Functional Analysis Exercise Class Week: December 4 8 Deadline to hand in the homework: your exercise class on week January 5. Exercises with solutions ) Let H, K be Hilbert spaces, and A : H K be a linear operator. Show that and x max{ y, x : y H, y } max{ y, x : y H, y }, x H, A sup{ y, Ax : x H, x, y K, y } sup{ y, Ax : x H, x, y K, y }. Solution: If x or A then all the claims are trivial, hence for the rest we assume that x and A. By the Cauchy-Schwarz inequality, y, x y x x when y, and hence sup{ y, x : y H, y } sup{ y, x : y H, y } x. On the other hand, choosing y : x/ x, we get y, x x x, x x x the suprema in both expressions above are attained and give x. Consequently, sup{ y, Ax : x H, x, y K, y } sup {sup{ y, Ax : y K, y } : x H, x } sup { Ax : x H, x } A, and the other expression follows the same way. x, and hence ) Show that there exists an inner product on l p : l p N, K) such that x p x, x for every x l p if and only if p. Hint: Use the characterization via the parallelogram identity.) Solution: For every i N, let e i j) : δ i,j. Let i, j N be such that i j. Then e i e j, and e i + e j + e i e j p + p.
Hence, e i + e j + e i e j e i + e j ) p 4 p. This shows that for p, there is no such inner product. For p, define x, y : xi)yi). i N This sum is finite by the Hölder inequality with p q ), and it is easily seen to give a sesquilinear form on l. Obviously, x : x, x i N xi) x, as required. 3) Let {x i } i I be a set of non-zero pairwise orthogonal vectors in an inner product space. Show that {x i } i I is linearly independent. Solution: Assume that there exist i,..., i r and λ,..., λ r K such that r j λ jx ij. For every k,..., r, r r r λ j x ij, x ik λ j xij, x ik λ j δ k,j x ik λ k x ik, j and hence λ... λ r. j 4) Show that if x,..., x n, are pairwise orthogonal vectors in an inner product space i.e., x i, x j for i j) then n n x i x i. i i j Solution: We have n n x i x i, i i n n n x i x i, x j δ i,j x i i i,j i,j n x i. i 5) Let H be a Hilbert space, and {x n } n N H be an orthogonal system, i.e., x n, x m for every n m. Show that x n is convergent x n < +, and then x n x n. Show that the if the x n are not pairwise orthogonal then + x n < + does not imply the convergence of + x n, and even if the infinite sum is covergent, the identity + x n + x n need not hold.
Solution: For every N N, let s N : N x n. Let N, M N, and assume without loss of generality that N M. Then s N s M M nn+ x n M nn+ x n. Hence, s N ) n N is convergent it is Cauchy + x n < +, where the first equivalence is due to the completeness of the space. Assume that the limit s : lim N + x n : + x n exists. Since the norm is continuous, we have x n lim s N N N + lim s N lim x n x n, N + N + where we used Exercise 4) for the third equality. Consider now a non-zero vector x, and let x n : x, n N. Then n x n x + However, for N < M, we have s N s M M nn+ n < +. x n x M nn+ n x M nn+ ). n Keeping N fixed, the above expression converges to + as M +. Hence the sequence s N : N x n is not Cauchy, and thus not convergent, either. Even if + x n is convergent, the identity + x n + x n need not hold, as one can check with just two non-orthogonal vectors i.e., x, x but x n n > ). 6) Let {e n } n N be an orthonormal system in a Hilbert space H, and let c n K, n N. Show that c n x n is convergent c n < +, and then c n x n c n. Solution: Let x n : c n e n ; then x n, x m δ n,m c n, and the statement is immediate from Exercise 5). 3
7) Let {e n } n N {} be an orthonormal system in a Hilbert space H, and for every t, ), let x t : n t n e n..) Show that the above sum is convergent, and the set {x t } t,) is linearly independent. Conclude that every infinite-dimensional Hilbert space is at least of continuum dimension. Hint: Connect the problem to the Vandermonde determinant.) Solution: With c n : t n, n N {}, we have n c n n t n t, and hence the convergence of.) follows from Exercise 6). t,..., t r, ) and λ,..., λ r K such that r r ) λ i x ti λ i t n i e n. i n i Assume that there exist Then + r ) λ i t n i e n, e m i r λ i t m i i for every m N. Writing it in matrix form for m,..., r, we get... λ t t... t r λ t t... t r λ 3.... t r t r... t r r λ r The matrix above is the so-called Vandermonde matrix, the determinant of which is i<j r t j t i ) if all the t i are different. Hence, λ... λ r. This shows that {x t } t,) is linearly independent. Assume that H is a Hilbert space of infinite algebraic dimension. Then there exists a linearly independent set {x n } n N of countably infinite cardinality. Applying the Gram- Schmidt orthogonalization process to this set, we get an orthonormal system {e n } n N. From this we can construct a linearly independent set {x t } t,) by the above procedure. Hence, the dimension of H is at least continuum. 8) Let e n, n N, be an orthonormal system in a Hilbert product space, and for every finite subset I N, let H I be the subspace generated by e i, i I. Show that if x + c ne n then its projection onto H I is given by i I c ie i. 4
Solution: Obviously, e i, i I, is an orthonormal basis in H I. According to what was shown in the lecture, the projection P I x of x onto H I is given by P I x i I e i x, e i. By the continuity of the inner product, + x, e i c n e n, e i c n e n, e i from which the statement follows. 9) Let f, g : π fx)gx) dx for every f, g C[, ], R). a) Show that.,. is an inner product on C[, ], R). b) Show that { π, π cos nx, } sin nx : n N π is an orthonormal system in C[, ], R). c) Determine π min cos x α sinx) β cosx) ) dx, α,β R c n δ n,i c i, and give the values of α, β where the minimum is reached. Hint: Use the addition formulas sinx±y) sin x cos y±cos x sin y, cosx+y) cos x cos y sin x sin y.) Solution: a) The linearity of the integral yields that.,. is sesquilinear, and it is hermitian, as Finally, f, g f, f π π fx)gx) dx fx) dx, π gx)fx) dx g, f. and it is equal to if and only if f, due to continuity. 5
b) Let e x) π, e n x) : π cos nx, and e n+ x) : π sin nx, n N. We only show that e n+, e m+ δ n,m ; the proof for the other cases goes the same way. Note that we have and hence sinnx) sinmx) cosn m)x cosn + m)x, π e n+, e m+ sinnx) sinmx) dx π π cosn m)x cosn + m)x dx π π cosn m)x dx π δ n,m, as required. c) Note that we have π m : min cos x α sinx) β cosx) ) dx min f αe 5 + βe 4 ), α,β R α,β R where fx) : cos x. By definition, the minimum is attained for the α, β such that αe 5 + βe 4 is the projection P f of f onto the subspace spanned by e 4, e 5, i.e., Note that α f, e 5, β f, e 4. and hence cos x) + cosx) π π e + e 4, α f, e 5, β f, e 4 Moreover, for the optimal α, β we have π, P f e 4. π min cos x α sinx) β cosx) ) dx α,β R f P f π e π 4 e π. 6
Homework with solution ) Let H be an inner product space. Show that a sequence x n ) n N converges to an x H if and only if i) x n x as n + ; ii) for every y H, y, x n y, x as n +. Solution: Convergence of x n to x implies i) and ii) due to the continuity of the norm and the inner product. Assume now that i) and ii) hold. Then x n x x n + x x n, x x, x n n + x + x x, x x, x, i.e., x n x as n +. ) Let C[, ], K) be equipped with the inner product f, g : fx)gx) dx. Find the projection of the function fx) : sin x to the subspace of all polynomial functions of degree at most. Using this result, find the optimal a, b where min a,b K is attained. sin πx ) a bx Hint: Construct an orthonormal basis in the given subspace.) Solution: Let f x), f x) x, x [, ]. Our aim is to construct an orthonormal basis {e, e } in span{f, f }. Note that f has norm, and we can choose e : f. To obtain f, we use Gram-Schmidt orthogonalization, that yields f f, e e )x) x x dx x, and f f, e e e. Let e : f / f, where f x ) dx x x + ) dx 4 3 + 4, and hence e x) x ). Then {e, e } is an ONB in span{f, f }. The projection P of any function g onto span{f, f } is given by P g g, e e + g, e e gt) dt + x ) [,] For g α x) sin αx we get sin αt dt [ cos αt] /α cos α)/α, [,] t sin αt dt cos α α + sin α α, [,] 7 gt) [,] t ) dt.
and hence P g α )x) cos α α In particular, for α we get α + sin α α cos α + x 6). α + sin cos P sin)x) cos + x 6) 4 + cos 6 sin + x 6 + sin 6 cos ), ) and for α π/, P g π/ )x) π + x 6)4 π π 8π 4 π 48 π + x. π We have min a,b K sin πx ) a bx min gπ/ af bf a,b K min g span{f,f } and the mimimum is reached at P g π/. Thus, the optimal a, b are given by gπ/ g, a 8π 4 π, b 48 π π. Scores: e : point, e : 3 points, correct application of the projection formula: 3 points, correct formula for the projection of sin: point, correct formula for the projection of g π/ : point, optimal a, b: points. 8