Exercise Solutions to Functional Analysis

Exercise Solutions to Functional Analysis Note: References refer to M. Schechter, Principles of Functional Analysis Exersize that. Let φ,..., φ n be an orthonormal set in a Hilbert space H. Show n f n α φ f (f, φ )φ = Solution. Since f H, we now that f V H, where V is some finite dimensional subspace of H, say dim V = m n. We may assume that V contains φ, φ 2,..., φ n. Let {φ,..., φ n, φ n+,..., φ m } be an orthonormal basis for V (this is possible by extending {φ,..., φ n } to an orthonormal basis via Gram-Schmidt). Now write f = m β φ. Then we have m f (f, φ )φ = β φ Hence = n f 2 (f, φ )φ = =n+ m =n+ since φ = for m. On the other hand, f α φ = (β α )φ + Hence It is clear now that It follows that We re done. n f 2 α φ = (β α ) 2 + β 2 m =n+ m =n+ β φ n f 2 n α φ f 2 (f, φ )φ n f n α φ f (f, φ )φ β 2

Exersize 2. Let c denote the set of all elements (α, α 2,... ) l such that {α n } is a convergent sequence, and let c 0 be the set of all such elements for which α n 0 as n. Show that c and c 0 are Banach spaces. Solution 2. It is nown that l is a Banach space. The linear structure, as well as the norm, of c and c 0 is inherited from that of l. To show completeness of c and c 0, it will suffice to show that c and c 0 are closed in l. We show that c is closed in l and c 0 is closed in c. Now, suppose that {x } is a sequence in c converging to some x l. Then, for any given ɛ > 0, we now that there exists 0 such that Hence for all j we have sup x j 0 x j < ɛ j 3 } We also now that, since {x j0 j N x j 0 x j < ɛ 3 exists N N such that for all n m N, we have From this we deduce: is a convergent sequence, it is Cauchy, so there x n 0 x m 0 < ɛ 3 x n x m x n x n 0 + x n 0 x m 0 + x m 0 x m < ɛ for all n m N. This shows that {x} is Cauchy, hence convergent. Thus x c. This shows that c contains all its limit points, hence closed. So c, being a closed subset of a complete space, must itself be complete. Similarly, let us suppose that {x } is a sequence in c 0 that converges to x l. Then for all ɛ > 0, there exists 0 such that Hence for all j we have sup x j 0 x j < ɛ j 2 x j 0 x j < ɛ 2 On the other hand, since x 0 c 0, there exists N N such that for all n N, we have x n 0 < ɛ 2 Hence for all n N we have x n x n 0 + x n 0 x n < ɛ So x = { x j} converges to 0, hence x c 0. So c 0 is complete. Exersize 3. Show that the norm of an element is never negative. Solution 3. Suppose that x < 0. But then 0 = x x x + x < 0. Can t be! Exersize 4. If x, y are are elements of a Hilbert space H and satisfy x + y = x + y, show that either x = cy or y = cx for some scalar c 0.

Solution 4. Well, if either x = 0 or y = 0, the result holds vacuously. So assume that neither x nor y is equal to zero. We may assume that x, y V H, with dim V = m <. If m =, we re done. If not, we may tae for a basis of V the set {x, x 2,..., x m } of orthonormal vectors where x = x/ x. This can be done via Gram-Schmidt. Write y = α x + + α m x m. Then we have x + (α x + + α m x m ) = x + α x + + α m x m From this we immediately deduce 2 (x, α x + + α m x m ) = 2 x α x + + α m x m By orthonormality of {x,..., x m }, we get from above: α = α 2 + + α2 m From this we immediately deduce that Hence we must have We re done. α 2 = = α m = 0 y = α x = α x x Exersize 5. Show that in a normed vector space v j = v j + j=n+ Solution 5. In a normed vector space the notion of the limit maes sense, and in fact all standard results regarding algebra of limits hold. In particular we have, by definition (for n fixed) v j = lim v j = lim On the other hand for n we have v j = v j + Hence v j = lim v j + j=n+ v j = v j v j, n j=n+ v j + lim Exersize 6. For f = (α,..., α n ) R n, show that and f 0 = max α f = α = v j j=n+ are norms. Is R n complete with respect to either of them? v j = v j + j=n+ v j

Solution 6. It is easy to see that () and (2) of the definition of a norm on p. 7 hold trivially in both cases. So we only verify the triangle inequality. Write f = (α,..., α n ) and g = (β,..., β n ). Then we have f + g 0 = max α + β max ( α + β ) which follows from the triangle inequality of on R. On the other hand, it isn t hard to see that max In the second case, f + g = ( α + β ) max α + β = α + max β = f 0 + g 0 ( α + β ) = f + g = Now, we now that R n is complete with respect to the standard Euclidean norm. But all norms on a finite dimensional vector space are equivalent. So R n is complete with respect to the two norms above. More directly, it isn t hard to see that: f 0 α 2 + + α2 n and f α 2 + + α2 n Exersize 7. If F is a bounded linear functional on a normed vector space X, show that F = sup x F (x) = sup x = F (x) Solution 7. By definition we have F (x) F = sup x 0 x But ( ) F (x) sup x 0 x = sup x F = sup F (x) x 0 x x = On the other hand, F (x) sup F (x) = sup x = x 0 x sup F (x) x x sup F (x) sup F (x) x x = This verifies the claim. Exersize 8. A functional F (x) is called additiveif F (x + y) = F (x) + F (y). If F is additive, show that F (αx) = αf (x) for all rational α. Solution 8. Well, F (0) = F (0 + 0) = F (0) + F (0). So F (0) = 0. By definition, for any n > 0 an integer, we have Hence we must have F (nx) = nf (x). nx = x + x + + x }{{} n - times

Now, for m > 0, F (x) = F (mx/m) = mf (x/m). So F (x/m) = F (x)/m. Now it immediately follows that for n 0, m > 0, we have F (nx/m) = nf (x)/m. Finally, 0 = F (x x) = F (x + ( x)) = F (x) + F ( x). So F ( x) = F (x). This completes the proof. Exersize 9. Show that an additive functional is continuous everywhere if it is continuous at one point. Solution 9. Suppose that F is an additive function which is continuous at some x 0. We shall show that F is bounded. That is, there exists C > 0 such that for all x we have F (x) C x Since F is continuous at x 0, we now that, for arbitrary ɛ > 0 fixed, there exists a δ > 0 such that for all x x 0 δ we have F (x) F (x 0 ) ɛ. Now tae y 0. If y is not rational, we can always pic y /2 < M < y rational sufficiently close to y. Set x = x 0 + δ 2M y Then it is clear that x x 0 < δ Hence we have (also since F is additive): F (x x 0 ) ɛ By the previous problem, since δ/2m is rational, we must have ( ) ɛ F (x x 0 ) = δ F 2M y = δ F (y) 2M Hence F (y) ɛ 2ɛ 2M < δ δ y This construction wors for all y, with 2ɛ δ being independent of y. Hence F is bounded, thus continuous everywhere. Exersize 0. Show that the set I ω is compact. Solution 0. Since l is a metric space, we now that a subset of a metric space is compact if and only if it is sequentially compact. So we show that I ω is sequentially compact: that is, every sequence has a convergent subsequence. Let x(i, j) i,j N be a sequence in l such that x(i, j) /j for all i, j. Clearly for each fixed j, x(i, j) has a subsequence x(i, j) that converges. Now let y (i, j) be a subsequence of x(i, j) such that y (i, ) ỹ. Let y 2 (i, j) be a subsequence of y (i, j), such that y 2 (i, 2) ỹ 2. In general, let y n (i, j) be a subsequence of y n (i, j), such that y n (i, n) ỹ n. Now let z(i, j) = y i (i, j). We claim that z(i, j), (ỹ, ỹ 2,... ) I ω and that z(i, j) (ỹ, ỹ 2,... ) as i. Clearly, by construction, for each fixed i, z(i, j) /j, since it is a germ of the sequence x(i, j) j N. Since ỹ j is the limit of a subsequence of x(i, j) i N, and x(i, j) /j for all i, we must have also ỹ j /j. Hence (ỹ j ) j N I ω. Now suppose that ɛ > 0 is fixed. Let N be so large that /N < ɛ/2. Let M, M 2,..., M N be such that for all M i, y i (, i) ỹ i < ɛ. Let M = max {max i {M i }, N}. Then observe that for all M, z(, j) ỹ j < ɛ when

j N, and z(, j) y j z(, j) + y j /N + /N = 2/N < ɛ when j > N. Hence we have sup z(, j) ỹ j < ɛ for all > M j So, as claimed, z(i, j) (ỹ j ) in l -norm. This proves sequential compactness of I ω, and hence compactness. Exersize. Let M be a totally bounded subset of a normed vector space X. Show that for each ɛ > 0, M has a finite ɛ-net N M. Solution. We proceed by contradiction. Suppose that M is totally bounded, and for some fixed ɛ > 0, there is no finite ɛ-net N contained entirely in M. Then given x M, there exists x 2 M such that x 2 x > ɛ. Similarly there must exist x 3 M such that x i x 3 > ɛ, with i =, 2. In general, we can continue this process and obtain x n M such that x n x i > ɛ with i =, 2,..., n. Let S = {x, x 2,... }. However, M is totally bounded, so there exists a finite ɛ/2-net of M, say given by {w, w 2,..., w } such that the balls B(w i, ɛ/2), with i =, 2,..., cover M. Since there are infinitely many distinct terms in the set S M, for some i j, x i, x j S and x i, x j B(w l, ɛ/2) for some l. But then x i x j ɛ/2, which cannot be by construction of S. This is a contradiction. Exersize 2. Prove that if M is finite dimensional, then we can tae θ = in Lemma 4.7 Solution 2. Let θ n = (n )/n < for n 2. We now then that there exists x n M c, such that d(x n, M) θ n and x n = (in some fixed norm on X: X has many admissible norms, but all are equivalent). Since X is finite-dimensional and (x n ) is a bounded sequence, we now that x n has a convergent subsequence, say x n x X. But then x = (this is obvious from the continuity of the norm, and the fact that x n = for all n) and d(x, M) (n )/n for all n. Hence d(x, M). We re done. Exersize 3. Show that if X is infinite dimensional and K is one-to-one operator in K(X), then K I cannot be in K(X). Solution 3. Let S X be the unit sphere (that is, the set of all vectors of unit norm). For convenience, let Sp(x, x 2,..., x ) denote the space spanned by the vectors x i, i =, 2,...,. Now, pic x S. Clearly Sp(x ) X, so there exists x 2 S such that x x 2 > /2. Now, obviously Sp(x, x 2 ) X. So there exists x 3 S such that x x 3, x 2 x 3 > /2. Continue this process, at n th step selecting x n S such that x i x n > /2 with i =, 2,..., n. This is possible since Sp(x, x 2,..., x n ) X (since X is infinite dimensional). Now, the sequence (x n ) n N is bounded. Keep this in mind... Now suppose that A = K I is compact. But then there exists a subsequence (x n ) such that Ax n converges; so in particular is Cauchy. Let, for convenience, y = x n. Thus, for ɛ < /2 there exists N N such that for all n m N we have Ay n Ay m < ɛ Hence y m y n Ky m Ky n (y m y n ) (Ky m Ky n ) = (Ky n y n ) (Ky m y m ) < ɛ

But then y m y n < ɛ + Ky m y n Now, K is compact, so there exists a subsequence z = y n such that Kz converges, so in particular is Cauchy. Hence for N large enough, ɛ + Kz m z n < /2 for all n m N. So z m z n < ɛ + Kz m z n < /2 which cannot be, since z m z n > /2 by construction of the sequence (x n ), of which (z n ) is a subsequence. So A cannot be compact. This completes the proof. Exersize 4. Let X be a vector space which can be made into a Banach space by introducing either of two different norms. Suppose that convergence with respect to the first norm always implies convergence with respect to the second norm. Show that the norms are equivalent. Solution 4. Let and 2 denote the two norms. It will suffice to show only that there exists C > 0 such that for all x X nonzero one has x C x 2 for then a similar argument will establish the existence of D > 0 such that x 2 D x from which equivalence of the two norms could be deduced easily. Suppose, for a contradiction, that for all n N, there exists x n X nonzero such that x n > n x n 2 But then we must have x n 2 < x n n Hence x n / x n 0 in the 2 -norm. But then, by hypothesis, x n / x n 0 in the -norm. This cannot be, since x n / x n = for all n. This is a contradiction. We re done. Exersize 5. Suppose A B(X, Y ), K K(X, Y ), where X, Y are Banach spaces. If R(A) R(K), show that A K(X, Y ). Solution 5. Since both A and K are linear operators, we now that R(A), R(K) Y are vector subspaces of Y. Now, we also now that since K is compact, R(K) is finite-dimensional (see next problem below). But then A : X R(A) R(K) is a bounded operator into a finite dimensional space. Hence A is compact. To see this: let (x n ) be a bounded sequence in X. Since A is bounded, A(x n ) is a bounded sequence in R(A). Since R(A) is finite-dimensional, A(x n ) must have a convergent subsequence. We re done. Exersize 6. Suppose X, Y are Banach spaces and K K(X, Y ). If R(K) = Y, show that Y is finite dimensional. Solution 6. Let X = X/N(K). Now, N(K) is certainly closed in X. Define the norm on X/N X : X R by x X = inf z x z N(K)

where is the norm on X. It isn t hard to verify that the norm above is welldefined, and actually is a norm. Let K : X Y be the homomorphism (of vector spaces) induced by the homomorphism K - that is, K([x]) = K(x). Then this induced homomorphism is actually a vector space isomorphism (fundamental theorem of vector space homomorphisms - isn t hard to verify). But more is true - K is compact! To see this, observe that since N(K) is closed, for all x / N(K), there exists z N(K) such that [x] X = x z. Now, let us suppose that [x ], [x 2 ],... is a sequence in X that is bounded, say by C > 0. But then there exist z, z 2, N(K) such that x n z n = [x n ] X < C for all n. Hence the sequence (x n z n ) is bounded in X. Now, for all n, K([x n ]) = K(x n ) = K(x n ) 0 = K(x n ) K(z n ) = K(x n z n ) since z n N(K). But then, by compactness of K, there is a subsequence K(x n ) that converges (in Y ). Hence K([x n ]) converges in Y, which is what we wished to show. Now, observe that K is compact, onto Y and is -to-. Hence K is bounded (hence closed), onto Y and is -to-. But then there exists C > 0 such that for all x X, (*) x X C K( x) Y We re now ready to prove that Y is finite dimensional. We ll proceed by contradiction. Suppose that dim Y =. But then there exists a sequence (y n ) in Y such that y n Y = for all n, and for all i j, one has y n y m Y > /2. See solution 4 above for construction of such a sequence. Clearly then (y n ) has no convergent subsequence in Y. Since K is -to-, let x n = K (y n ). But then by (*) above, (x n ) is a bounded sequence in X, and, as noted above, its image, namely (y n ), does not have a convergent subsequence. This contradicts compactness of K. Exersize 7. Show that if X is an infinite dimensional normed vector space, then there is a sequence {x n } such that x n =, x n x m for m n. Solution 7. For convenience, S and Sp will be used as in solution 4. Throughout, all our vectors are taen to be nonzero. Now, let x X. But then there exists y X such that dim Sp(x, y ) = 2. But then, according to solution 3, there exists x 2 Sp(x, y ) such that x 2 S and x x 2. Proceed inductively. At the n th step, since X is infinite dimensional, there exists y n X such that dim Sp(x, x 2,..., x n, y n ) = n +. So there exists x n+ S such that x i x n+ for i n. Thus (x n ) is the sequence we need. Exersize 8. Show that every finite-dimensional vector space can be made into a strictly convex normed vector space. Solution 8. Observe that both, R n and C n are strictly convex vector spaces (under the standard Euclidean norm). This follows easily from the Cauchy-Schwartz inequality, since both are actually Hilbert spaces. Now, if dim V = n, where V is a vector space over K = R, C, then V is isomorphic, as a vector space, to K n. Suppose that f : K n V realizes this isomorphism (for instance, if {β,..., β n }

is a basis for V, we could tae f(e j ) = β j ), where {e,..., e n } is the standard basis for K n, and extend f linearly to all of K n ). Now define a norm on V via v = E(f (v)), where E is the standard Euclidean norm on K n. Then V is strictly convex under. Exersize 9. Let V be a vector space having n linearly independent elements v,..., v n such that every element v V can be expressed in the form (4.2). Show that dim V = n. Solution 9. Obviously dim V n, since V is spanned by n vectors. Say u,..., u m also span V. Then we can write v = c u + c 2 u 2 + + c m u m... v n = c nu + c 2 nu 2 + + c m n u m We also now that each u j is a linear combination of v,..., v n. Hence we may write v = c n dj v j + c 2 n dj 2 v j + + c m n dj mv j Or, equivalently,. v n. =. c n n dj v j + c 2 n n dj 2 v j + + c m n n dj mv j 0 = (c d )v + n j=2 c d j v j + n c2 d j 2 v j + + n cm d j mv j. 0. =. n c nd j v j + n c2 nd j 2 v j + + n cm n d j mv j + (c m n d n m )v n If m < n, then the system above clearly has at least one nontrivial solution in c j i dl, contradicting linear independence of v,..., v n. Hence m n. Thus dim V n. We conclude that dim V = n. Exersize 20. Let V, W be subspaces of a Hilbert space. If dim V < and dim V < dim W, show that there is a u W such that u = and (u, v) = 0 for all v V. Solution 20. Suppose that u,..., u n is a basis for V. We may assume, without loss of generality, that {u,..., u n } is an orthonormal basis (perform Gramm- Schmidt). Now, obviously V W, so there exists w W such that w / V. Hence the set {u,..., u n, w} is linearly independent. Perform Gramm-Schidt orthonormalization process on w to obtain the set {u, u 2,..., u n, u} of orthonormal vectors. Then u / V, u = and (u, u j ) = 0 for j n. Then, obviously by linearity of the inner product, (v, u) = 0 for all v V. We re done. Exersize 2. For X, Y Banach spaces, let A be an operator in B(X, Y ) such that R(A) is closed and infinite-dimensional. Show that A is not compact. Solution 2. Well, since Y is a Banach space and R(A) is closed, R(A) is also a Banach space. Let Ỹ = R(A). Then A B(X, Ỹ ) and is onto. Now use solution 7 above. Exersize 22. Suppose X is a Banach space consisting of finite linear combinations of a denumerable set of elements. Show that dim X <.

Solution 22. We prove here the contrapositive. So, suppose that X is infinite dimensional. Suppose S = {x, x 2,..., x n,... } is a basis for X in the sense that for any N and any choice of i, i 2,..., i N, x i,..., x i are linearly independent and every element of X is a finite linear combination of elements of S. We may assume, without loss of generality, that x j =, for j N (simply normalize x j : x j = x j / x j, by abuse of notation). Now let s n = 2 j x j Then the sequence (s n ) is easily seen to be Cauchy (it is bounded above by partial sums of a geometric series), but does not converge, since 2 j x j cannot be written as a linear combination of finitely many elements of S, by linear independence of the elements of S. Hence X is not Banach. This proves the contrapositive of the original claim. Exersize 23. Show that every linear functional on a finite dimensional normed vector space is bounded. Solution 23. Let K = R or C. Let K on K n be the norm given by (α,..., α n ) = α j Suppose now that f is a linear functional on K n. Then we have f((α,..., α n )) = α f((, 0,..., 0)) + + α n f((0,..., 0, )) hence f((α,..., α n )) = α j f((δ j,..., δ nj )) max { f((δ j,..., δ nj )) } α j j n = max j n { f((δ j,..., δ nj )) } (α,..., α n ) so f is bounded. Now, suppose that X is an n-dimensional vector space over K. Then X is isomorphic, as a vector space, to K n. Let φ : V K n realize this isomorphism. Define a norm on V by v V = φ(v). Then it is easily seen that every functional g on V is bounded with respect to this norm, with g V = g φ, where g φ is viewed as a functional on K n. But all norms on V are equivalent, hence g is bounded with respect to any norm on V. Exersize 24. If M is a subspace of a Banach space X, then we define codimm = dim X/M If M is closed, show that codimm = dim M and codimm = dim M.

Solution 24. For convenience, set X = X/M. Suppose now that dim X = n <. Let [β ],..., [β n ] be a basis for X. Then, as is easily verified, β,..., β n are linearly independent in X. In fact, we may write X = Sp(β,..., β n ) M Let X and X denote the spaces of functionals on X and X, respectively. Consider given by φ : X X φ(f)(β j ) = f([β j ]) and extending linearly to all of X by setting φ(f)(x) = 0 for all x / Sp(β,..., β n ). Then clearly φ is a vector space homomorphism, and φ( X ) M. Since dim X = dim X = n, and N(φ) = 0, we conclude that dim φ( X ) = n, so dim M n. On the other hand, if g M, then g is completely determined by its action on β,..., β n. Let g be the functional on X given by g([β j ]) = g(β j ) Then, obviously, φ( g) = g. So in fact φ is a vector space isomorphism from X to M. Hence dim M = dim X = n. If dim X =, then for all n N there exists a subspace V n X of dimension n. Restricting φ to V n shows that dim M n. This shows that dim M =. One shows similarly that codimm = dim M. Exersize 25. If M, N are subspaces of a Banach space X and codimn < dim M, show that M N {0}. Solution 25. We prove the contrapositive. Suppose that M N = {0}. Assume first that dim M = m <. Let β,..., β m a basis for M. By assumption, β,..., β m / N, and [β ],..., [β m ] are linearly independent when viewed as elements of X/N. Hence Sp([β ],..., [β m ]) X/N is a subspace of X/N of dimension m. So codimn dim M. Suppose that dim M =. But then for any n N there exists a subspace V n of dimension n in M, for which the above argument can be carried out. Hence again codimn n for all n N, so codimn = dim M =. This proves the contrapositive. Exersize 26. If A B(X, Y ), R(A) is dense in Y and D is dense in X, show that A maps D onto a dense subset of Y. Solution 26. Why do we need the assumption R(A) is dense in X? We now that A is bounded, hence continuous from X to Y, when X and Y are viewed as topological spaces (the topologies are induced by the norms). But a continuous map maps dense sets to dense sets. To see this, suppose f : U V is continuous, with U, V topological spaces. Say D U is dense. Let v V and N an open neighborhood about v in V. By continuity of f, f (N) is open in U. By density of D, there exists d D such that d f (N). Hence f(d) f(d) and f(d) N. This shows that f(d) is dense in V. We re done. Exersize 27. Show that every infinite dimensional normed vector space has an unbounded linear functional.

Solution 27. Let S = {x, x 2,... x n,... } X be linearly independent. We can assume, without loss of generality, that x j = after normalizing each x j. Let V be the subspace of X consisting of all finite linear combinations of elements of S. Define f j on V by { jc if x = cxj f j (x) = 0 otherwise Let f be the functional on V defined by f = f is well-defined, since for each x V, f is a finite sum. Now, f is unbounded on V (it is unbounded on the unit ball in V ). To see this, let x = x + x 2 + + x n. Extend f to all of X by defining f(x) = 0 when x / V. Exersize 28. Let F, G be linear functionals on a vector space V, and assume that F (v) = 0 = G(v) = 0, v V. Show that there is a scalar C such that G(v) = CF (v), v V. Solution 28. From the hypothesis it is evident that N(F ) N(G). We claim that codimn(f ). So, suppose that codimn(f ) > 0 and suppose that V is a vectorspace over the field K. Then F : V K is a homomorphism of vector spaces which is surjective. But then f j F : V/N(F ) K is an isomorphism of vector spaces, where F is induced by F in a natural way (i.e., F ([v]) = F (v)). Hence dim V/N(F ) = dim K =. From this it follows that F is completely determined by its action on β / N(F ). If N(F ) = V, then there is nothing to prove. Otherwise, since N(F ) N(G) and, by the argument above, codimn(g). If N(G) = V, then again we re done (tae C = 0). If not, then there exists β / N(F), N(G). Hence both F and G are completely determined by their action on β. Since K is a field and F (β) 0, we may tae C = G(β)/F (β). Exersize 29. If {x } is a sequence of elements in a normed vector space X and {α } is a sequence of scalars, show that a necessary and sufficient condition for the existence of an x X satisfying is that holds for each n and scalars β,..., β n. x (x ) = α and x = M, β α M β x Solution 29. Suppose there exists such a functional x. Let β,..., β n be scalars, fixed. Let x = β x + + β n x n X. Applying x to x, we obtain x (x) = x β j x j = β j α j

Since x = M, we immediately get from above β α = x (x) M x = M β x Let now V be the vector subspace of X consisting of all finite linear combinations of {x }. Suppose that β α M β x holds for each n and scalars β,..., β n. Define x (x ) = α and extend linearly to V. Then extend x to all of X by defining x (y) = 0 if y / V. However, the condition that β α M β x holds for each n and scalars β,..., β n does not imply that x = M. Indeed, if K > M, then one has β α K β x holds for each n and scalars β,..., β n. Exersize 30. If a normed vector space X has a subspace M such that M and X/M are complete, show that X is a Banach space. Solution 30. Exersize 3. If X, Y are normed vector spaces and B(X, Y ) is complete, show that Y is complete. Solution 3. Let {y n } be a Cauchy sequence in Y. Fix nonzero x 0 in X. Define A n (cx 0 ) = cy n and for all x / Sp(x 0 ), let A(x) = 0. Clearly then A n B(X, Y ) and A n = y n / x 0 (the norms are taen in respective spaces, of course). Now, observe that (A n A m )(x 0 ) = A n (x 0 ) A m (x 0 ) y n y m / x 0 So {A n } is Cauchy in B(X, Y ), hence A n A B(X, Y ). But then A n (x 0 ) A(x 0 ) in Y. Hence y n A(x 0 ) in Y. So Y is complete, hence Banach. Exersize 32. Prove that if {T n } is a sequence in B(X, Y ) such that lim T n x exists for each x X, then there is a T B(X, Y ) such that T n x T x for all x X. Solution 32. Set T x = lim T n x. First observe that T is linear. Indeed, T (cx+y) = lim T n (cx+y) = c lim T n (x)+lim T n (y) = ct x+t y. Next observe that T is bounded. Suppose first that T n (x) T (x) uniformly for all x S, where S is the unit sphere in X. If this is the case, then there exists N N such that for all x of unit norm, T x T n x < for all n N, x S so T x < T N x +, x S

hence T x T N + for all x S So T is bounded. If convergence on S is not necessarily uniform, then T is not necessarily bounded. Consider the following example. Let S = {e, e 2,... }, where e j = (x j ) : N {0, } given by xj = δ j. We can visualize each e j as an -tuple, having in j th position and 0 everywhere else. Let V be the (infinite-dimensional) vector space over R formed by all finite linear combinations of elements of S. Endow V with the norm m c e j + c 2 e j2 + + c m e jm = c Define T j : V R by T j (e i ) = jδ ji, and extend T j linearly to V. Then each T j is a linear functional on V. Define T n = T Then each T n is obviously a linear functional on V. Let T = lim T n = T n n T is well-defined, since for each x V, T (v) is a finite sum (recall that v is a finite linear combination of elements of S). Further, each T n is bounded. T, however, is unbounded! Indeed, T (e + e 2 + + e n ) = T j (e j ) = n(n + ) j = 2 n= On the other hand, e + e 2 + + e n = n So T (e + e 2 + + e n ) e + e 2 + + e n = n + 2 which cannot be bounded by a constant. Exersize 33. Let F, G be linear functionals on a vector space V, and assume that F (v) = 0 = G(v) = 0, v V. Show that there is a scalar C such that G(v) = CF (v), v V. Solution 33. From the hypothesis it is evident that N(F ) N(G). We claim that codimn(f ). So, suppose that codimn(f ) > 0 and suppose that V is a vectorspace over the field K. Then F : V K is a homomorphism of vector spaces which is surjective. But then F : V/N(F ) K is an isomorphism of vector spaces, where F is induced by F in a natural way (i.e., F ([v]) = F (v)). Hence dim V/N(F ) = dim K =. From this it follows that F is completely determined by its action on β / N(F ). If N(F ) = V, then there is nothing to prove. Otherwise, since N(F ) N(G) and, by the argument above, codimn(g). If N(G) = V, then again we re done

(tae C = 0). If not, then there exists β / N(F), N(G). Hence both F and G are completely determined by their action on β. Since K is a field and F (β) 0, we may tae C = G(β)/F (β). Exersize 34. If {x } is a sequence of elements in a normed vector space X and {α } is a sequence of scalars, show that a necessary and sufficient condition for the existence of an x X satisfying is that holds for each n and scalars β,..., β n. x (x ) = α and x = M, β α M β x Solution 34. Suppose there exists such a functional x. Let β,..., β n be scalars, fixed. Let x = β x + + β n x n X. Applying x to x, we obtain x (x) = x β j x j = β j α j Since x = M, we immediately get from above β α = x (x) M x = M β x Let now V be the vector subspace of X consisting of all finite linear combinations of {x }. Suppose that β α M β x holds for each n and scalars β,..., β n. Define x (x ) = α and extend linearly to V. Then extend x to all of X by defining x (y) = 0 if y / V. However, the condition that β α M β x holds for each n and scalars β,..., β n does not imply that x = M. Indeed, if K > M, then one has β α K β x holds for each n and scalars β,..., β n. Exersize 35. If a normed vector space X has a subspace M such that M and X/M are complete, show that X is a Banach space. Solution 35. Exersize 36. If X, Y are normed vector spaces and B(X, Y ) is complete, show that Y is complete.

Solution 36. Let {y n } be a Cauchy sequence in Y. Fix nonzero x 0 in X. Define A n (cx 0 ) = cy n and for all x / Sp(x 0 ), let A(x) = 0. Clearly then A n B(X, Y ) and A n = y n / x 0 (the norms are taen in respective spaces, of course). Now, observe that (A n A m )(x 0 ) = A n (x 0 ) A m (x 0 ) y n y m / x 0 So {A n } is Cauchy in B(X, Y ), hence A n A B(X, Y ). But then A n (x 0 ) A(x 0 ) in Y. Hence y n A(x 0 ) in Y. So Y is complete, hence Banach. Exersize 37. Prove that if {T n } is a sequence in B(X, Y ) such that lim T n x exists for each x X, then there is a T B(X, Y ) such that T n x T x for all x X. Solution 37. Set T x = lim T n x. First observe that T is linear. Indeed, T (cx+y) = lim T n (cx+y) = c lim T n (x)+lim T n (y) = ct x+t y. Next observe that T is bounded. Suppose first that T n (x) T (x) uniformly for all x S, where S is the unit sphere in X. If this is the case, then there exists N N such that for all x of unit norm, T x T n x < for all n N, x S so T x < T N x +, x S hence T x T N + for all x S So T is bounded. If convergence on S is not necessarily uniform, then T is not necessarily bounded. Consider the following example. Let S = {e, e 2,... }, where e j = (x j ) : N {0, } given by xj = δ j. We can visualize each e j as an -tuple, having in j th position and 0 everywhere else. Let V be the (infinite-dimensional) vector space over R formed by all finite linear combinations of elements of S. Endow V with the norm m c e j + c 2 e j2 + + c m e jm = c Define T j : V R by T j (e i ) = jδ ji, and extend T j linearly to V. Then each T j is a linear functional on V. Define T n = T Then each T n is obviously a linear functional on V. Let T = lim T n = T n n T is well-defined, since for each x V, T (v) is a finite sum (recall that v is a finite linear combination of elements of S). Further, each T n is bounded. T, however, is unbounded! Indeed, T (e + e 2 + + e n ) = T j (e j ) = n(n + ) j = 2 On the other hand, n= e + e 2 + + e n = n

So T (e + e 2 + + e n ) e + e 2 + + e n = n + 2 which cannot be bounded by a constant.