We call V a Banach space if it is complete, that is to say every Cauchy sequences in V converges to an element of V. (a (k) a (l) n ) (a (l)

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1 Queston 1 Part () Suppose (a n ) =. Then sup{ a N} =. But a for all, so t follows that a = for all, that s to say (a n ) =. Let (a n ) c and α C. Then αa = α α for each, so α(a n ) = (αa n ) = sup{ α a N} = α sup{ a N} = α (a n ). Let (a n ), (b n ) c. Then Part () (a n ) +(b n ) = sup{ a +b N} sup{ a + b j, j N} = (a n ) + (b n ). We call V a Banach space f t s complete, that s to say every Cauchy sequences n V converges to an element of V. Part () Let (a (k) n ) k=1 be a Cauchy sequence n c. Fx N. Then for any k, l a (k) a (l) (a (k) n ) (a (l) n ) so t follows that (a (k) ) k=1 s a Cauchy sequence n C. But the complex numbers C are complete, so the sequence (a (k) ) k=1 con- a n as verges to some lmt, a. We clam that (a n ) c, and (a (k) n n. We wll prove the second of these frst. Let ε >. Snce the sequence (a (k) n that we have N such that a (k) a (l) whenever k, l N. Thus s we let l, we see a (k) ) k=1 s Cauchy, for each N, t follows (a (k) n ) (a (l) n ) < ε 2 a (a (k) n ) (a n ) ε 2 < ε whenever k N. So (a (k) n a n as n as requred. Fnally, we need to check that (a n ) c. Agan, let ε >. By the above, pck K such that (a (K) n ) (a n ) < ε 2 that a (K) n < ε 2 whenever n N. Now let n N. Then a n a n a (K) n. Snce (a(k) n so a n as n, and (a n ) c as we need. + a (K) n < ε 2 + ε 2 = ε ) c we have N N such 1

2 Part (v) Let V be a Banach space. We call A V closed f for any sequence (a n ) n A whch converges to a lmt v V, we have v A. Part (v) (a) Any sequence n c that converges n l s Cauchy. By the above, Cauchy sequences n c have lmts n c. Therefore c s closed. (b) Defne v n c by v n = (1, 12,..., 1n,,,... ). Let v = (1, 12, 13,... ). Observe v n v = 1 n+1, so the sequence (v n) has lmt v. But v c. So c s not closed. (c) The space c N s a fnte-dmensonal subspace of l, and s therefore closed. 2

3 Queston 2 Part () (a) Call T a bounded lnear map f there s a constant M such that T x x for all x V. We defne the norm T = T x sup x V \{} x (b) We call T open f t maps open sets to open sets. The open mappng theorem asserts that a surjectve bounded lnear map between Banach spaces s open. (c) Let c be the set of bounded sequences (a n ), whch are eventually zero. Defne T : c c by T ((a n )) = (a n /n). Then T s certanly a surjectve bounded lnear map. But T 1 ((a n )) = (na n ) whch does not defne a bounded lnear map. It follows that T 1 s not contnuous. By the formulaton of contnuty n terms of open sets, T s not open. Part () (a) We defne V to be the vector space of bounded lnear maps V R, wth norm as defned above. We want to show that V s complete. Let (f n ) be a Cauchy sequence n V. Let x V. If x =, then f n (x) = for all n, and f n (x) as n. So suppose x. Let ε >. Then, as (f n ) s Cauchy, we have N N such that f m f n < ε/ x whenever m, n N. Hence, for m, n N, we have f m (x) f n (x) f m f n x < ε. So the sequence (f n (x)) s a Cauchy sequence n R, for any x V. Cauchy sequences converge n R, so let f(x) be the lmt as n. We clam that f V, and f n f as n. Snce each f n s lnear, the algebra of lmts tells us that f s lnear. Observe f m f n f m f n for all m, n N. Thus, snce the sequence (f n ) s Cauchy, the sequence ( f n ) s a Cauchy sequence n R. Let f n C as n. Then for any x V, we have lm sup n x V \{} f n (x) x = C 3

4 whch means, when we rearrage f = f(x) sup x V \{} x = C In partcular, f s bounded. Now let ε >. Snce (f n ) s Cauchy, we have N such that f m (x) f n (x) ε 2 x for all x V, whenever m, n N. Let m. Then f(x) f n (x) ε 2 x for all x V, whenever m, n N. It follows that f n f < ε whenever n N, and we are done. (b) Let W be a subspace of a normed vector space V. Let f W. Then we have F V such that F W = f and F = f. (c) Let v V. Observe for all f V. It follows that τ(v) v. τ(v)(f) = f(v) f v If v =, then τ(v) =, so τ(v) = v. So suppose τ(v). Let W be the one-dmensonal vector space spanned by the vector v. Defne f: W K by f(αv) = α v α K. Then by the Hahn-Banach theorem, we have F V such that F = f, and F (v) = f(v) = 1. We see that f(αv) F = f = sup α α v = 1. Hence τ(v) τ(v)(f ) = F (v) = v. Combnng the two nequaltes, we are done. 4

5 Queston 3 Part () Let X be a compact metrc space. Let A be a untal subalgebra of C(X) whch separates ponts. Then A s dense n C(X). Part () Let m > n. The formulae cos((m + n)x) = cos(mx) cos(nx) sn(mx) sn(nx) let us wrte cos((m n)x) = cos(mx) cos(nx) + sn(mx) sn(nx) cos(mx) cos(nx) = 1 (cos((m + n)x) + cos((m n)x) 2 that s as as a lnear combnaton of cos(mx) and cos(nx). It follows that A s an algebra. Allowng n =, we have that 1 A, so A s a untal algebra. If x, y [, π], then for x y we have cos x cos y. Hence A separates ponts. By the Stone-Weerstrass theorem, A s dense n C[, π]. Part () Let M (C[, π], ) be dense, where s the supremum norm. Let be the Hlbert space norm on L 2 [, π]. Let f, g C[, π]. Then f g 2 = (f g) 2 π f g 2 f g π f g. It follows that M s also dense n the space (C[, π], ). But (C[, π], ) s tself dense n L 2 [, π]. So M s also dense n L 2 [, π]. Part (v) By above, the span of the set {e, e 1, e 2,...} s dense n L 2 [, π]. We must check that the set s orthonormal. Let m > n >. Then e m, e n = 2 π Let m >. Then cos(mx) cos(nx) dx = 1 π e m, e = 2 π 5 cos((m+n)x) dx+ 1 π cos(mx) dx = cos((m n)x) dx =.

6 Observe Fnally, for m > e 2 = 1 π dx = 1. e m 2 = 2 π cos 2 (mx) dx = 1. Part (v) We have so for n = : For n > : The formulae α = 1 π α n = 2 π α n = e n, f sn x dx = 2 π cos(nx) sn x dx sn((m + 1)x) = sn(mx) cos(x) + cos(mx) sn(x) let us wrte sn((m 1)x) = (mx) cos(x) cos(mx) sn(x) so that s cos(mx) sn(x) = 1 (sn((m + 1)x) sn((m 1)x) 2 α n = 1 sn((n + 1)x) sn((n 1)x) dx 2π α n = By Parseval s law ( 2 1 π n 1 1 ) 2 2 = n + 1 π(n2 1). α n 2 = f 2 = n= sn 2 x dx =

7 Queston 4 Part () (a) (b) Let S n = 1 + x + + x n. Observe, for n > m Spectrum(x) = {λ C x λ s not nvertble}. S n S m = x m x n x m+1 1 x as m, n, snce x < 1. So (S n ) s a Cauchy sequence, meanng t converges n norm to some S A by completeness. Now (1 x)s n = (1 + x + x 2 + x n ) (x + x x n + x n+1 ) = 1 x n+1 so 1 (1 x)s n = x n+1 x n+1. Let n. Snce x < 1, we see (1 x)s =, and so (1 x)s =. In the same way, S(1 x) =, so 1 x s nvertble, wth (1 x) 1 = S. (c) We call U untary f U U = UU = I. Observe that f U s untary, then for any v V, In partcular, U = 1. Uv 2 = Uv, Uv = U Uv, v = v, v = v 2. Let λ C and suppose λ > 1. Then U/λ < 1, so I Uλ s nvertble by part (b). Multplyng by λ, we see that U λi s nvertble, so λ Spectrum(U). In the same way, f λ < 1, then I λu s nvertble, and multplyng by U, we see that U λi s nvertble, so λ Spectrum(U). Therefore f λ Spectrum(U), then λ = 1. 7

8 Part () (a) Observe x n x n and the seres x n n= converges. So the seres n= converges absolutely, and therefore (snce A s a Banach space) also converges to an element e x A. Now By the bnomal theorem exp(x) exp(y) = (x + y) n = n k= x n n n= k= x k k! y n k (n k)! k!(n k)! xk y n k Therefore n (x + y) n exp(x) exp(y) = n= k= = exp(x + y) (b) Observe (T ) = T as T s self-adjont. So By the above exp(t ) ((T ) ) n = n= = ( T ) n n= = exp( T ). exp(t ) exp( T ) = exp( T ) exp(t ) = exp() = I. Therefoe, by defnton, exp(t ) s untary. 8

9 Queston 5 Part () If V and W are normed vector spaces, we call a lnear map T : V W compact f the mage T [B(, 1)] has compact closure. Part () Let (x n ) be a bounded sequence n V. Say x n < M for all n. Then the sequence (x n /M) s contaned n the unt ball B(, 1) n V. Hence the sequence (T (x n )/M) s contaned n the compact space T [B(, 1)], and so has a convergent subsequence, T (x nk )/M. It follows that the sequence (T x n ) has a convergent subsequence (T x nk ). Part () Let T : V W be a bounded lnear map wth fnte-dmensonal mage. Then, f v B(, 1), we have v < 1, so T v < T. It follows that the mage T [B(, 1)] s bounded, so the closure T [B(, 1)] s also bounded. So T [B(, 1)] s a closed bounded subset of a fnte-dmensonal normed vector space. By the Hene-Borel theorem, ths set must be compact, meanng T s a compact operator. Part (v) Observe T x m T x n 2 = T x m T x n, T x m T x n = T T x m T T x n, x m x n Snce the sequence (x n ) s bounded, we have a constance C such that x n C for all n. By the trangle nequalty, x m x n 2C for all m and n. Hence, by the Cauchy-Schwarz nequalty T x m T x n 2 2C T T x m T x n. Snce the sequence (T T x n ) converges, t must be a Cauchy sequence. Let ε >. Then we have N such that f m, n N, then T T x m T x n < ε 2 /2C. By the above, t follows that T x m T x n < ε for m, n N, so the sequence (T x n ) s Cauchy as desred. Part (v) Let (x n ) be a bounded sequence n H. Then as K K s compact, (K Kx n ) has a convergent subsequence (K Kx nk ). By the above, the sequence (Kx nk ) s Cauchy, and therefore converges. Thus (Kx n ) has a convergent subsequence, and therefore K s compact. 9

10 Part (v) Let K be compact. The product of a compact operator wth another operator s compact. Therefore K K s compact, and the converse holds. 1