solutions to graduate functional analysis - Little Dumb · PDF fileShow that an additive...

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

Exersize 1. Let φ1, . . . , φn be an orthonormal set in a Hilbert space H. Showthat ∥∥∥∥∥f −

n∑k=1

αkφk

∥∥∥∥∥ ≥∥∥∥∥∥f −

n∑k=1

(f, φk)φk

∥∥∥∥∥Solution 1. Since f ∈ H, we know that f ∈ V ⊂ H, where V is some finitedimensional subspace of H, say dimV = m ≥ n. We may assume that V containsφ1, φ2, . . . , φn. Let {φ1, . . . , φn, φn+1, . . . , φm} be an orthonormal basis for V (thisis possible by extending {φ1, . . . , φn} to an orthonormal basis via Gram-Schmidt).

Now write f =∑mk βkφk. Then we have

f −n∑k

(f, φk)φk =

m∑k=n+1

βkφk

Hence ∥∥∥∥∥f −n∑k

(f, φk)φk

∥∥∥∥∥2

=

m∑k=n+1

β2k

since ‖φk‖ = 1 for 1 ≤ k ≤ m. On the other hand,

f −n∑k

αkφk =

n∑k

(βk − αk)φk +

m∑k=n+1

βkφk

Hence ∥∥∥∥∥f −n∑k

αkφk

∥∥∥∥∥2

=

n∑k

(βk − αk)2 +

m∑k=n+1

β2k

It is clear now that ∥∥∥∥∥f −n∑k

αkφk

∥∥∥∥∥2

≥

∥∥∥∥∥f −n∑k

(f, φk)φk

∥∥∥∥∥2

It follows that ∥∥∥∥∥f −n∑k

αkφk

∥∥∥∥∥ ≥∥∥∥∥∥f −

n∑k

(f, φk)φk

∥∥∥∥∥We’re done. �

1

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Exercise Solutions to Functional Analysis

Exersize 2. Let c denote the set of all elements (α1, α2, . . . ) ∈ l∞ such that {αn}is a convergent sequence, and let c0 be the set of all such elements for which αn → 0as n→∞. Show that c and c0 are Banach spaces.

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

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

supj|xjk0 − x

j | < ε

3

Hence for all j we have

|xjk0 − xj | < ε

3

We also know that, since{xjk0

}j∈N

is a convergent sequence, it is Cauchy, so there

exists N ∈ N such that for all n ≥ m ≥ N , we have

|xnk0 − xmk0 | <

ε

3

From this we deduce:

|xn − xm| ≤ |xn − xnk0 |+ |xnk0 − x

mk0 |+ |x

mk0 − x

m| < ε

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 closedsubset of a complete space, must itself be complete.

Similarly, let us suppose that {xk} is a sequence in c0 that converges to x ∈ l∞.Then for all ε > 0, there exists k0 such that

supj|xjk0 − x

j | < ε

2

Hence for all j we have

|xjk0 − xj | < ε

2On the other hand, since xk0 ∈ c0, there exists N ∈ N such that for all n ≥ N , wehave

|xnk0 | <ε

2Hence for all n ≥ N we have

|xn| ≤ |xnk0 |+ |xnk0 − x

n| < ε

So x ={xj}

converges to 0, hence x ∈ c0. So c0 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‖+ | − 1| ‖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.

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Solution 4. Well, if either x = 0 or y = 0, the result holds vacuously. So assumethat neither x nor y is equal to zero. We may assume that x, y ∈ V ⊂ H, withdimV = m < ∞. If m = 1, we’re done. If not, we may take for a basis of V theset {x1, x2, . . . , xm} of orthonormal vectors where x1 = x/ ‖x1‖. This can be donevia Gram-Schmidt. Write y = α1x1 + · · ·+ αmxm. Then we have

‖x1 + (α1x1 + · · ·+ αmxm)‖ = ‖x1‖+ ‖α1x1 + · · ·+ αmxm‖From this we immediately deduce

2|(x1, α1x1 + · · ·+ αmxm)| = 2 ‖x1‖ ‖α1x1 + · · ·+ αmxm‖By orthonormality of {x1, . . . , xm}, we get from above:

|α1| =√α21 + · · ·+ α2

m

From this we immediately deduce that

α2 = · · · = αm = 0

Hence we must have

y = α1x1 =α1

‖x‖x

We’re done. �

Exersize 5. Show that in a normed vector space∞∑j=1

vj =

n∑j=1

vj +

∞∑j=n+1

vj

Solution 5. In a normed vector space the notion of the limit makes sense, and infact all standard results regarding algebra of limits hold. In particular we have, bydefinition (for n fixed)

∞∑j=1

vj = limk→∞

k∑j=1

vj = limk→∞,k≥n

k∑j=1

vj

On the other hand for k ≥ n we have

k∑j=1

vj =n∑j=1

vj +k∑

j=n+1

vj

Hence

∞∑j=1

vj = limk→∞

n∑j=1

vj +

k∑j=n+1

vj

=

n∑j=1

vj + limk→∞

k∑j=n+1

vj =

n∑j=1

vj +

∞∑j=n+1

vj

�

Exersize 6. For f = (α1, . . . , αn) ∈ Rn, show that

‖f‖0 = maxk|αk|

and

‖f‖1 =

n∑k=1

|αk|

are norms. Is Rn complete with respect to either of them?


Solution 6. It is easy to see that (11) and (12) of the definition of a norm on p.7 hold trivially in both cases. So we only verify the triangle inequality.

Write f = (α1, . . . , αn) and g = (β1, . . . , βn). Then we have

‖f + g‖0 = maxk|αk + βk| ≤ max

k(|αk|+ |βk|)

which follows from the triangle inequality of | · | on R. On the other hand, it isn’thard to see that

maxk

(|αk|+ |βk|) ≤ maxk|αk|+ max

k|βk| = ‖f‖0 + ‖g‖0

In the second case,

‖f + g‖1 =

n∑k=1

|αk + βk| ≤n∑k=1

(|αk|+ |βk|) = ‖f‖1 + ‖g‖1

Now, we know that Rn is complete with respect to the standard Euclidean norm.But all norms on a finite dimensional vector space are equivalent. So Rn is completewith respect to the two norms above.

More directly, it isn’t hard to see that:

‖f‖0 ≤√α21 + · · ·+ α2

n

and

‖f‖1 ≤√α21 + · · ·+ α2

n

�

Exersize 7. If F is a bounded linear functional on a normed vector space X, showthat

‖F‖ = sup‖x‖≤1

|F (x)| = sup‖x‖=1

|F (x)|

Solution 7. By definition we have

‖F‖ = supx 6=0

|F (x)|‖x‖

But

supx6=0

|F (x)|‖x‖

= supx 6=0

∣∣∣∣F ( x

‖x‖

)∣∣∣∣ = sup‖x‖=1

|F (x)|

On the other hand,

sup‖x‖=1

|F (x)| = supx 6=0

|F (x)|‖x‖

≥ sup‖x‖≤1

|F (x)|‖x‖

≥ sup‖x‖≤1

|F (x)| ≥ sup‖x‖=1

|F (x)|

This verifies the claim. �

Exersize 8. A functional F (x) is called additiveif F (x+ y) = F (x) + F (y). If Fis 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

nx = x+ x+ · · ·+ x︸︷︷︸n - times

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


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

|F (x)| ≤ C ‖x‖Since F is continuous at x0, we know that, for arbitrary ε > 0 fixed, there exists aδ > 0 such that for all ‖x− x0‖ ≤ δ we have |F (x)− F (x0)| ≤ ε. Now take y 6= 0.If ‖y‖ is not rational, we can always pick ‖y‖ /2 < M < ‖y‖ rational sufficientlyclose to ‖y‖. Set

x = x0 +δ

2My

Then it is clear that

‖x− x0‖ < δ

Hence we have (also since F is additive):

|F (x− x0)| ≤ ε

By the previous problem, since δ/2M is rational, we must have

ε ≥ |F (x− x0)| =∣∣∣∣F ( δ

2My

)∣∣∣∣ =δ

2M|F (y)|

Hence

|F (y)| ≤ ε

δ2M <

2ε

δ‖y‖

This construction works for all y, with 2εδ being independent of y. Hence F is

bounded, thus continuous everywhere. �

Exersize 10. Show that the set Iω is compact.

Solution 10. Since l∞ is a metric space, we know that a subset of a metric space iscompact if and only if it is sequentially compact. So we show that Iω is sequentiallycompact: 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)| ≤ 1/j for all i, j. Clearlyfor each fixed j, x(i, j) has a subsequence x(ik, j) that converges. Now let y1(i, j)be a subsequence of x(i, j) such that y1(i, 1) → y1. Let y2(i, j) be a subsequenceof y1(i, j), such that y2(i, 2) → y2. In general, let yn(i, j) be a subsequence ofyn−1(i, j), such that yn(i, n) → yn. Now let z(i, j) = yi(i, j). We claim thatz(i, j), (y1, y2, . . . ) ∈ Iω and that z(i, j)→ (y1, y2, . . . ) as i→∞.

Clearly, by construction, for each fixed i, |z(i, j)| ≤ 1/j, since it is a germ ofthe sequence x(i, j)j∈N. Since yj is the limit of a subsequence of x(i, j)i∈N, andx(i, j) ≤ 1/j for all i, we must have also |yj | ≤ 1/j. Hence (yj)j∈N ∈ Iω.

Now suppose that ε > 0 is fixed. Let N be so large that 1/N < ε/2. LetM1,M2, . . . ,MN be such that for all k ≥ Mi, |yi(k, i) − yi| < ε. Let M =max {maxi {Mi} , N}. Then observe that for all k ≥ M , |z(k, j) − yj | < ε when


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

supj|z(k, j)− yj | < ε for all k > M

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

Exersize 11. Let M be a totally bounded subset of a normed vector space X.Show that for each ε > 0, M has a finite ε-net N ⊂M .

Solution 11. We proceed by contradiction. Suppose that M is totally bounded,and for some fixed ε > 0, there is no finite ε-net N contained entirely in M . Thengiven x1 ∈ M , there exists x2 ∈ M such that ‖x2 − x1‖ > ε. Similarly there mustexist x3 ∈ M such that ‖xi − x3‖ > ε, with i = 1, 2. In general, we can continuethis process and obtain xn ∈M such that ‖xn − xi‖ > ε with i = 1, 2, . . . , n−1. LetS = {x1, x2, . . . }. However, M is totally bounded, so there exists a finite ε/2-net ofM , say given by {w1, w2, . . . , wk} such that the balls B(wi, ε/2), with i = 1, 2, . . . , kcover M . Since there are infinitely many distinct terms in the set S ⊂ M , forsome i 6= j, xi, xj ∈ S and xi, xj ∈ B(wl, ε/2) for some 1 ≤ l ≤ k. But then‖xi − xj‖ ≤ ε/2, which cannot be by construction of S. This is a contradiction. �

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

Solution 12. Let θn = (n− 1)/n < 1 for n ≥ 2. We know then that there existsxn ∈M c, such that d(xn,M) ≥ θn and ‖xn‖ = 1 (in some fixed norm on X: X hasmany admissible norms, but all are equivalent). Since X is finite-dimensional and(xn) is a bounded sequence, we know that xn has a convergent subsequence, sayxnk→ x ∈ X. But then ‖x‖ = 1 (this is obvious from the continuity of the norm,

and the fact that ‖xn‖ = 1 for all n) and d(x,M) ≥ (nk − 1)/nk for all nk. Henced(x,M) ≥ 1. We’re done. �

Exersize 13. Show that if X is infinite dimensional and K is one-to-one operatorin K(X), then K − I cannot be in K(X).

Solution 13. Let S ⊂ X be the unit sphere (that is, the set of all vectors ofunit norm). For convenience, let Sp(x1, x2, . . . , xk) denote the space spanned bythe vectors xi, i = 1, 2, . . . , k. Now, pick x1 ∈ S. Clearly Sp(x1) 6= X, so thereexists x2 ∈ S such that ‖x1 − x2‖ > 1/2. Now, obviously Sp(x1, x2) 6= X. So thereexists x3 ∈ S such that ‖x1 − x3‖ , ‖x2 − x3‖ > 1/2. Continue this process, at nth

step selecting xn ∈ S such that ‖xi − xn‖ > 1/2 with i = 1, 2, . . . , n − 1. This ispossible since Sp(x1, x2, . . . , xn−1) 6= X (since X is infinite dimensional). Now, thesequence (xn)n∈N is bounded. Keep this in mind...

Now suppose that A = K − I is compact. But then there exists a subsequence(xnk

) such that Axnkconverges; so in particular is Cauchy. Let, for convenience,

yk = xnk. Thus, for ε < 1/2 there exists N ∈ N such that for all n ≥ m ≥ N we

have

‖Ayn −Aym‖ < ε

Hence

‖ym − yn‖−‖Kym −Kyn‖ ≤ ‖(ym − yn)− (Kym −Kyn)‖ = ‖(Kyn − yn)− (Kym − ym)‖ < ε


But then

‖ym − yn‖ < ε+ ‖Kym − yn‖Now, K is compact, so there exists a subsequence zk = ynk

such thatKzk converges,so in particular is Cauchy. Hence for N large enough, ε + ‖Kzm − zn‖ < 1/2 forall n ≥ m ≥ N . So

‖zm − zn‖ < ε+ ‖Kzm − zn‖ < 1/2

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

So A cannot be compact. This completes the proof. �

Exersize 14. Let X be 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 and ‖·‖2 denote the two norms. It will suffice to show onlythat there exists C > 0 such that for all x ∈ X nonzero one has

‖x‖1 ≤ C ‖x‖2for then a similar argument will establish the existence of D > 0 such that

‖x‖2 ≤ D ‖x‖1from which equivalence of the two norms could be deduced easily.

Suppose, for a contradiction, that for all n ∈ N , there exists xn ∈ X nonzerosuch that

‖xn‖1 > n ‖xn‖2But then we must have

‖xn‖2‖xn‖1

<1

n

Hence xn/ ‖xn‖1 → 0 in the ‖·‖2-norm. But then, by hypothesis, xn/ ‖xn‖1 → 0in the ‖·‖1-norm. This cannot be, since ‖xn/ ‖xn‖1‖1 = 1 for all n. This is acontradiction. We’re done. �

Exersize 15. 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 15. Since bothA andK are linear operators, we know thatR(A), R(K) ⊂Y are vector subspaces of Y . Now, we also know 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 seethis: let (xn) be a bounded sequence in X. Since A is bounded, A(xn) is a boundedsequence in R(A). Since R(A) is finite-dimensional, A(xn) must have a convergentsubsequence. We’re done. �

Exersize 16. Suppose X,Y are Banach spaces and K ∈ K(X,Y ). If R(K) = Y ,show that Y is finite dimensional.

Solution 16. Let X = X/N(K). Now, N(K) is certainly closed in X. Define thenorm on X/N

‖·‖X : X → R by ‖x‖X = infz∈N(K)

‖z − x‖


where ‖·‖ is the norm on X. It isn’t hard to verify that the norm above is well-defined, and actually is a norm.

Let K : X → Y be the homomorphism (of vector spaces) induced by the ho-

momorphism K - that is, K([x]) = K(x). Then this induced homomorphism isactually a vector space isomorphism (fundamental theorem of vector space homo-

morphisms - 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 [x1], [x2], . . . is a sequence in Xthat is bounded, say by C > 0. 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 in X. 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 of K, there is a subsequence K(xnk)

that converges (in Y ). Hence K([xnk]) converges in Y , which is what we wished to

show.Now, observe that K is compact, onto Y and is 1-to-1. Hence K is bounded

(hence closed), onto Y and is 1-to-1. 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 contradic-tion.

Suppose that dimY =∞. But then there exists a sequence (yn) in Y such that‖yn‖Y = 1 for all n, and for all i 6= j, one has ‖yn − ym‖Y > 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 = K−1(yn). But then by (*) above,

(xn) is a bounded sequence in X, and, as noted above, its image, namely (yn), does

not have a convergent subsequence. This contradicts compactness of K. �

Exersize 17. Show that if X is an infinite dimensional normed vector space, thenthere is a sequence {xn} such that ‖xn‖ = 1, ‖xn − xm‖ ≥ 1 for m 6= n.

Solution 17. For convenience, S and Sp will be used as in solution 4. Throughout,all our vectors are taken to be nonzero.

Now, let x1 ∈ X. But then there exists y1 ∈ X such that dimSp(x1, y1) = 2.But then, according to solution 3, there exists x2 ∈ Sp(x1, y1) such that x2 ∈ S and‖x1 − x2‖ ≥ 1. Proceed inductively. At the nth step, since X is infinite dimensional,there exists yn ∈ X such that dimSp(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 dimV = 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 {β1, . . . , βn}


is a basis for V , we could take f(ej) = βj), where {e1, . . . , en} is the standardbasis for Kn, and extend f linearly to all of Kn). Now define a norm on V via‖v‖ = E(f−1(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, . . . , vn such that every element v ∈ V can be expressed in the form (4.12). Showthat dimV = n.

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

v1 = c11u1 + c21u2 + · · ·+ cm1 um...

......

vn = c1nu1 + c2nu2 + · · ·+ cmn um

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

v1 = c11∑nj=1 d

j1vj + c21

∑nj=1 d

j2vj + · · ·+ cm1

∑nj=1 d

jmvj

......

...

vn = c1n∑nj=1 d

j1vj + c2n

∑nj=1 d

j2vj + · · ·+ cmn

∑nj=1 d

jmvj

Or, equivalently,

0 = (c11d11 − 1)v1 +

∑nj=2 c

11dj1vj +

∑nj=1 c

21dj2vj + · · ·+

∑nj=1 c

m1 d

jmvj

......

...

0 =∑nj=1 c

1nd

j1vj +

∑nj=1 c

2nd

j2vj + · · ·+

∑n−1j=1 c

mn d

jmvj + (cmn d

nm − 1)vn

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

contradicting linear independence of v1, . . . , vn. Hence m ≥ n. Thus dimV ≥ n.We conclude that dimV = n. �

Exersize 20. Let V,W be subspaces of a Hilbert space. If dimV < ∞ anddimV < dimW , show that there is a u ∈ W such that ‖u‖ = 1 and (u, v) = 0 forall v ∈ V .

Solution 20. Suppose that u1, . . . , un is a basis for V . We may assume, with-out loss of generality, that {u1, . . . , un} is an orthonormal basis (perform Gramm-Schmidt). Now, obviously V 6= W , so there exists w ∈W such that w /∈ V . Hencethe set {u1, . . . , un, w} is linearly independent. Perform Gramm-Schidt orthonor-malization process on w to obtain the set {u1, u2, . . . , un, u} of orthonormal vectors.Then u /∈ V , ‖u‖ = 1 and (u, uj) = 0 for 1 ≤ j ≤ n. Then, obviously by linearityof the inner product, (v, u) = 0 for all v ∈ V . We’re done. �

Exersize 21. For X,Y Banach spaces, let A be an operator in B(X,Y ) such thatR(A) is closed and infinite-dimensional. Show that A is not compact.

Solution 21. Well, since Y is a Banach space and R(A) is closed, R(A) is also a

Banach space. Let Y = R(A). Then A ∈ B(X, Y ) and is onto. Now use solution 7above. �

Exersize 22. Suppose X is a Banach space consisting of finite linear combinationsof a denumerable set of elements. Show that dimX <∞.


Solution 22. We prove here the contrapositive. So, suppose that X is infinitedimensional. Suppose S = {x1, x2, . . . , xn, . . . } is a basis for X in the sense that forany k ∈ N and any choice of i1, i2, . . . , ik ∈ N, xi1 , . . . , xik are linearly independentand every element of X is a finite linear combination of elements of S. 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 =

n∑j=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 of S, 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 = R or C. Let ‖·‖K on Kn be the norm given by

‖(α1, . . . , αn)‖ =

n∑j=1

|αj |

Suppose now that f is a linear functional on Kn. Then we have

f((α1, . . . , αn)) = α1f((1, 0, . . . , 0)) + · · ·+ αnf((0, . . . , 0, 1))

hence

|f((α1, . . . , αn))| =

∣∣∣∣∣∣n∑j=1

αjf((δ1j , . . . , δnj))

∣∣∣∣∣∣ ≤ max1≤j≤n

{‖f((δ1j , . . . , δnj))‖}n∑j=1

|αj |

= max1≤j≤n

{‖f((δ1j , . . . , δnj))‖} ‖(α1, . . . , αn)‖

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 Kn. Let φ : V → Kn 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−1∥∥, where g◦φ−1

is viewed as a functional on Kn. But all norms on V are equivalent, hence g isbounded with respect to any norm on V . �

Exersize 24. If M is a subspace of a Banach space X, then we define

codimM = dimX/M

If M is closed, show that codimM = dimM◦ and codimM◦ = dimM .


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

Let [β1], . . . , [βn] be a basis for X. Then, as is easily verified, β1, . . . , βn are linearlyindependent in X. In fact, we may write

X = Sp(β1, . . . , βn)⊕M

Let X∗ and X∗ denote the spaces of functionals on X and X, respectively. Consider

φ : X∗ → X∗

given byφ(f)(βj) = f([βj ])

and extending linearly to all of X by setting φ(f)(x) = 0 for all x /∈ Sp(β1, . . . , βn).

Then clearly φ is a vector space homomorphism, and φ(X∗) ⊂M◦. Since dim X∗ =

dimX = n, and N(φ) = 0, we conclude that dimφ(X∗) = n, so dimM◦ ≥ n.On the other hand, if g ∈M◦, then g is completely determined by its action on

β1, . . . , βn. Let g be the functional on X given by

g([βj ]) = g(βj)

Then, obviously, φ(g) = g. So in fact φ is a vector space isomorphism from X∗ to

M◦. Hence dimM◦ = dim X∗ = n.If dim X =∞, then for all n ∈ N there exists a subspace Vn ⊂ X of dimension

n. Restricting φ to Vn shows that dimM◦ ≥ n. This shows that dimM◦ =∞.One shows similarly that codimM◦ = dimM . �

Exersize 25. If M,N are subspaces of a Banach space X and codimN < dimM ,show that M ∩N 6= {0}.

Solution 25. We prove the contrapositive. Suppose that M ∩ N = {0}. As-sume first that dimM = m < ∞. Let β1, . . . , βm a basis for M . By assumption,β1, . . . , βm /∈ N , and [β1], . . . , [βm] are linearly independent when viewed as ele-ments of X/N . Hence

Sp([β1], . . . , [βm]) ⊂ X/Nis a subspace of X/N of dimension m. So codimN ≥ dimM .

Suppose that dimM =∞. But then for any n ∈ N there exists a subspace Vn ofdimension n in M , for which the above argument can be carried out. Hence againcodimN ≥ n for all n ∈ N, so codimN = dimM =∞.

This proves the contrapositive. �

Exersize 26. If A ∈ B(X,Y ), R(A) is dense in Y and D is dense in X, show thatA maps D onto a dense subset of Y .

Solution 26. Why do we need the assumption R(A) is dense in X? We knowthat A is bounded, hence continuous from X to Y , when X and Y are viewed astopological spaces (the topologies are induced by the norms). But a continuousmap 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 openneighborhood about v in V . By continuity of f , f−1(N) is open in U . By densityof D, there exists d ∈ D such that d ∈ f−1(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 anunbounded linear functional.


Solution 27. Let S = {x1, x2, . . . xn, . . . } ⊂ X be linearly independent. We canassume, without loss of generality, that ‖xj‖ = 1 after normalizing each xj . Let Vbe the subspace of X consisting of all finite linear combinations of elements of S.Define fj on V by

fj(x) =

{jc if x = cxj0 otherwise

Let f be the functional on V defined by

f =

∞∑j=1

fj

f is well-defined, since for each x ∈ V , f is a finite sum. Now, f is unbounded onV (it is unbounded on the unit ball in V ). To see this, let x = x1 + x2 + · · ·+ xn.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 assumethat F (v) = 0 =⇒ G(v) = 0, v ∈ V . Show that there is a scalar C such thatG(v) = CF (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 field K. Then F : V → K is a homomorphism of vector spaceswhich 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 dimV/N(F ) = dimK = 1.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) ≤ 1. If N(G) = V , then again we’re done(take C = 0). If not, then there exists β /∈N(F), N(G). Hence both F and G arecompletely determined by their action on β. Since K is a field and F (β) 6= 0, wemay take C = G(β)/F (β). �

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

x′(xk) = αk and ‖x′‖ = M,

is that ∣∣∣∣∣n∑1

βkαk

∣∣∣∣∣ ≤M∥∥∥∥∥n∑1

βkxk

∥∥∥∥∥holds for each n and scalars β1, . . . , βn.

Solution 29. Suppose there exists such a functional x′. Let β1, . . . , βn be scalars,fixed. Let x = β1x1 + · · ·+ βnxn ∈ X. Applying x′ to x, we obtain

x′(x) = x′

n∑j=1

βjxj

=

n∑j=1

βjαj


Since ‖x′‖ = M , we immediately get from above∣∣∣∣∣∣n∑j=1

βkαk

∣∣∣∣∣∣ = |x′(x)| ≤M ‖x‖ = M

∥∥∥∥∥∥n∑j=1

βkxk

∥∥∥∥∥∥Let now V be the vector subspace of X consisting of all finite linear combinations

of {xk}. Suppose that ∣∣∣∣∣n∑1

βkαk

∣∣∣∣∣ ≤M∥∥∥∥∥n∑1

βkxk


Define x′(xk) = αk and extend linearly to V . Then extend x′ to all of X bydefining x′(y) = 0 if y /∈ V .

However, the condition that∣∣∣∣∣n∑1

βkαk

∣∣∣∣∣ ≤M∥∥∥∥∥n∑1

βkxk

∥∥∥∥∥holds for each n and scalars β1, . . . , βn does not imply that ‖x′‖ = M . Indeed, ifK > M , then one has ∣∣∣∣∣

n∑1

βkαk

∣∣∣∣∣ ≤ K∥∥∥∥∥n∑1

βkxk

∥∥∥∥∥holds for each n and scalars β1, . . . , βn. �

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

Solution 30. �

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

Solution 31. Let {yn} be a Cauchy sequence in Y . Fix nonzero x0 in X. DefineAn(cx0) = cyn and for all x /∈ Sp(x0), let A(x) = 0. Clearly then An ∈ B(X,Y )and ‖An‖ = ‖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 then An(x0) →A(x0) in Y . Hence yn → A(x0) in Y . So Y is complete, hence Banach. �

Exersize 32. Prove that if {Tn} is a sequence in B(X,Y ) such that limTnx existsfor each x ∈ X, then there is a T ∈ B(X,Y ) such that Tnx→ Tx for all x ∈ X.

Solution 32. Set Tx = limTnx. First observe that T is linear. Indeed, T (cx+y) =limTn(cx+y) = c limTn(x)+limTn(y) = cTx+Ty. Next observe that T is bounded.

Suppose first that Tn(x) → T (x) uniformly for all x ∈ S, where S is the unitsphere in X. If this is the case, then there exists N ∈ N such that for all x of unitnorm,

‖Tx− Tnx‖ < 1 for all n ≥ N, x ∈ Sso

‖Tx‖ < TNx+ 1, x ∈ S


hence‖Tx‖ ≤ ‖TN‖+ 1 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 = {e1, e2, . . . }, where ej = (xjk) : N → {0, 1} given by xjk = δkj . We can

visualize each ej as an ∞-tuple, having 1 in jth 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

‖c1ej1 + c2ej2 + · · ·+ cmejm‖ =

m∑1

|ck|

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

Tn =

n∑1

Tk

Then each Tn is obviously a linear functional on V . Let

T = limn→∞

Tn =

∞∑n=1

Tn

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

|T (e1 + e2 + · · ·+ en)| =

∣∣∣∣∣∣n∑j=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 scalar C such thatG(v) = CF (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 field K. Then F : V → K is a homomorphism of vector spaceswhich 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 dimV/N(F ) = dimK = 1.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) ≤ 1. If N(G) = V , then again we’re done


(take C = 0). If not, then there exists β /∈N(F), N(G). Hence both F and G arecompletely determined by their action on β. Since K is a field and F (β) 6= 0, wemay take C = G(β)/F (β). �

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

x′(xk) = αk and ‖x′‖ = M,

is that ∣∣∣∣∣n∑1

βkαk

∣∣∣∣∣ ≤M∥∥∥∥∥n∑1

βkxk


Solution 34. Suppose there exists such a functional x′. Let β1, . . . , βn be scalars,fixed. Let x = β1x1 + · · ·+ βnxn ∈ X. Applying x′ to x, we obtain

x′(x) = x′

n∑j=1

βjxj

=

n∑j=1

βjαj

Since ‖x′‖ = M , we immediately get from above∣∣∣∣∣∣n∑j=1

βkαk

∣∣∣∣∣∣ = |x′(x)| ≤M ‖x‖ = M

∥∥∥∥∥∥n∑j=1

βkxk

∥∥∥∥∥∥Let now V be the vector subspace of X consisting of all finite linear combinations

of {xk}. Suppose that ∣∣∣∣∣n∑1

βkαk

∣∣∣∣∣ ≤M∥∥∥∥∥n∑1

βkxk


Define x′(xk) = αk and extend linearly to V . Then extend x′ to all of X bydefining x′(y) = 0 if y /∈ V .

However, the condition that∣∣∣∣∣n∑1

βkαk

∣∣∣∣∣ ≤M∥∥∥∥∥n∑1

βkxk

∥∥∥∥∥holds for each n and scalars β1, . . . , βn does not imply that ‖x′‖ = M . Indeed, ifK > M , then one has ∣∣∣∣∣

n∑1

βkαk

∣∣∣∣∣ ≤ K∥∥∥∥∥n∑1

βkxk

∥∥∥∥∥holds for each n and scalars β1, . . . , βn. �

Exersize 35. If a normed vector space X has a subspace M such that M andX/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, showthat Y is complete.


Solution 36. Let {yn} be a Cauchy sequence in Y . Fix nonzero x0 in X. DefineAn(cx0) = cyn and for all x /∈ Sp(x0), let A(x) = 0. Clearly then An ∈ B(X,Y )and ‖An‖ = ‖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 then An(x0) →A(x0) in Y . Hence yn → A(x0) in Y . So Y is complete, hence Banach. �

Exersize 37. Prove that if {Tn} is a sequence in B(X,Y ) such that limTnx existsfor each x ∈ X, then there is a T ∈ B(X,Y ) such that Tnx→ Tx for all x ∈ X.

Solution 37. Set Tx = limTnx. First observe that T is linear. Indeed, T (cx+y) =limTn(cx+y) = c limTn(x)+limTn(y) = cTx+Ty. Next observe that T is bounded.

Suppose first that Tn(x) → T (x) uniformly for all x ∈ S, where S is the unitsphere in X. If this is the case, then there exists N ∈ N such that for all x of unitnorm,

‖Tx− Tnx‖ < 1 for all n ≥ N, x ∈ Sso

‖Tx‖ < TNx+ 1, x ∈ Shence

‖Tx‖ ≤ ‖TN‖+ 1 for all x ∈ SSo T is bounded.

If convergence on S is not necessarily uniform, then T is not necessarily bounded.Consider the following example.

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

visualize each ej as an ∞-tuple, having 1 in jth 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

‖c1ej1 + c2ej2 + · · ·+ cmejm‖ =

m∑1

|ck|

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

Tn =

n∑1

Tk

Then each Tn is obviously a linear functional on V . Let

T = limn→∞

Tn =

∞∑n=1

Tn

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

|T (e1 + e2 + · · ·+ en)| =

∣∣∣∣∣∣n∑j=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.


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