Graduate Functional Analysis Problem Solutions w

7/26/2019 Graduate Functional Analysis Problem Solutions w

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Note: References refer to M. Schechter, Principles of Functional Analysis

Exersize 1. Let 1, . . . , n be an orthonormal set in a Hilbert space H. Show

that fn

k=1

kk

f

nk=1

(f, k)k

Solution 1. Since f H, we know that f V H, where V is some finitedimensional subspace ofH, say dim V =m n. We may assume thatV contains1, 2, . . . , n. Let{1, . . . , n, n+1, . . . , m} be an orthonormal basis for V (thisis possible by extending {1, . . . , n} to an orthonormal basis via Gram-Schmidt).

Now write f=m

k kk. Then we have

fnk

(f, k)k=m

k=n+1

kk

Hence fnk

(f, k)k

2

=m

k=n+1

2k

sincek= 1 for 1 k m. On the other hand,

fnk

kk =

nk

(k k)k+m

k=n+1

kk

Hence fnk

kk

2

=

nk

(k k)2 +

mk=n+1

2k

It is clear now that fnk

kk

2

fnk

(f, k)k

2

It follows that fnk

kk

f

nk

(f, k)k

Were done.

1

More solutions manual at www.DumbLittleDoctor.com

Thanks to the William's work!





Exercise Solutions to Functional Analysis


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Exersize 2. Letc denote the set of all elements (1, 2, . . . ) l such that {n}is a convergent sequence, and let c0 be the set of all such elements for which n 0as n . Show thatc and c

0are Banach spaces.

Solution 2. It is known thatl is a Banach space. The linear structure, as wellas the norm, ofc and c0 is inherited from that of l. To show completeness ofcand c0, it will suffice to show that c and c0 are closed in l. We show that c isclosed in l andc0 is closed in c.

Now, suppose that{xk}is a sequence in c converging to somex l. Then, forany given >0, we know that there exists k0 such that

supj

|xjk0 xj |0 an integer, we have

nx= x + x + + x n - times

Hence we must have F(nx) = nF(x).






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Now, for m > 0, F(x) = F(mx/m) = mF(x/m). So F(x/m) = F(x)/m. Nowit 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 iscontinuous at one point.

Solution 9. Suppose thatFis an additive function which is continuous at somex0. We shall show that Fis bounded. That is, there exists C >0 such that for allx we have

|F(x)| Cx

SinceF is continuous at x0, we know that, for arbitrary >0 fixed, there exists a >0 such that for all x x0 we have |F(x) F(x0)| . Now takey = 0.Ify is not rational, we can always pick y /2 < M < y rational sufficientlyclose to y. Set

x= x0+ 2M

y

Then it is clear that

x x0<

Hence we have (also since F is additive):

|F(x x0)|

By the previous problem, since /2Mis rational, we must have

|F(x x0)|=

F

2My

= 2M|F(y)|Hence

|F(y)|

2M 0 is fixed. Let N be so large that 1/N < /2. LetM1, M2, . . . , M N be such that for all k Mi, |y

i(k, i) yi| < . Let M =max {maxi {Mi} , N}. Then observe that for all k M, |z(k, j) y

j | < when






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j N, and |z(k, j) yj | |z(k, j)| + |yj | 1/N+ 1/N = 2/N < when j > N.Hence we have

supj |z(k, j) yj

|< for all k > M

So, as claimed, z(i, j) (yj) in l-norm. This proves sequential compactness ofI, and hence compactness.

Exersize 11. Let M be a totally bounded subset of a normed vector space X.Show that for each >0, Mhas a finite-netNM.

Solution 11. We proceed by contradiction. Suppose thatMis totally bounded,and for some fixed >0, there is no finite -netN contained entirely in M. Thengiven x1 M, there exists x2 M such that x2 x1> . Similarly there mustexist x3 M such that xi x3 > , with i = 1, 2. In general, we can continuethis process and obtainxn Msuch that xn xi> withi = 1, 2, . . . , n1. LetS= {x1, x2, . . . }. However,M is totally bounded, so there exists a finite/2-net of

M, say given by{w1, w2, . . . , wk} such that the ballsB(wi, /2), withi = 1, 2, . . . , kcover M. Since there are infinitely many distinct terms in the set S M, forsome i = j, xi, xj S and xi, xj B(wl, /2) for some 1 l k. But thenxi xj /2, which cannot be by construction ofS. This is a contradiction.

Exersize 12. Prove that if M is finite dimensional, then we can take = 1 inLemma 4.7

Solution 12. Let n = (n 1)/n 1/2. Now, obviouslyS p(x1, x2)=X. So thereexists x3 S such that x1 x3 , x2 x3> 1/2. Continue this process, at n

th

step selecting xn S such that xi xn > 1/2 with i = 1, 2, . . . , n 1. This ispossible sinceSp(x1, x2, . . . , xn1)=X(sinceXis infinite dimensional). Now, thesequence (xn)nN is bounded. Keep this in mind...

Now suppose that A = K I is compact. But then there exists a subsequence(xnk) such that Axnk converges; so in particular is Cauchy. Let, for convenience,yk = xnk . Thus, for < 1/2 there exists N N such that for all n m N wehave

Ayn Aym<

Hence

ym ynKym Kyn (ym yn) (Kym Kyn)= (Kyn yn) (Kym ym)<






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But thenym yn< + Kym yn

Now,Kis compact, so there exists a subsequencezk = ynksuch thatKzkconverges,so in particular is Cauchy. Hence forN large enough, + Kzm zn < 1/2 forall n m N. So

zm zn< + Kzm zn< 1/2

which cannot be, since zm zn > 1/2 by construction of the sequence (xn), ofwhich (zn) is a subsequence.

SoA cannot be compact. This completes the proof.

Exersize 14. LetXbe a vector space which can be made into a Banach space byintroducing either of two different norms. Suppose that convergence with respect tothe first norm always implies convergence with respect to the second norm. Showthat the norms are equivalent.

Solution 14. Let 1 and2 denote the two norms. It will suffice to show onlythat there exists C >0 such that for all x Xnonzero one has

x1 Cx2

for then a similar argument will establish the existence ofD >0 such that

x2 D x1

from which equivalence of the two norms could be deduced easily.Suppose, for a contradiction, that for all n N, there exists xn X nonzero

such thatxn1 > n xn2

But then we must havexn2

xn10. But then there exist z1, z2, N(K) such that

xn zn= [xn]X < C

for all n. Hence the sequence (xn zn) is bounded inX. Now, for all n,

K([xn]) = K(xn) = K(xn) 0 = K(xn) K(zn) = K(xn zn)

since zn N(K). But then, by compactness ofK, there is a subsequence K(xnk)

that converges (inY). Hence K([xnk ]) converges inY, which is what we wished toshow.

Now, observe that K is compact, onto Y and is 1-to-1. Hence K is bounded(hence closed), onto Yand is 1-to-1. But then there exists C >0 such that for allx X,

(*) xX CK(x)

Y

Were now ready to prove that Y is finite dimensional. Well proceed by contradic-tion.

Suppose that dim Y =. But then there exists a sequence (yn) in Y such thatynY = 1 for all n, and for all i = j, one has yn ymY > 1/2. See solution4 above for construction of such a sequence. Clearly then (yn) has no convergent

subsequence in Y. Since K is 1-to-1, let xn = K1(yn). But then by (*) above,

(xn) is a bounded sequence in X, and, as noted above, its image, namely (yn), doesnot have a convergent subsequence. This contradicts compactness ofK.

Exersize 17. Show that ifXis an infinite dimensional normed vector space, thenthere is a sequence {xn}such that xn= 1, xn xm 1 for m =n.

Solution 17. For convenience,Sand Spwill be used as in solution 4. Throughout,all our vectors are taken to be nonzero.

Now, let x1 X. But then there existsy1 X such that dim Sp(x1, y1) = 2.But then, according to solution 3, there exists x2 Sp(x1, y1) such thatx2 Sandx1 x2 1. Proceed inductively. At then

th step, sinceXis infinite dimensional,there exists yn X such that dim Sp(x1, x2, . . . , xn, yn) = n+ 1. So there existsxn+1 S such that xi xn+1 1 for 1 i n. Thus (xn) is the sequence weneed.

Exersize 18. Show that every finite-dimensional vector space can be made into astrictly convex normed vector space.

Solution 18. Observe that both, Rn and Cn are strictly convex vector spaces (un-der the standard Euclidean norm). This follows easily from the Cauchy-Schwartzinequality, since both are actually Hilbert spaces. Now, if dim V = n, where Vis a vector space over K = R,C, then V is isomorphic, as a vector space, to Kn.Suppose that f : Kn V realizes this isomorphism (for instance, if{1, . . . , n}






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is a basis for V, we could take f(ej) = j), where {e1, . . . , en} is the standardbasis for Kn, and extend f linearly to all ofKn). Now define a norm on V viav = E(f1(v)), where E is the standard Euclidean norm on Kn. Then V isstrictly convex under .

Exersize 19. Let V be a vector space having n linearly independent elementsv1, . . . , vnsuch that every elementv V can be expressed in the form (4.12). Showthat dim V =n.

Solution 19. Obviously dim V n, since V is spanned by n vectors. Sayu1, . . . , um also span V. Then we can write

v1 = c11u1+ c

21u2+ + c

m1 um

......

...vn = c

1nu1+ c

2nu2+ + c

mnum

We also know that each uj is a linear combination of v1, . . . , vn. Hence we may

writev1 = c

11

nj=1 d

j1vj+ c

21

nj=1 d

j2vj+ + c

m1

nj=1 d

jmvj

......

...

vn = c1n

nj=1 d

j1vj+ c

2n

nj=1 d

j2vj+ + c

mn

nj=1 d

jmvj

Or, equivalently,

0 = (c11d11 1)v1+

nj=2 c

11d

j1vj+

nj=1 c

21d

j2vj+ +

nj=1 c

m1 d

jmvj

......

...

0 =n

j=1 c1nd

j1vj+

nj=1 c

2nd

j2vj+ +

n1j=1 c

mnd

jmvj+ (c

mnd

nm 1)vn

Ifm < n, then the system above clearly has at least one nontrivial solution in cjidlk,

contradicting linear independence ofv1, . . . , vn. 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 < anddim V


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Solution 22. We prove here the contrapositive. So, suppose that X is infinitedimensional. SupposeS= {x1, x2, . . . , xn, . . . }is a basis for Xin the sense that foranyk Nand any choice ofi

1, i

2, . . . , i

k N, x

i1

, . . . , xik

are linearly independentand every element ofX is a finite linear combination of elements ofS. We mayassume, without loss of generality, that xj= 1, for j N (simply normalize xj :xj =xj/ xj, by abuse of notation). Now let

sn=nj=1

1

2jxj

Then the sequence (sn) is easily seen to be Cauchy (it is bounded above by partialsums of a geometric series), but does not converge, since

j=1

1

2jxj

cannot be written as a linear combination of finitely many elements ofS, by linearindependence of the elements of S. Hence X is not Banach. This proves thecontrapositive of the original claim.

Exersize 23. Show that every linear functional on a finite dimensional normedvector space is bounded.

Solution 23. Let K= Ror C. LetK onKn be the norm given by

(1, . . . , n)=n

j=1

|j |

Suppose now thatfis a linear functional on Kn. Then we have

f((1, . . . , n)) = 1f((1, 0, . . . , 0)) + + nf((0, . . . , 0, 1))

hence

|f((1, . . . , n))|=

nj=1

jf((1j, . . . , nj))

max1jn {f((1j , . . . , nj))}n

j=1

|j |

= max1jn

{f((1j , . . . , nj))} (1, . . . , n)

so fis bounded.Now, suppose that X is an n-dimensional vector space over K. Then X is

isomorphic, as a vector space, to Kn. Let : V Kn realize this isomorphism.Define a norm onV byvV =(v). Then it is easily seen that every functional

gon V is bounded with respect to this norm, withgV = g 1, whereg 1is viewed as a functional on Kn. But all norms on V are equivalent, hence g isbounded with respect to any norm onV .

Exersize 24. IfMis a subspace of a Banach space X, then we define

codimM= dim X/M

IfMis closed, show that codimM= dim M and codimM = dim M.






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Solution 24. For convenience, set X= X/M. Suppose now that dim X=n


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Solution 27. Let S= {x1, x2, . . . xn, . . . } Xbe linearly independent. We canassume, without loss of generality, that xj= 1 after normalizing eachxj . LetVbe the subspace ofXconsisting of all finite linear combinations of elements ofS.Definefj onV by

fj(x) =

jc if x= cxj0 otherwise

Letfbe the functional onV defined by

f=

j=1

fj

fis well-defined, since for each x V, f is a finite sum. Now, fis unbounded onV(it is unbounded on the unit ball in V). To see this, let x = x1+ x2+ + xn.Extendfto all ofXby definingf(x) = 0 when x /V.

Exersize 28. Let F, G be linear functionals on a vector space V, and assumethat F(v) = 0 = G(v) = 0, v V. Show that there is a scalarC such that

G(v) = C F(v), v V.

Solution 28. From the hypothesis it is evident that N(F) N(G). We claimthat codimN(F) 1. So, suppose that codimN(F)> 0 and suppose that V is avectorspace over the fieldK. ThenF :V Kis a homomorphism of vector spaceswhich is surjective. But then

F :V /N(F) K

is an isomorphism of vector spaces, where F is induced byFin a natural way (i.e.,F([v]) = F(v)). Hence dim V/N(F) = dim K= 1.

From this it follows that F is completely determined by its action on /N(F).

IfN(F) = V, then there is nothing to prove. Otherwise, since N(F) N(G) and,by the argument above, codimN(G) 1. If N(G) = V, then again were done(take C= 0). If not, then there exists /N(F), N(G). Hence both F and G arecompletely determined by their action on . Since K is a field and F()= 0, wemay take C= G()/F().

Exersize 29. If{xk} is a sequence of elements in a normed vector space X and{k}is a sequence of scalars, show that a necessary and sufficient condition for theexistence of anx X satisfying

x(xk) = k and x= M,

is that n

1kk

M

n

1kxk

holds for each n and scalars 1, . . . , n.

Solution 29. Suppose there exists such a functionalx. Let1, . . . , n be scalars,fixed. Let x = 1x1+ + nxn X. Applyingx

tox, we obtain

x(x) = x

n

j=1

jxj

= n

j=1

jj






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Sincex= M, we immediately get from above

n

j=1

kk =|x(x)| Mx= Mn

j=1

kxkLet nowVbe the vector subspace ofXconsisting of all finite linear combinations

of{xk}. Suppose that n1

kk

M

n1

kxk


Define x(xk) = k and extend linearly to V. Then extend x to all ofX by

definingx(y) = 0 ify /V.However, the condition that

n

1 kk Mn

1 kxkholds for each n and scalars 1, . . . , n does not imply that x

= M. Indeed, ifK > M, then one has

n1

kk

K

n1

kxk


Exersize 30. If a normed vector space X has a subspace M such that M andX/Mare complete, show that Xis a Banach space.

Solution 30.

Exersize 31. IfX, Y are normed vector spaces and B(X, Y) is complete, show

thatY is complete.

Solution 31. Let{yn} be a Cauchy sequence in Y . Fix nonzero x0 in X. DefineAn(cx0) = cyn and for all x / S p(x0), let A(x) = 0. Clearly thenAn B (X, Y)andAn= yn / x0(the norms are taken in respective spaces, of course). Now,observe that

(An Am)(x0)= An(x0) Am(x0) yn ym / x0

So {An} is Cauchy in B(X, Y), hence An A B(X, Y). But thenAn(x0) A(x0) in Y. Hence yn A(x0) in Y . SoY is complete, hence Banach.

Exersize 32. Prove that if{Tn}is a sequence inB(X, Y) such that lim Tnxexistsfor each x X, then there is aTB(X, Y) such that Tnx T x for all x X.

Solution 32. SetT x= lim Tnx. First observe thatTis linear. Indeed,T(cx+y) =lim Tn(cx+y) = c lim Tn(x)+lim Tn(y) = cT x+T y. Next observe that Tis bounded.

Suppose first that Tn(x) T(x) uniformly for all x S, where S is the unitsphere inX. If this is the case, then there exists N Nsuch that for all x of unitnorm,

T x Tnx< 1 for all n N, x S

so

T x< TNx + 1, x S






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henceT x TN + 1 for all x S

SoTis bounded.If convergence onSis not necessarily uniform, then Tis not necessarily bounded.Consider the following example.

Let S= {e1, e2, . . . }, where ej = (xjk) : N {0, 1} given by x

jk = kj . We can

visualize each ej as an-tuple, having 1 in jth position and 0 everywhere else.

LetVbe the (infinite-dimensional) vector space over R formed by allfinitelinearcombinations of elements ofS. Endow Vwith the norm

c1ej1 + c2ej2 + + cmejm=m1

|ck|

Define Tj :V R by Tj(ei) = jji, and extend Tj linearly to V . Then each Tjis a linear functional on V . Define

Tn=

n

1

Tk

Then eachTn is obviously a linear functional on V . Let

T = limn

Tn=n=1

Tn

T is well-defined, since for eachx V, T(v) is a finite sum (recall thatv is a finitelinear combination of elements ofS). Further, each Tn is bounded. T, however, isunbounded! Indeed,

|T(e1+ e2+ + en)|=

nj=1

Tj(ej)

=n

j=1

j =n(n + 1)

2

On the other hand,e1+ e2+ + en= n

So|T(e1+ e2+ + en)|

e1+ e2+ + en =

n + 1

2

which cannot be bounded by a constant.

Exersize 33. Let F, G be linear functionals on a vector space V, and assumethat F(v) = 0 = G(v) = 0, v V. Show that there is a scalarC such thatG(v) = C F(v), v V.

Solution 33. From the hypothesis it is evident that N(F) N(G). We claimthat codimN(F) 1. So, suppose that codimN(F)> 0 and suppose that V is avectorspace over the fieldK. ThenF :V Kis 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 byFin a natural way (i.e.,F([v]) = F(v)). Hence dim V/N(F) = dim K= 1.

From this it follows that F is completely determined by its action on /N(F).IfN(F) = V, then there is nothing to prove. Otherwise, since N(F) N(G) and,by the argument above, codimN(G) 1. If N(G) = V, then again were done






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(take C= 0). If not, then there exists /N(F), N(G). Hence both F and G arecompletely determined by their action on . Since K is a field and F()= 0, wemay take C= G()/F().

Exersize 34. If{xk} is a sequence of elements in a normed vector space X and{k}is a sequence of scalars, show that a necessary and sufficient condition for theexistence of anx X satisfying

x(xk) = k and x= M,

is that n1

kk

M

n1

kxk


Solution 34. Suppose there exists such a functionalx. Let1, . . . , n be scalars,fixed. Let x = 1x1+ + nxn X. Applyingx tox, we obtain

x(x) = x

n

j=1

jxj

= n

j=1

jj

Sincex= M, we immediately get from aboven

j=1

kk

=|x(x)| Mx= M

nj=1

kxk

Let nowVbe the vector subspace ofXconsisting of all finite linear combinations

of{xk}. Suppose that n

1kk

M

n

1kxk

holds for each n and scalars 1, . . . , n.Define x(xk) = k and extend linearly to V. Then extend x

to all ofX bydefiningx(y) = 0 ify /V.

However, the condition thatn1

kk

M

n1

kxk

holds for each n and scalars 1, . . . , n does not imply that x = M. Indeed, ifK > M, then one has

n1

kk

K

n1

kxk


Exersize 35. If a normed vector space X has a subspace M such that M andX/Mare complete, show that Xis a Banach space.

Solution 35.

Exersize 36. IfX, Y are normed vector spaces and B(X, Y) is complete, showthatY is complete.






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Solution 36. Let{yn} be a Cauchy sequence in Y . Fix nonzero x0 in X. DefineAn(cx0) = cyn and for all x / S p(x0), let A(x) = 0. Clearly thenAn B (X, Y)andA

n= y

n / x

0(the norms are taken in respective spaces, of course). Now,

observe that

(An Am)(x0)= An(x0) Am(x0) yn ym / x0

So {An} is Cauchy in B(X, Y), hence An A B(X, Y). But thenAn(x0) A(x0) in Y. Hence yn A(x0) in Y . SoY is complete, hence Banach.

Exersize 37. Prove that if{Tn}is a sequence inB(X, Y) such that lim Tnxexistsfor each x X, then there is aTB(X, Y) such that Tnx T x for all x X.

Solution 37. SetT x= lim Tnx. First observe thatTis linear. Indeed,T(cx+y) =lim Tn(cx+y) = c lim Tn(x)+lim Tn(y) = cT x+T y. Next observe that Tis bounded.

Suppose first that Tn(x) T(x) uniformly for all x S, where S is the unitsphere inX. If this is the case, then there exists N Nsuch that for all x of unitnorm,

T x Tnx< 1 for all n N, x S

soT x< TNx + 1, x S

henceT x TN + 1 for all x S

SoTis bounded.If convergence onSis not necessarily uniform, then Tis not necessarily bounded.

Consider the following example.Let S= {e1, e2, . . . }, where ej = (x

jk) : N {0, 1} given by x

jk = kj . We can

visualize each ej as an-tuple, having 1 in jth position and 0 everywhere else.

LetVbe the (infinite-dimensional) vector space over R formed by allfinitelinearcombinations of elements ofS. Endow Vwith the norm

c1ej1 + c2ej2 + + cmejm=m1

|ck|

Define Tj :V R by Tj(ei) = jji, and extend Tj linearly to V . Then each Tjis a linear functional on V . Define

Tn=

n1

Tk

Then eachTn is obviously a linear functional on V . Let

T = limn

Tn=

n=1

Tn

T is well-defined, since for eachx V, T(v) is a finite sum (recall thatv is a finitelinear combination of elements ofS). Further, each Tn is bounded. T, however, isunbounded! Indeed,

|T(e1+ e2+ + en)|=

nj=1

Tj(ej)

=n

j=1

j =n(n + 1)

2

On the other hand,e1+ e2+ + en= n






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So|T(e1+ e2+ + en)|

e1+ e2+ + en =

n + 1

2

which cannot be bounded by a constant.





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Graduate Functional Analysis Problem Solutions w