗ (1 ⊗ g) (φg ⊗ 1) ⊗ (1 ⊗ g−1). The equation R∆R−1 = τ ∆ results in an isomorphism between U ⊗ V and V ⊗ U for D(G)-bimodules U and V, given by flip follows by the action of R. If G is non-abelian, then this is non-commutative and non-co-commutative. The point of defining this is that the representations... |
have to change the multiplication. Consider Oq(M2(k)) defined by X12X11 = qX11X12 X22X12 = qX12X22 X21X11 = qX11X21 X22X21 = qX21X22 X21X12 = X12X21 X11X22 − X22X11 = (q−1 − q)X12X21. We define the quantum determinant = X11X22 − q−1X12X21 = X22X11 − qX12X21. det q Then Then we define ∆(det q ) = det q, ⊗ det q ε(det q ) ... |
s, t ∈ {1, 2, 3}, we let Rst be the invertible map which acts like R on the sth and tth components, and identity on the other. So for example, R12(e1 ⊗ e2 ⊗ v) = m i,j e ⊗ em ⊗ v. Definition (Yang–Baxter equation). R satisfies the quantum Yang–Baxter equation (QYBE ) if and the braided form of QYBE (braid equation) if R... |
Example. The quantum plane Oq(k2) can be written as SR(V ) with R(e1 ⊗ e2) = qe2 ⊗ e1 R(e1 ⊗ e1) = e1 ⊗ e1 R(e2 ⊗ e1) = q−1e1 ⊗ e2 R(e2 ⊗ e2) = e2 ⊗ e2. Generally, given a V which is finite-dimensional as a vector space, we can identify (V ⊗ V )∗ with V ∗ ⊗ V ∗. We set E = V ⊗ V ∗ ∼= Endk(V ) ∼= Mn(k). We define R13 and... |
Gelfand-Kirillov dimension, 39 Gerstenhaber algebra, 64 Gerstenhaber bracket, 63 Goldie rank, 48 Goldie’s theorem, 5, 49 graded algebra, 33 homogeneous components, 33 graded derivation, 63 graded ideal, 33 graded Jacobi identity, 64 graded Lie algebra, 64 group algebra, 15 Hattori–Stallings trace map, 28 Hilbert basis... |
65 Schur’s lemma, 13 semi-direct product, 55 semisimple algebra, 9 separable algebra, 53 separated filtration, 32 separating idempotent, 53 simple algebra, 9 simple module, 7 skew fields, 3 split extension, 56 star product, 59 equivalence, 60 trivial, 60 submodule essential, 44 trace of projective, 28 trivial star produ... |
– Constant functions are continuous. – The function f (x) = x is continuous (take δ = ε). The definition of continuity of a function looks rather like the definition of convergence. In fact, they are related by the following lemma: Lemma. The following two statements are equivalent for a function f : A → R. – f is conti... |
]. Then an → 0 and f (an) → 1 = f (0). We can use this sequence definition as the definition for continuous functions. This has the advantage of being cleaner to write and easier to work with. In particular, we can reuse a lot of our sequence theorems to prove the analogous results for continuous functions. Lemma. Let A ... |
� ⇒ |f (y) − f (a)| < η. Therefore |y − a| < δ ⇒ |g(f (y)) − g(f (a))| < ε. There are two important theorems regarding continuous functions — the maximum value theorem and the intermediate value theorem. Theorem (Maximum value theorem). Let [a, b] be a closed interval in R and let f : [a, b] → R be continuous. Then f i... |
� = |f (s)| in the definition of continuity, we can find δ > 0 such that ∀y, |y − s| < δ ⇒ f (y) < 0. Then s + δ/2 ∈ A, so s is not an upper bound. Contradiction. 2 exists in Numbers and Sets). √ If f (s) > 0, by the same argument, we can find δ > 0 such that ∀y, |y − s| < δ ⇒ f (y) > 0. So s − δ/2 is a smaller upper boun... |
. So f is a surjection. So it is a bijection and hence invertible. Let g be the inverse. Let y ∈ [c, d] and let ε > 0. Let x = g(y). So f (x) = y. Let u = f (x − ε) and v = f (x + ε) (if y = c or d, make the obvious adjustments). Then u < y < v. So we can find δ > 0 such that (y − δ, y + δ) ⊆ (u, v). Then |z − y| < δ ⇒ ... |
= [a, b] 26 4 Continuous functions IA Analysis I Proof. We prove that version 1 ⇒ version 2. Suppose A satisfies the conditions of v2. Let A = {x ∈ [a, b] : [a, x] ⊆ A}. Then a ∈ A. If x ∈ A with x = b, then [a, x] ⊆ A. So ∃y > x such that [a, y] ⊆ A. So ∃y > x such that y ∈ A. If ∀ε > 0, ∃y ∈ (x − ε, x] such that [a, ... |
∃δ > 0 such that |y − x| < δ ⇒ f (y) > 0. So y ∈ A. Hence by continuous induction, b ∈ A. Contradiction. Now we prove that continuous functions in closed intervals are bounded. Theorem. Let [a, b] be a closed interval in R and let f : [a, b] → R be continuous. Then f is bounded. Proof. Let f : [a, b] be continuous. Le... |
b] by open intervals has a finite subcover. We say closed intervals are compact (cf. Metric and Topological Spaces). 27 4 Continuous functions IA Analysis I Proof. Let {Iγ : γ ∈ Γ} be a cover of [a, b] by open intervals. Let A = {x : [a, x] can be covered by finitely many of the Iγ}. Then a ∈ A since a must belong to so... |
, we can find x1, · · ·, xn such that [a, b] ⊆ n 1 (xi − δxi, xi + δxi ). But f is bounded in each interval (xi − δxi, xi + δxi) by |f (xi)| + 1. So it is bounded on [a, b] by max |f (xi)| + 1. 28 5 Differentiability IA Analysis I 5 Differentiability In the remainder of the course, we will properly develop calculus, and p... |
as x → a. It follows also that f (x) → as x → a iff f (xn) → for every sequence (xn) in A with xn → a. The previous limit theorems of sequences apply here as well Proposition. If f (x) → and g(x) → m as x → a, then f (x) + g(x) → + m, f (x)g(x) → m, and f (x) m if g and m don’t vanish. g(x) → 5.2 Differentiation Similar... |
Proposition. f (x + h) = f (x) + hf (x) + o(h). We can interpret this as an approximation of f (x + h): f (x + h) = f (x) + hf (x) linear approximation + o(h) error term. And differentiability shows that this is a very good approximation with small o(h) error. Conversely, we have Proposition. If f (x + h) = f (x) + hf ... |
f (x) + g(x)) + o(h) f g(x + h) = f (x + h)g(x + h) = [f (x) + hf (x) + o(h)][g(x) + hg(x) + o(h)] = f (x)g(x) + h[f (x)g(x) + f (x)g(x)] + o(h)[g(x) + f (x) + hf (x) + hg(x) + o(h)] + h2f (x)g(x) error term By limit theorems, the error term is o(h). So we can write this as = f g(x) + h(f (x)g(x) + f (x)g(x)) + o(h). L... |
(h). We want to show that the error term is o(h), i.e. it divided by h tends to 0 as h → 0. But ε1(h)g(f (x)) → 0, f (x)+ε1(h) is bounded, and ε2(hf (x)+hε1(h)) → 0 because hf (x) + hε1(h) → 0 and ε2(0) = 0. So our error term is o(h). 31 5 Differentiability IA Analysis I We usually don’t write out the error terms so exp... |
at x with derivative f (x) g(x) − f (x) g(x) g(x)2 = f (x)g(x) − f (x)g(x) g(x)2. Lemma. If f is differentiable at x, then it is continuous at x. Proof. As y → x, f (y) − f (x) y − x → f (x). Since, y − x → 0, f (y) − f (x) → 0 by product theorem of limits. So f (y) → f (x). So f is continuous at x. Theorem. Let f : [a... |
0, since g is continuous, g(y + k) → g(y). So h → 0. So g(y + k) − g(y) k → [f (x)]−1 = [f (g(y)]−1. Example. Let f (x) = x1/2 for x > 0. Then f is the inverse of g(x) = x2. So f (x) = 1 g(f (x)) = 1 2x1/2 = 1 2 x−1/2. Similarly, we can show that the derivative of x1/q is 1 q x1/q−1. Then let’s take xp/q = (x1/q)p. By... |
f (x + h) − f (x) ≤ 0 by maximality of f (x). By considering both sides as we take the limit h → 0, we know that f (x) ≤ 0 and f (x) ≥ 0. So f (x) = 0. 33 5 Differentiability IA Analysis I Corollary (Mean value theorem). Let f be continuous on [a, b] (a < b), and differentiable on (a, b). Then there exists x ∈ (a, b) su... |
(1) ≥ 2, or else we can find x ∈ (0, 1) such that 2 ≤ f (x) = f (1) − f (0) 1 − 0 = f (1). 34 5 Differentiability IA Analysis I Theorem (Local version of inverse function theorem). Let f be a function with continuous derivative on (a, b). Let x ∈ (a, b) and suppose that f (x) = 0. Then there is an open interval (u, v) c... |
� (a, b) such that f (k)(xk) = 0. Since f (k)(a) = 0, we can find xk+1 ∈ (a, xk) such that f (k+1)(xk+1) = 0 by Rolle’s theorem. So the result follows by induction. Corollary. Suppose that f and g are both differentiable on an open interval containing [a, b] and that f (k)(a) = g(k)(a) for k = 0, 1, · · ·, n − 1, and als... |
f (b) = Q(b) + (b − a)n n! f (n)(x). That is, Therefore Alternatively, f (b) = f (a) + (b − a)f (a) + · · · + (b − a)n−1 (n − 1)! f (n−1)(a) + (b − a)n n! f (n)(x). Setting b = a + h, we can rewrite this as Theorem (Taylor’s theorem with the Lagrange form of remainder). f (a + h) = f (a) + hf (a) + · · · + hn−1 (n − 1... |
x) = f (0) + f (0)x + f (2)(0) 2! x2 + · · · + f (n−1)(0) (n − 1)! xn−1 + f (n)(u) n! xn. for some u between 0 and x. This equals to f (x) = n−1 k=0 xk k! + f (n)(u) n! xn. We must show that the remainder term f (n)(u) that x is fixed, but u can depend on n. n! xn → 0 as n → ∞. Note here But we know that f (n)(u) = f (u... |
. A linear map in C is something in the form x → bx, which can only be a dilation or rotation, not reflections or other weird things. Example. f (z) = |z| is also not differentiable. If it were, then |z|2 would be as well (by the product rule). So would |z|2 z = ¯z when z = 0 by the quotient rule. At z = 0, it is certain... |
Definition (Radius of convergence). The radius of convergence of a power series anzn is R = sup |z| : anzn converges. {z : |z| < R} is called the circle of convergence.1. If |z| < R, then anzn converges. If |z| > R, then anzn diverges. When |z| = R, the series can converge at some points and not the others. Example. ∞ ... |
an| ≥ 1 for infinitely many n. Therefore |anzn| ≥ 1 for infinitely many n. So anzn does not converge. Example. The radius of convergence of So lim sup n|an| = 1 2. So 1/ lim sup n|an| = 2. zn 2n is 2 because n|an| = 1 2 for every n. But often it is easier to find the radius convergence from elementary methods such as the ... |
0 ∞ ∞ bn. an cn = n=0 n=0 Proof. We first show that a rearrangement of cn converges absolutely. Hence it converges unconditionally, and we can rearrange it back to cn. Consider the series (a0b0) + (a0b1 + a1b1 + a1b0) + (a0b2 + a1b2 + a2b2 + a2b1 + a2b0) + · · · (∗) Let SN = N n=0 an, TN = N n=0 bn, UN = N n=0 |an|, VN ... |
n−2 (n − 2)! + · · · + · 1 z2wn−2 + · · · + zn zn n! n n n=0 ∞ = (z + w)n by the binomial theorem n=0 = ez+w. Note that if (cn) is the convolution of (an) and (bn), then the convolution of (anzn) and (bnzn) is (cnzn). Therefore if both anzn and bnzn converge absolutely, then their product is cnzn. Note that we have now... |
sin 0 = 0, and d dx sin x = cos x ≤ 1. So for x ≥ 0, sin x ≤ x, “by the mean value theorem”. Also, cos 0 = 1, and d dx cos x = − sin x, which, for x ≥ 0, is greater than −x. From this, it follows that when x ≥ 0, cos x ≥ 1 − x2 2 comes from “integrating” −x, (or finding a thing whose derivative is −x)). 2 (the 1 − x2 C... |
� n=1 nanzn−1. n=0 an z is We first prove some (seemingly arbitrary and random) lemmas to build up the proof of the above statement. They are done so that the final proof will not be full of tedious algebra. Lemma. Let a and b be complex numbers. Then bn − an − n(b − a)an−1 = (b − a)2(bn−2 + 2abn−3 + 3a2bn−4 + · · · + (n... |
n−1 converges, by the ratio test. So n|anzn−1| converges, Corollary. Under the same conditions, ∞ n=2 n 2 anzn−2 converges absolutely. Proof. Apply Lemma above again and divide by 2. Theorem. Let anzn be a power series with radius of convergence R. For |z| < R, let ∞ ∞ f (z) = anzn and g(z) = nanzn−1. Then f is differe... |
3 Hyperbolic trigonometric functions Definition (Hyperbolic sine and cosine). We define cosh z = sinh z = ez + e−z 2 ez − e−z 2 = 1 + = z + z2 2! z3 3! + + z4 4! z5 5! + + z6 6! z7 7! + · · · + · · · Either from the definition or from differentiating the power series, we get that Proposition. d dz d dz cosh z = sinh z sinh... |
] f (x), mi = inf x∈[xi−1−xi] f (x). The upper sum is the total area of the red rectangles, while the lower sum is the total area of the black rectangles: y · · · · · · a x1 x2 x3 xi xi+1 · · · b x Definition (Refining dissections). If D1 and D2 are dissections of [a, b], we say that D2 refines D1 if every point of D1 is ... |
ary. Let D1 and D2 be two dissections of [a, b]. Then UD1f ≥ LD2 f. Proof. Let D be the least common refinement (or indeed any common refinement). Then by lemma above (and by definition), UD1f ≥ UDf ≥ LDf ≥ LD2 f. Finally, we can define the integral. Definition (Upper, lower, and Riemann integral). The upper integral is The... |
We know that the integral is b2−a2 2. So we put each term of the sum into the form i −x2 x2 2 i−1 plus some error terms. = = = n i=1 x2 i 2 − x2 i−1 2 + x2 i 2 − xi−1xi + x2 i−1 2 1 2 1 2 n i=1 (x2 i − x2 i−1 + (xi − xi−1)2) (b2 − a2) + 1 2 n (xi − xi−1)2. i=1 Definition (Mesh). The mesh of a dissection D is maxi(xi+1 ... |
function is Lebesgue-integrable. Using the definition to show integrability is often cumbersome. Most of the time, we use the Riemann’s integrability criterion, which follows rather immediately from the definition, but is much nicer to work with. Proposition (Riemann’s integrability criterion). This is sometimes known a... |
xi−1,xi] f (x), and similarly for inf, we have UD(λf ) = λUDf LD(λf ) = λLDf. So if we choose D such that UDf − LDf < ε/λ, then UD(λf ) − LD(λf ) < ε. So the result follows from Riemann’s integrability criterion. Proposition. Let f : [a, b] → R be integrable. Then −f is integrable, and b a −f (x) dx = − b a f (x) dx. P... |
= f (x) dx + b a b a f (x) dx + b a g(x) dx. g(x) dx. So the upper and lower integrals are equal, and the result follows. So we now have that b a (λf (x) + µg(x)) dx = λ b a f (x) dx + µ b a g(x) dx. We will prove more “obvious” results. Proposition. Let f, g : [a, b] → R be integrable, and suppose that f (x) ≤ g(x) f... |
s integrability criterion. Combining these two propositions, we get that if |f (x) − g(x)| ≤ C, for every x ∈ [a, b], then b a f (x) dx − b a g(x) dx ≤ C(b − a). Proposition (Additivity property). Let f : [a, c] → R be integrable, and let b ∈ (a, c). Then the restrictions of f to [a, b] and [b, c] are Riemann integrabl... |
f ) + UD2(f ) = UD(f ) ≤ UD(f ) ≤ c a f (x) dx + ε. f (x) dx ≥ LD1(f ) + LD2 (f ) = LD(f ) ≥ LD(f ) ≥ c a f (x) dx − ε. Since ε is arbitrary, it follows that b a f (x) dx + c b f (x) dx = c a f (x) dx. The other direction is left as an (easy) exercise. Proposition. Let f, g : [a, b] → R be integrable. Then f g is integ... |
. So the result follows. 53 7 The Riemann Integral IA Analysis I Theorem. Every continuous function f on a closed bounded interval [a, b] is Riemann integrable. Proof. wlog assume [a, b] = [0, 1]. n, 2 Suppose the contrary. Let f be non-integrable. This means that there exists some ε such that for every dissection D, U... |
small, the difference between the upper sum and the lower sum can be arbitrarily small by uniform continuity. Thus to prove the above theorem, we just have to show that continuous functions on a closed bounded interval are uniformly continuous. Theorem (non-examinable). Let a < b and let f : [a, b] → R be continuous. T... |
)) ≤ ε f (b) − f (a) n i=1 = ε. (f (xi) − f (xi−1)) Pictorially, we see that the difference between the upper and lower sums is total the area of the red rectangles. y x To calculate the total area, we can stack the red areas together to get something f (b)−f (a) and height f (b) − f (a). So the total area is just ε. of... |
, 1] is integrable. – g(x) = x x ≤ 1 x2 + 1 x > 1 defined on [0, 1] is integrable. Corollary. Every piecewise continuous and bounded function on [a, b] is integrable. Proof. Partition [a, b] into intervals I1, · · ·, Ik, on each of which f is (bounded and) continuous. Hence for every Ij with end points xj−1, xj, f is in... |
intervals [xi−1, xi] are subdivided in D. 56 7 The Riemann Integral IA Analysis I For each interval that gets chopped up, the upper sum decreases by at most b−a n · 2C. Therefore UDn f − UDf ≤ b − a n 2C · m. Pick n such that 2Cm(b − a)/n < ε 2. Then So UDnf − UDf < ε 2. UDn f − b a f (x) dx < ε. This is true whenever ... |
f (a). g(x) = x a f (t) dt. Then d dx Since g(x) − f (x) = 0, g(x) − f (x) must be a constant function by the mean value theorem. We also know that g(x) = f (x) = (f (x) − f (a)). g(a) = 0 = f (a) − f (a) So we must have g(x) = f (x) − f (a) for every x, and in particular, for x = b. Theorem (Fundamental theorem of ca... |
You will be asked to find it in the example sheet. Using the fundamental theorem of calculus, we can easily prove integration by parts: Theorem (Integration by parts). Let f, g : [a, b] → R be integrable such that everything below exists. Then b a f (x)g(x) dx = f (b)g(b) − f (a)g(a) − b a f (x)g(x) dx. Proof. By the f... |
Note that the form of the integral remainder is rather weird and unexpected. How could we have come up with it? We might start with the fundamental 59 7 The Riemann Integral IA Analysis I theorem of algebra and integrate by parts. The first attempt would be to integrate 1 to t and differentiate f (t) to f (2)(t). So we ... |
We can think of “integration by parts” as what you get by integrating the product rule, and “integration by substitution” as what you get by integrating the chain rule. 7.2 Improper integrals It is sometimes sensible to talk about integrals of unbounded functions or integrating to infinity. But we have to be careful an... |
decreasing (the right hand inequality is valid only for n ≥ 2). It follows that N +1 N N f (x) dx ≤ f (n) ≤ f (x) dx + f (1) 1 n=1 1 So if the integral exists, then f (n) is increasing and bounded above by ∞ 1 f (x) dx, so converges. 1 f (x) dx is unbounded. Then N n=1 f (n) If the integral does not exist, then N is u... |
y| in R. This is known as a norm. It is useful to define and study this structure in an abstract setting, as opposed to thinking about Rn specifically. This leads to the general notion of normed spaces. Definition (Normed space). Let V be a real vector space. A norm on V is a function ∥ · ∥ : V → R satisfying (i) ∥x∥ ... |
|p1/p. It is, however, not trivial to check the triangle inequality, and we will not do this. We can show that as p → ∞, ∥x∥p → ∥x∥∞, which justifies our notation above. We also have some infinite dimensional examples. Often, we can just extend our notions on Rn to infinite sequences with some care. We write RN for the... |
b a |f | dx. – We can define L2 similarly by ∥f ∥L2 = ∥f ∥2 = 1 2 f 2 dx b a – In general, we can define Lp for p ≥ 1 by ∥f ∥Lp = ∥f ∥p = 1 p f p dx b a – Finally, we have L∞ by ∥f ∥L∞ = ∥f ∥∞ = sup |f |... This is also called the uniform norm, or the supremum norm. Later, when we define convergence for general normed... |
= 1 x = 0.5 0 x ̸= 0.5, then f is integrable with integral 0, but is not identically zero. So we cannot expand our vector space to be too large. To define Lp properly, we need some more sophisticated notions such as Lebesgue integrability and other fancy stuff, which will be done in the IID Probability and Measure cou... |
is the ball with respect to ∥ · ∥. Actual proof of equivalence is on the second example sheet. 28 4 Rn as a normed space IB Analysis II Example. Consider R2. Then the norms ∥ · ∥∞ and ∥ · ∥2 are equivalent. This is easy to see using the ball picture: where the blue ones are the balls with respect to ∥ · ∥∞ and the red... |
two inequalities, one holds and one does not. Is it possible for both inequalities to not hold? The answer is yes. This is an exercise on the second example sheet as well. This is all we are going to say about Lipschitz equivalence. We are now going to define convergence, and study the consequences of Lipschitz equiva... |
under different norms. This is, again, on the second example sheet. We have some easy facts about convergence: Proposition. Let (V, ∥ · ∥) be a normed space. Then (i) If xk → x and xk → y, then x = y. 30 4 Rn as a normed space IB Analysis II (ii) If xk → x, then axk → ax. (iii) If xk → x, yk → y, then xk + yk → x + y.... |
such that convergence under the norm is equivalent to pointwise convergence, i.e. pointwise convergence is not normable. In fact, it is not even metrizable. However, we will not prove this. We’ll now generalize the Bolzano-Weierstrass theorem to Rn. Theorem (Bolzano-Weierstrass theorem in Rn). Any bounded sequence in ... |
∥e(k) − 0∥∞ = 1 for all k. Also, this is bounded but does not have a convergent subsequence for the same reasons. We know that all finite dimensional vector spaces are isomorphic to Rn as vector spaces for some n, and we will later show that all norms on finite dimensional spaces are equivalent. This means every finit... |
∥) is complete if every Cauchy sequence converges to an element in V. We’ll start with some easy facts about Cauchy sequences and complete spaces. Proposition. Any convergent sequence is Cauchy. Proof. If xk → x, then ∥xk − xℓ∥ ≤ ∥xk − x∥ + ∥xℓ − x∥ → 0 as k, ℓ → ∞. Proposition. A Cauchy sequence is bounded. Proof. By... |
, then (x(k) ) is a Cauchy sequence of real numbers for j j → xj ∈ R each j ∈ {1, · · ·, n}. By the completeness of the reals, we know that xk for some x. So xk → x = (x1, · · ·, xn) since convergence in Rn is equivalent to componentwise convergence. Note that the spaces ℓ1, ℓ2, ℓ∞ are all complete with respect to the ... |
Lebesgue measure zero. So they disagree on at most a set of Lebesgue measure zero. Example. Let V = {(xn) ∈ RN : xj = 0 for all but finitely many j}. Take the supremum norm ∥ · ∥∞ on V. This is a subspace of ℓ∞ (and is sometimes denoted ℓ0). Then (V, ∥ · ∥∞) is not complete. We define x(k) = (1, 1 k, 0, 0, · · · ) for... |
If E′ = (0, 1) ∪ {2}. Then the set of limit points of E′ is still [0, 1]. 34 4 Rn as a normed space IB Analysis II There is a nice result characterizing whether a set contains all its limit points. Proposition. Let E ⊆ V. Then E contains all of its limit points if and only if V \ E is open in V. Using this proposition... |
∈ E). (⇐) Suppose V \ E is open. Let y ∈ V \ E. Since V \ E is open, there is some r such that Br(y) ⊆ V \ E. By the lemma, y is not a limit point of E. So all limit points of E are in E. 4.3 Sequential compactness In general, there are two different notions of compactness — “sequential compactness” and just “compactne... |
xkj. By closedness of K, we know that the limit is in K. So K is compact. 4.4 Mappings between normed spaces We are now going to look at functions between normed spaces, and see if they are continuous. Let (V, ∥ · ∥), (V ′, ∥ · ∥′) be normed spaces, and let E ⊆ K be a subset, and f : E → V ′ a mapping (which is just a... |
) ∈ Bε(f (y)), or equivalently, So done. |f (yk) − f (y)| < ε. 36 4 Rn as a normed space IB Analysis II (⇐) If f is not continuous at y, then there is some ε > 0 such that for any k, we have B 1 k (y) ̸⊆ f −1(Bε(f (y))). Choose yk ∈ B 1 f (y)∥ ≥ ε, contrary to the hypothesis. k (y) \ f −1(Bε(f (y))). Then yk → y, yk ∈ ... |
Applying this fact to F = f (K) gives the desired result, and similarly for infimum. Finally, we will end the chapter by proving that any two norms on a finite dimensional space are Lipschitz equivalent. The key lemma is the following: Lemma. Let V be an n-dimensional vector space with a basis {v1, · · ·, vn}. Then fo... |
show that an arbitrary norm ∥ · ∥ is equivalent to ∥ · ∥2, we have to show that for any ∥x∥, we have We can divide by ∥x∥2 and obtain an equivalent requirement: a∥x∥2 ≤ ∥x∥ ≤ b∥x∥2. a ≤ x ∥x∥2 ≤ b. We know that any x/∥x∥2 lies in the unit sphere S = {x ∈ V : ∥x∥2 = 1}. So we want to show that the image of ∥ · ∥ is bou... |
continuous. So there is some x0 ∈ S such that ∥x0∥ = inf x∈S ∥x∥ = a, say. Since ∥x∥ > 0, we know that ∥x0∥ > 0. So ∥x∥ ≥ a∥x∥2 for all x ∈ V. ∥x∥ − ∥y∥ The key to the proof is the compactness of the unit sphere of (V, ∥ · ∥). On the other hand, compactness of the unit sphere also characterizes finite dimensionality. ... |
been using? Recall we define xk → x to mean ∥xk − x∥ → 0 as a sequence in R. What is ∥xk − x∥ really about? It is measuring the distance between xk and x. So what we really need is a measure of distance. To do so, we can define a distance function d : V ×V → R by d(x, y) = ∥x−y∥. Then we can define xk → x to mean d(xk... |
and that d(x, y) = 0 implies x = y. Finally, to show that a constant sequence has a limit, suppose xk = x for all k ∈ N. Then we know that d(x, xk) = d(x, x) should tend to 0. So we must have d(x, x) = 0 for all x. We will use these properties to define metric spaces. 5.1 Preliminary definitions Definition (Metric spa... |
let’s check the triangle inequality for h. We’ll use a general fact that for numbers a, c ≥ 0, b, d > 0 we have ⇔ ≤ ≤. Based on this fact, we can start with d(x, y) ≤ d(x, z) + d(z, y). Then we obtain d(x, y) 1 + d(x, y) So done. ≤ = ≤ d(x, z) + d(z, y) 1 + d(x, z) + d(z, y) d(x, z) 1 + d(x, z) + d(z, y) + d(z, y) 1 +... |
�ε)(∃K)(∀k > K) d(xk, x) < ε. Alternatively, this says that given any ε, for sufficiently large k, we get xk ∈ Bε(x). Again, Br(a) is the open ball centered at a with radius r, defined as Br(a) = {x ∈ X : d(x, a) < r}. Proposition. The limit of a convergent sequence is unique. Proof. Same as that of normed spaces. Note... |
). Let (X, d) be a metric space. The topology on (X, d) is the collection of open subsets of X. We say it is the topology induced by the metric. Definition (Topological notion). A notion or property is said to be a topological notion or property if it only depends on the topology, and not the metric. We will introduce ... |
δi > 0 with Bδi(x) ⊆ Vi. Take δ = min{δ1, · · ·, δn}. So Bδ(x) ⊆ Vi for all i. So Bδ(x) ⊆ V. So V is open. α Vα = U. So U is open. (iii) ∅ satisfies the definition of an open subset vacuously. X is open since for any x, B1(x) ⊆ X. This theorem is not important in this course. However, this will be a key defining proper... |
its limit points. So it is closed. 44 5 Metric spaces IB Analysis II 5.3 Cauchy sequences and completeness Definition (Cauchy sequence). Let (X, d) be a metric space. A sequence (xn) in X is Cauchy if (∀ε)(∃N )(∀n, m ≥ N ) d(xn, xm) < ε. Proposition. Let (X, d) be a metric space. Then (i) Any convergent sequence is Ca... |
of an incomplete metric on Rn. We start by defining h : Rn → Rn by h(x) = x 1 + ∥x∥, where ∥ · ∥ is the Euclidean norm. We can check that this is injective: h(x) = h(y), taking the norm gives if ∥x∥ 1 + ∥x∥ = ∥y∥ 1 + ∥y∥. 45 5 Metric spaces IB Analysis II So we must have ∥x∥ = ∥y∥, i.e. x = y. So h(x) = h(y) implies x... |
that “goes to infinity” in the usual norm will be Cauchy in this norm, but we have nothing at infinity for it to converge to. Suppose we have a complete metric space (X, d). We know that we can form arbitrary subspaces by taking subsets of X. When will this be complete? Clearly it has to be closed, since it has to inc... |
not compact. It follows from definition that compactness is a topological property, since it is defined in terms of convergence, and convergence is defined in terms of open sets. The following theorem relates completeness with compactness. Theorem. All compact spaces are complete and bounded. Note that X is bounded if... |
N X = Bε(xi). It is easy to check that being totally bounded implies being bounded. We then have the following strengthening of the previous theorem. i=1 47 5 Metric spaces IB Analysis II Theorem. (non-examinable) Let (X, d) be a metric space. Then X is compact if and only if X is complete and totally bounded. Proof. ... |
ε for all j = 1, · · ·, n. Again, this exists, or else n i=1 Bε(yi) covers X. Then clearly the sequence (yn) is not Cauchy. So done. In IID Linear Analysis, we will prove the Arzel`a-Ascoli theorem that characterizes the compact subsets of the space C([a.b]) in a very concrete way, which is in some sense a strengtheni... |
as d(x, y) = min{1, |x − y|}, d′(x, y) = |x − y|. Now consider the function f : (X, d) → (X ′, d′) defined by f (x) = x. We can then check that this is uniformly continuous but not Lipschitz. Note that the statement that metrics d and d′ are Lipschitz equivalent is equivalent to saying the two identity maps i : (X, d)... |
= 1 n but d′(f (xn), f (yn)) > ε. n, there is some xn, yn such that d(xn, yn) < 1 By compactness of X, (xn) has a convergent subsequence (xni) → x. Then we also have yni → x. So by continuity, we must have f (xni) → f (x) and f (yni) → f (x). But d′(f (xni), f (yni)) > ε for all ni. This is a contradiction. In the pro... |
ary. A function f : (X, d) → (X ′, d′) is continuous if f −1(V ) is open in X whenever V is open in X ′. Proof. Follows directly from the equivalence of (i) and (iii) in the theorem above. 50 5 Metric spaces IB Analysis II 5.6 The contraction mapping theorem If you have already taken IB Metric and Topological Spaces, t... |
A counterexample is to be found on example sheet 3. Proof. We first focus on the case where f itself is a contraction. Uniqueness is straightforward. By assumption, there is some 0 ≤ λ < 1 such that d(f (x), f (y)) ≤ λd(x, y) for all x, y ∈ X. If x and y are both fixed points, then this says d(x, y) = d(f (x), f (y)) ... |
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