Solutions: Problem Set 3 Math 201B, Winter 2007 Problem 1. Prove that an infinite-dimensional Hilbert space is a separable metric space if and only if it has a countable orthonormal basis. Solution. If H is a finite-dimensional Hilbert space with orthonormal basis then {e n 1 n d}, { d } D = c n e n c n = q n + ir n with q n, r n Q n=1 is a countable dense subset of H. If H is an infinite-dimensional Hilbert space with countable orthonormal basis {e n n N}, then { N } D = c n e n N N, c n = q n + ir n with q n, r n Q n=1 is a countable dense subset. Thus, H is separable if it has a countable orthonormal basis. Suppose that H has an uncountable orthonormal basis, and let D be a dense subset of H. E = {e α α A}, The orthonormality of E implies that if α β, then e α e β 2 = e α 2 + e β 2 = 2. The open balls B 2/2 e α) are therefore disjoint, and, since D is dense, each ball contains at least one point x α D, say. The map α x α is a one-to-one map of A into D, so the cardinality of D is greater than or equal to the cardinality of A. It follows that no dense subset of H is countable, so H is not separable.
Problem 2. Prove that if M is a dense linear subspace of a separable Hilbert space H, then H has an orthonormal basis consisting of elements in M. Solution. If H is finite-dimensional, then every linear subspace is closed. Thus, the only dense linear subspace of H is H itself, and the result follows from the fact that H has an orthonormal basis. Suppose that H is infinite-dimensional. Since H is separable, it has a countable dense subset {x n n N}, which need not be a subset of M. Since M is dense in H, for each n N, there exists a sequence x mn ) in M such that x mn x n as m. The set {x mn m, n N} is then a countable subset of M that is dense in H. Let D be a subset of M that is dense in H, and let B = {x n n N} be a maximal linearly independent subset of D. Then the linear span of B, meaning all finite linear combinations of elements of B, contains D so it is dense in H. The closed linear span of B is therefore equal to H. Gram-Schmidt orthonormalization of B gives an orthonormal set E = {e n n N} whose closed linear span is equal to that of B, meaning that E is an orthonormal basis of H. Moreover, since each x n M and each e n E is a finite linear combination of {x 1,..., x n }, it follows that e n M. Thus, E is an orthonormal basis of H consisting of elements of M.
Problem 3. Define the Legendre polynomials P n by P n x) = 1 d n. 2 n n! n a) Compute the first four Legendre polynomials, P 0 x), P 1 x), P 2 x), P 3 x). b) Show that the Legendre polynomials are orthogonal in L 2 [, 1]). c) Show that the Legendre polynomials are obtained by Gram-Schmidt orthogonalization of the monomials {1, x, x 2,...} in L 2 [, 1]). d) Show that P n x) 2 = 2 2n + 1. e) Show that the Legendre polynomial P n is an eigenfunction of the differential operator L = d ) 1 x 2 d with eigenvalue λ n = nn + 1), meaning that LP n = λ n P n. f) Compute the polynomial qx) of degree 2 that is closest to e x on [, 1], in the sense that { } e x qx) 2 = min e x fx) 2 fx) = ax 2 + bx + c. Solution. a) The first few Legendre polynomials are P 0 x) = 1, P 1 x) = x, P 2 x) = 3 2 x2 1 2, P 3 x) = 5 2 x3 3 2 x, P 4x) = 35 8 x4 15 4 x2 + 3 8. To prove that P m is orthogonal to P n, where we may assume m < n without loss of generality, it suffices to prove that x m is orthogonal to P n for every m < n. It then follows by linearity that every polynomial of degree m < n is orthogonal to P n, including, in particular, P m.
Integrating by parts m-times, we compute that x m, P n = 1 1 2 n n! = )m 2 n n! x m dn d m n m xm ) dn m n m [ = )m m! d n m x 2 1 ) ] 1 n 2 n n! n m = 0. All of the boundary terms vanish at x = ±1 because, for 1 k n, the polynomial d n k n k has a factor x 2 1) k. Hence x m P n for m < n, and P m P n for m n. c) The linear subspace of polynomials of degree n has dimension n+1. The orthogonal complement of the polynomials of degree n 1 in the space of polynomials of degree n is equal to 1, and therefore {P n } is a basis of the orthogonal complement. The Gram-Schmidt orthogonalization of the monomials gives a polynomial of degree n in this complement, so it gives the Legendre polynomials up to normalization. d) Integrating by parts as in a), we compute that P n, P n = 1 1 2 n n!) 2 = )n 2 n n!) 2 = )n 2n)! 2 n n!) 2 d n n d 2n Using integration by parts again, we get d n n 2n. x 2 1 ) n = x 1) n x + 1) n
= n x 1) n x + 1) n+1 n + 1 ) n nn 1)... 1 1 = x + 1) 2n n + 1)n + 2)... 2n) n!) 2 2 2n+1 = 2n)!2n + 1). Using this integral in the expression for the inner product of P n, and simplifying the result we get e) We write D = d/. product gives P n 2 = P n, P n = 2 2n + 1 D n fg) = Leibnitz s rule for the nth derivative of a n k=0 In particular, since D k x m = 0 for k > m, D n xf) = xd n f + nd n f, ) n D k f D n k g. k D n x 2 f ) = x 2 D n f + 2nxD n f + nn 1)D n 2 f. Let ux) = x 2 1) n. Then x 2 1 ) Du = 2nxu. We apply D n+1 to this equation, and use Leibnitz s rule to expand the derivatives, which gives x 2 1 ) D n+2 u + n + 1) 2xD n+1 u + n + 1)nD n u = 2nxD n+1 u + 2nn + 1)D n u. After simplification we obtain that x 2 1 ) D n+2 u + 2xD n+1 u nn + 1)D n u = 0, which implies that x 2 1 ) D 2 P n + 2xDP n nn + 1)P n = 0. This equation is equivalent to LP n = λ n P n.
f) By the projection theorem, the closest polynomial of degree N to f L 2 [, 1]) is the one such that the error f q is orthogonal to the linear space of polynomials of degree N, meaning that N q = c n P n n=1 where the {c n n = 0, 1, 2,...} are the Fourier coefficients of f with respect to the Legendre polynomials {P n n = 0, 1, 2,...}, c n = P n, f P n 2. If fx) = e x, then we compute that P 0, f = = [e x ] 1 = e 1 e, 1 e x P 1, f = xe x = [xe x e x ] 1 = 2 e, 3 P 2, f = 2 x2 1 ) e x 2 [ 3 = 2 x2 e x 3xe x + 3e x 1 2 ex = e 7 e. Using the normalization of the Legendre polynomials from d), we find that the closest quadratic polynomial q to e x in L 2 [, 1]) is qx) = 1 e 1 ) + 3 ) 2 x + 5 e 7 ) 3 2 e 2 e 2 e 2 x2 1 ) 2 = 3 e 11 ) + 3 4 e e x + 15 e 7 ) x 2. 4 e ] 1
Problem 4. Define the Hermite polynomials H n by a) Define H n x) = ) n e x2 dn e x2). n φ n x) = e x2 /2 H n x). Show that {φ n n = 0, 1, 2,...} is an orthogonal set in L 2 R). b) Show that the nth Hermite function φ n is an eigenfunction of the linear operator with eigenvalue Solution. H = d2 2 + x2 λ n = 2n + 1. a) It is sufficient to show that φ n is orthogonal to e x2 /2 x m for each m < n, since then φ n is orthogonal to every function of the form e x2 /2 p m, where p m is a polynomial of degree m < n, and hence in particular to φ m. Integrating by parts m-times, and using the fact that p e x2 /2 0 as x for every polynomial p, we compute that e x2 /2 x m, φ n which proves the result. = ) n = ) m+n m! = ) m+n m! = 0, e x2) n x m dn d n m e x2) [ n m d n m e x2)] n m b) Let A = d + x, A = d + x.
We show below that dh n = 2nH n = H n+1 + 2xH n. 1) Using this result, we compute that ) d ) Aφ n = + x e x2 /2 H n = e x2 /2 dh n = 2ne x2 /2 H n = 2nφ n, and A φ n = d = e x2 /2 ) ) + x e x2 /2 H n ) dh n + 2xH n = e x2 /2 H n+1 = φ n+1, The product rule xf) = xf + f implies that Hence AA = d x = x d + 1. ) d + x d ) + x = d2 + d 2 x x d + x2 = H + 1, so H = AA 1.
It follows that Hφ n = AA 1) φ n = AA φ n φ n = Aφ n+1 φ n = 2n + 1)φ n φ n = 2n + 1) φ n. Finally, we prove 1). First, using the product rule, we get dh n = ) n d dn x2 [e e x2)] n = ) n dn+1 x2 e e x2) + ) n 2xe x2 dn e x2) n+1 n = H n+1 + 2xH n. 2) Second, carrying out one differentiation and using the Leibnitz formula for the nth derivative of a product, we get d n+1 e x2) = dn 2xe x2) n+1 n = 2x dn n e x2) 2n dn n e x2). Multiplying this equation by ) n+1 e x2 and using the definition of the Hermite polynomials, we get the recurrence relation H n+1 = 2xH n 2nH n. Using this equation to eliminate H n+1 from 2), we find that dh n = 2nH n. Remark. It follows from the Weierstrass approximation theorem that both the Legendre polynomials and the Hermite functions are complete orthonormal sets, and hence they provide orthonormal bases of L 2 [, 1]) and L 2 R), respectively.