Real and Complex Analysis, 3rd Edition, W.Rudin Elementary Hilbert Space Theory

Transcription

1 Real and Complex Analysis, 3rd Edition, W.Rudin Chapter 4 Elementary Hilbert Space Theory Yung-Hsiang Huang. It s easy to see M (M ) and the latter is a closed subspace of H. Given F be a closed subspace containing M, then F M and hence (M ) (F ). Since F is closed, H = F F. Given z (F ), z = x + y with x F and y F. Since = (z, y) = (y, y), y =. Hence (M ) (F ) F. So (M ) is the smallest closed subspace containing M, that is, M! (even for non-closed subspace). Gram-Schmidt Process. 3. The case p < is separable since C(T ) is dense in L p (T ) and by Weierstrass theorem the set of trigonometric polynomials with rational coefficients is dense in (C(T ), ) and hence in (C(T ), p ). Next, consider F = {f h (x) = χ (, )(x + h), h (, 4 )}, then given f g F, f g =. This shows the dense subset of L ([, ), m) = L (T ) must be uncountable. 4. ( ) There exists a countable dense set {w n } in H. We may choose a subset {w nj } that are linearly independent by ic out the linear dependent element inductively and assume they are orthonormal by Gram-Schmidt process. Note that span{w nj } will be dense in H since its span contains {w n }. By Theorem 4.8, we find a countable maximal orthonormal system span Q {w nj }. Note this proof does NOT rely on Hausdorff maximality principle. ( ) Let {u n } be tha maximal orthonormal system. By theorem 4.8, span{u n } is dense in H. So does span Q {u n } which is countable.. We assume M H, that is, L. Then there is a unique y H such that Lx = (x, y) for all x H. Then M = {x : (x, y) = } = Y, where Y is the span of {y} which is a closed one-dimensional subspace. So M = (Y ) = Y by Exercise. Last Modified: 6/8/ Department of Math., National Taiwan University. d4@ntu.edu.tw

2 6. (a) {u n } is norm-bounded by and closed (since every pair of distinct elements has distance ), but the open cover {B(u n, )} has no finite subcover. (b) (, δ n compact. < ) By Bozano-Weierstrass theorem, it s enough to show S is sequentially Since for each sequence c n with c n δ n, the partial sums of c n u n form a Cauchy sequence in H, by completeness of H, c n u n exists. Moreover its norm is n c n. Given {x j = c j nu n } S. Since for each n, c j n δ n for all j, by diagonal process, there is a subsequence, still denoted by x j, such that c j n c n for each n. Define x = n c nu n, clearly it belongs to S. Since δ n < and x j x = n c j n c n M c j n c n + n= n=m+ We may choose M large enough so that the second term is small uniformly in j and then pic large J such that for all j > J the first term is always small. Therefore, there is a convergent subsequence of any sequence in S. (c) ( ) Let x j = j n= δ nu n S, then x j = j n= δ n. Since S is compact, there is M > such that M x j = j n= δ n for all j, that is, M n= δ n. (d) Suppose H is locally compact. Since we now the set of all balls forms the local base of every Banach space, there is B r () K which is a compact neighborhood of. So B r () is compact, which is a contradiction by applying the argument of (a) to rescaled {ru n }. Remar.. For (d), there is a general result states that every locally compact topological vector space is finite dimensional. See Rudin [, Theorem.]. 7. This is a special case of Exercise 6.4. See my remars made there. 4δ n. If (a n ) l, then there are < n < n < such that n a j >. Then for each Define b n = an c c := + n + a j > if n < n n + and if n n. Then b n = + n + a n c = <. an b n = + n + a n c =.

3 8. Also see Folland [3, Exercise.6]. 9. Apply Bessel s ineqality with the orthonormal set {sin nx, cos nx} n= to χ A L [, ].. By Hint, E = E + E where sin n x ± on E ± respectively. Then LDCT and Exercise 9 implies that = lim E ± sin n x = E ± ± = ± m(e ± ).. This exercise shows that the convexity assumption can NOT be dropped from Theorem 4. without other restrictions. The desired E = { n+e n n} where e n (i) = δn, i the Kronecer delta. On the other hand, the existence and uniqueness assertions in Theorem 4. are not true for every Banach spaces, see Exercise.4-.. st proof. Using Stirling Formula, we see = c = c ( + cos t ) c dt = ( t) t dt = c / (cos w) dw = c e c Γ( + ) Γ( + ) ( z ) dz ( + ). nd proof. lim sup c 3 4 since = c c ( + cos t / ) dt = c / (cos w) dw = c ( )z dz = c ( ) > c 4 3. ( z ) dz I DO NOT have any good idea to get the lower bound with a simpler method than Stirling. 3. This exercise also appears in Baby Rudin [4, Exercise 8.9]. Note this is related to the notion of equidistributed sequence and ergodic theorem. See Stein-Shaarchi [6, Section 4.]. By direct computation, this is valid for all f = e nx (n =,, ) and hence for all trigonometric polynomials. We pass the validity to all continuous functions by Weierstrass approximation theorem. 3

4 Remar.. This is valid for all step function χ ( a, b) by approximate it from above and below by continuous function. And hence being valid for simple functions of Riemann upper and lower sum of Riemann integrable function. Finally, it s valid for all Riemann integrable function. However, this theorem is not true for Lebesgue integrable function f = χ {(nα)}, where (nα) is the fractional part of nα, since the right-hand integral is zero but the left-hand sum is always. 4. Due to the special structure of minimal problem, it s enough to consider the real a, b, c and real-valued function g. Consider Hilbert space L ([, ], dx) with inner product (f, h) = f(x)h(x) dx. Let A be the span of {, x, x } which is the same as span of the orthonormal 3 set {, x, 8 (3x )}. The corresponding Fourier series of x 3 is 3x. Thus, (x 3 3 x) A, inf a,b,c x 3 a bx cx dx = that is, minima exists and equal to 8. Moreover, 7 sup g A, g = x 3 g(x) dx = x 3 3 x dx = 8 7, sup (g, x 3 ) L = sup (g, x g A, g = g A, g = x) L 7, by Cauchy-Schwarz. Note that the supreme is attained by g(x) = 8 (x 3 3x). 7. Due to the special structure of minimal problem, it s enough to consider the real a, b, c. Consider Hilbert space L ([, ], e x/ dx) with inner product (f, h) = f(x)h(x)e x dx. Let A be the span of {, x, x } which equals to span of the orthonormal set {, x, (x 4x+)}. The corresponding Fourier series of x 3 is 9x 8x + 6. Thus, x 3 (9x 8x + 6) A, inf a,b,c x 3 a bx cx e x dx = that is, minima exists and equal to 36. Next, we try to maximize where g is subject to the restrictions g(x) e x =, g(x)e x = x 3 g(x)e x dx x 3 (9x 8x + 6) e x dx = 36, xg(x)e x = x g(x)e x =. That is, g A, g =. Note that it s enough to consider real-valued g and sup g A, g = x 3 g(x)e x dx = sup (g, x 3 ) L = sup (g, x 3 (9x 8x + 6)) L 6, g A, g = g A, g = by Cauchy-Schwarz. Moreover, the supreme is attained by g(x) = 6 [x3 (9x 8x + 6)]. 4

5 6. This is a general result for the computations of Exercise Uniform limit on R of members of X are called almost periodic. See Stein [7, Problem 4.], Besicovitch[] and Bohr []. 9. References [] Abram Samoilovitch Besicovitch. Almost periodic functions, volume 9. Dover publications New Yor, 94. [] Harald Bohr. Almost Periodic Functions. Chelsea Publishing Company, Chapman and Hall, 947. [3] Gerald B Folland. Real analysis: Modern Techniques and Their Applications. John Wiley & Sons, nd edition, 999. [4] Walter Rudin. Principles of Mathematical Analysis, volume 3. McGraw-Hill New Yor, 3rd edition, 976. [] Walter Rudin. Functional analysis. McGraw-Hill, Inc., New Yor, nd edition, 99. [6] Elias M Stein and Rami Shaarchi. Fourier Analysis : An Introduction, volume. Princeton University Press, 3. [7] Elias M Stein and Rami Shaarchi. Real analysis: Measure theory Integration, and Hilbert Spaces. Princeton University Press,.