Davidson K.R., Donsig A.P. Real Analysis with Real Applications

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7.5 Orthogonal Expansions in Inner Product Spaces 203

Thus e is orthogonal to M. In particular, {e, e

: i ≥ 1} is orthonormal, which

contradicts the maximality of E. Since E is maximal, it follows that M = H.

Parseval’s Theorem shows that the closed span of an orthonormal set consists

of all `

combinations of this orthonormal set. So every vector x ∈ H may be

expressed as

∞

i=1

. The coefﬁcients are unique because this expression for x

implies that

hx, e

i = lim

n→∞



i=1

, e

i=1

, e

i = α

Exercises for Section 7.5

A. When does equality hold in Equation (7.5.3)?

B. Let x = (x

)

∞

n=1

and y = (y

)

∞

n=1

be elements of `

(a) Show that

n=1

| ≤ kxkkyk. HINT: Schwarz inequality.

(b) Hence prove that

∞

n=1

converges absolutely.

C. If M is a closed subspace of a Hilbert space H, deﬁne the orthogonal complement

of M to be M

⊥

= {x : hx, mi = 0 for all m ∈ M}.

(a) Show that every vector in H can be written uniquely as x = m + y, where m ∈ M

and y ∈ M

⊥

. Moreover, kxk

= kmk

+ kyk

(b) Show that M = (M

⊥

)

⊥

D. Let P be a projection on an inner product space V . Prove that the following are

equivalent:

(a) P is an orthogonal projection.

(b) kvk

= kP vk

+ kv − P vk

for all v ∈ V .

(d) hP v, wi = hv, P wi for all v, w ∈ V .

HINT: For (c) =⇒ (d), show that not (d) implies there are vectors v = P v and

w = (I −P )w such that hv, wi > 0. Compute kv −twk

−kvk

for small t > 0. For

(d) =⇒ (a), show that hP v, (I − P )wi = 0.

E. Show that the orthogonal projection P onto a subspace M is unique.

HINT: If P and Q are orthogonal projections onto M , apply Q to x = P x+ (I −P )x.

F. Formulate and prove a precise version of the following statement “A separable inﬁnite-

dimensional Hilbert space with an orthonormal basis {e

: n ≥ 1} behaves like `

.”

HINT: Look at the ﬁnite-dimensional statement, Corollary 7.4.8.

G. Let M and N be closed subspaces of a Hilbert space H.

(a) Show that (M ∩ N )

⊥

+ N

⊥

(b) Let {e

: n ≥ 1} be an orthonormal basis for `

. Let M = span{e

: n ≥ 1} and

N = span{e

2n−1

+ ne

: n ≥ 1}. Show that M + N is not closed.

204 Normed Vector Spaces

H. Consider the functions f

in C[−π, π] given by

(x) =











0 −π ≤ x ≤ 0

nx 0 ≤ x ≤

≤ x ≤ π.

(a) Show that f

converges in the L

norm to the characteristic function

of (0, π]. In

particular, (f

)

∞

n=1

is an L

-Cauchy sequence.

(b) Show that k

− hk

> 0 for every function h in C[−π, π].

norm.

I. Let {r

: n ≥ 1} be a list of the rational points in [0, 1]. Deﬁne an inner product on

C[0, 1] by hf, gi

∗

∞

n=1

−n

f(r

)g(r

(a) Show that this is indeed a positive deﬁnite inner product.

(b) Why is this not an inner product on the vector space of all bounded functions on

[0, 1]?

HINT: Find a Cauchy sequence converging to the characteristic function of {0}.

7.6. Finite-Dimensional Normed Spaces

In this section, we will study the ﬁnite-dimensional subspaces of an arbitrary

normed vector space. Many of the nice properties of R

carry over to this setting.

A particularly important result is Theorem 7.6.5, which will be very useful in the

chapter on approximation.

7.6.1. LEMMA. If {v

, v

, . . . , v

} is a linearly independent set in a normed

vector space (V, k · k), then there exist positive constants 0 < c < C so that for all

a = (a

, . . . , a

) ∈ R

, we have

ckak

≤

i=1

≤ Ckak

PROOF. By the triangle inequality and the Schwarz inequality (4.1.1),

i=1

≤

i=1

|kv

≤

i=1

1/2

i=1

1/2

= Ckak

where C =

i=1

1/2

7.6 Finite-Dimensional Normed Spaces 205

Deﬁne a function N on R

by N(a) =

i=1

. Observe that N is a con-

tinuous function since

|N(x) − N(y)| =

i=1

−

i=1

≤

i=1

− y

≤ Ckx − yk

So N is Lipschitz and thus continuous.

Let S = {a ∈ R

: kak

= 1} be the unit sphere of R

. Since the set

, . . . , v

} is linearly independent, it follows that N(x) > 0 when x 6= 0. In

particular, N never vanishes on the compact set S. By the Extreme Value Theorem

(Theorem 5.4.4), N must achieve its minimum value c at some point on S; whence

c > 0. So if a is an arbitrary vector in R

, we obtain

i=1

= kak

N(a/kak

) ≥ ckak

FIGURE 7.3. Euclidean balls inside and outside the unit ball of V .

The effect of this theorem is that every ﬁnite-dimensional normed space has

the same topology as (R

, k·k

), in the sense that they have the “same” convergent

sequences, the “same” open sets, and so on. Let us make this more precise. If

, v

, . . . , v

} is a basis for an n-dimensional normed space V , deﬁne a linear

transformation from R

into V by T a =

i=1

. The map T carries R

one-to-

one and onto V by the deﬁnition of a basis. Since every element of V is a linear

combination of {v

, . . . , v

}, we can deﬁne the inverse map by T

−1

i=1

= a.

206 Normed Vector Spaces

7.6.2. COROLLARY. Suppose that V is an n-dimensional normed space with

basis {v

, v

, . . . , v

}. Then the maps T and T

−1

deﬁned previously are both Lip-

schitz, and therefore continuous. Thus a set A in R

is closed, bounded, open, or

compact if and only if T (A) is closed, bounded, open, or compact in V .

PROOF. We have for vectors x and y in R

kT x − T yk = N(x − y) ≤ Ckx − yk

So T is Lipschitz with constant C. Similarly, if u = T x and v = T y are typical

vectors in V ,

−1

u − T

−1

= kT

−1

(T x) − T

−1

(T y)k

= kx − yk

≤

kT x − T yk =

ku − vk.

So T

−1

is Lipschitz with constant 1/c. It follows that both T and T

−1

are continu-

ous.

This means that if x

is a sequence of vectors in R

converging to a point x,

then T x

converges to T x because of the continuity of T . And conversely, if v

is a sequence of vectors in V converging to a vector v, then T

−1

converges to

−1

v in R

. This just says that there is a direct correspondence between conver-

gent sequences in R

and V . Since closed and compact sets are deﬁned in terms

of convergent sequences, these sets correspond as well. Open sets are the comple-

ments of closed sets, so open sets correspond. If A ⊂ R

is bounded by L, the

Lipschitz condition shows that

kT ak = kT a − T 0k ≤ Ckak

≤ CL

for every a in A. So T (A) is bounded. Likewise, since T

−1

is Lipschitz, if B is a

subset of V bounded by L, it follows that T

−1

(B) is bounded by L/c in R

. ¥

Notice that we may conclude that closed and bounded sets also correspond.

Since the Heine–Borel Theorem shows that closed and bounded sets are compact

in R

, we can conclude that this is also true in all ﬁnite-dimensional normed spaces.

7.6.3. COROLLARY. A subset of a ﬁnite-dimensional normed vector space is

compact if and only if it is closed and bounded.

Another immediate consequence refers to the way a ﬁnite-dimensional sub-

space sits inside an arbitrary normed space. Arbitrary normed spaces are in general

not complete. However, ﬁnite-dimensional subspaces are, because we know the

Heine–Borel Theorem in R

7.6.4. COROLLARY. A ﬁnite-dimensional subspace of a normed vector space

is complete, and in particular it is closed.

PROOF. Let V be a normed vector space and let W be an n-dimensional subspace.

Let T be a linear invertible map from R

onto W as just constructed. Suppose that

7.6 Finite-Dimensional Normed Spaces 207

)

∞

k=1

is a Cauchy sequence in W . Then since T

−1

is Lipschitz, the sequence

= T

−1

for k ≥ 1 is Cauchy in R

. (Check this yourself!) Since R

complete (Theorem 4.2.5), the sequence x

converges to a vector x. Again by

Corollary 7.6.2, we see that w

must converge to w = T x. Thus W is complete.

In particular, all of the limit points of W lie in W , so W is closed. ¥

As an application of these corollaries, we prove the following result, which is

fundamental to approximation theory.

7.6.5. THEOREM. Let (V, k·k) be a normed vector space, and let W be a ﬁnite

dimensional subspace of V . Then for any v ∈ V , there is at least one closest point

∗

∈ W so that kv − w

∗

k = inf{kv − wk : w ∈ W }.

PROOF. Notice that the zero vector is in W and so

inf{kv − wk : w ∈ W } ≤ kv − 0k = kvk.

Let M = kvk. If w satisﬁes kv − wk ≤ kvk, then

kwk ≤ kw − vk + kvk ≤ M + M = 2M.

Thus if we deﬁne K := {w ∈ W : kwk ≤ 2M}, then

inf{kv − wk : w ∈ K} = inf{kv − wk : w ∈ W }.

We will show that K is compact. Clearly K is bounded by 2M. The norm

function is Lipschitz, and hence continuous. Thus any convergent sequence of

vectors in K will converge to a vector of norm at most 2M. And since W is

complete, this limit also lies in W ; whence the limit lies in K. This shows that K

is closed and bounded. By our corollary, it follows that K is compact.

Now deﬁne a function on K by f (w) = kv − wk. This function has Lipschitz

constant 1 since

|f(w) − f(x)| =

kv − wk − kv − xk

≤ kw − xk.

In particular, f is continuous on the compact set K. By the Extreme Value Theorem

(Theorem 5.4.4), f achieves its minimum at some point w

∗

in K. This is a closest

point to v in W . ¥

Exercises for Section 7.6

A. Let V be a ﬁnite-dimensional vector space with two norms k · k and ||| · |||. Show that

there are constants 0 < a < A such that akvk ≤ |||v||| ≤ Akvk for all v ∈ V .

B. Let T be the invertible linear map from Corollary 7.6.2.

(a) Use the Lipschitz property of T and T

−1

to show that T (B

(x)) contains a ball

about T x in V ; and that T

−1

(T x)) contains a ball about x in R

(b) Hence show directly that U is open if and only if T (U) is open.

C. Write out a careful proof of Corollary 7.6.3.

208 Normed Vector Spaces

D. Suppose that (w

)

∞

k=1

is a Cauchy sequence in a normed space W . If T : W → V is

a Lipschitz map into another normed space V , show that the sequence v

= T w

for

k ≥ 1 is Cauchy in V .

E. Show that for each integer n and each function f in C[a, b], there is a polynomial of

degree at most n that is closest to f in the max norm on C[a, b].

F. Let R

have the max norm kxk

∞

= max{|x

| : 1 ≤ i ≤ n}. Let K be the unit ball of

V and let v = (2, 0, . . . , 0). Find all closest points to v in K.

G. For strictly convex normed vector spaces (see Exercise 7.3.F), we have better approx-

imation results.

(a) Show that if W is a ﬁnite-dimensional subspace of a strictly convex normed vector

space V , then each point v ∈ V has a unique closest point in W .

(b) Show that R

with the standard Euclidean norm is strictly convex.

with the norm k(x, y)k

= |x| + |y| is not strictly convex.

(d) Find a subspace W of V = (R

, k · k

) so that every point in V that is not in W

has more than one closest point in W .

7.7. The L

norms

The L

norms on C[a, b] for 1 ≤ p < ∞ were deﬁned in Example 7.1.5 by

kfk

|f(x)|

1/p

The point of this section is to prove that these really are norms by proving the trian-

gle inequality, which is the only part of the deﬁnition not established in Section 7.1.

There are other variants of the L

norms that are handled in exactly the same

way. Here are two important examples.

7.7.1. EXAMPLES. (1) For 1 ≤ p < ∞, `

consists of the set of all inﬁnite

sequences a = (a

) such that

kak

∞

n=1

1/p

< ∞.

It is again easy to see that this measure is positive deﬁnite and homogeneous.

We can deduce the triangle inequality from the main example by identifying

the sequence a with the function f on [0, ∞) given by

f(x) = a

for n − 1 ≤ x < n, n ≥ 1.

This function has the property that kfk

= kak

. So we can deduce our inequality

from the main case. You probably noticed that this function is not continuous.

However, if we prove the result for piecewise continuous functions, we shall be

able to apply the result here. (Alternatively, we could doctor this function so that

we still obtain the correct integral and the function is continuous. But this is not as

natural.)

7.7 The L

norms 209

(2) Let w(x) be a strictly positive piecewise continuous function on [a, b]. Then

deﬁne a norm on C[a, b] or even on the space P C[a, b] of piecewise continuous

functions by

kfk

(w)

|f(x)|

w(x) dx

1/p

We call this the L

(w) norm. The standard case takes w(x) = 1. So if we prove

the result for L

(w), then it will follow for L

and for `

First we need an easy lemma from calculus.

7.7.2. LEMMA. Let A, B > 0. Then

1−t

≤ tA + (1 −t)B for all 0 < t < 1.

Moreover, equality holds for some (or all) t only if A = B.

PROOF. Since A > 0, write A = e

, where a = logA. Similarly, B = e

for

b = log B. Substituting these formulas for A and B, we have to prove that

(1−t)b

= e

ta+(1−t)b

≤ te

+ (1 − t)e

But the preceding inequality follows from the convexity of the exponential function

on R (see Figure 7.4). As the exponential function is convex (apply Exercise 6.2.J),

we are done. ¥

ta + (1 − t)b

+ (1 − t)e

exp(ta + (1 − t)b)

FIGURE 7.4. The convexity of e

We can now prove one of the most important inequalities of analysis, the

older inequality.

210 Normed Vector Spaces

7.7.3. H

OLDER’S INEQUALITY.

Let w be a positive function on an interval [a, b]. Let f ∈ L

(w) and g ∈ L

(w)

where 1 < p < ∞ and

= 1. Then

f(x)g(x)w(x) dx

≤ kfk

(w)

kgk

(w)

PROOF. We may assume that f and g are nonzero because both sides are easily

seen to be 0 otherwise. Let

A =

|f(x)|

kfk

(w)

and B =

|g(x)|

kgk

(w)

Then take t = 1/p, so that 1 −t = 1/q, and apply the lemma. We obtain

|f(x)|

kfk

(w)

|g(x)|

kgk

(w)

= A

1−t

≤ tA + (1 −t)B

|f(x)|

pkfk

(w)

|g(x)|

qkgk

(w)

We now multiply by w(x) and integrate from a to b.

f(x)g(x)

kfk

(w)

kgk

(w)

w(x) dx

≤

|f(x)|

kfk

(w)

|g(x)|

kgk

(w)

w(x) dx

≤

|f(x)|

kfk

(w)

w(x) dx +

|g(x)|

kgk

(w)

w(x) dx

= 1

Multiplying by kfk

(w)

kgk

(w)

gives the inequality. ¥

7.7.4. EXAMPLES.

(1)

f(x)g(x) dx ≤

f(x)

1/p

g(x)

1/q

for all continuous functions f and g on [a, b]. This is just the standard weight

w(x) = 1.

(2) If we take p = 2, then q = 2 and we obtain the Cauchy–Schwarz inequality,

Theorem 7.3.4, for L

(w).

7.7 The L

norms 211

(3) As in Example 7.7.1(1), take w(x) = 1 on [0, ∞) and consider f(x) = a

and

g(x) = b

on [n − 1, n] for n ≥ 1, where (a

) ∈ `

and (b

) ∈ `

. Then

∞

n=1

∞

f(x)g(x) dx ≤

f(x)

1/p

g(x)

1/q

∞

n=1

1/p

∞

n=1

1/q

Again, this reduces to the Cauchy–Schwarz inequality if p = q = 2.

The triangle inequality now follows by another trick.

7.7.5. MINKOWSKI’S INEQUALITY.

The triangle inequality holds for L

(w), that is,

|f(x) + g(x)|

w(x)dx

1/p

≤

|f(x)|

w(x)dx

1/p

|g(x)|

w(x)dx

1/p

PROOF. Notice that

= 1 −

p − 1

. So q =

p − 1

. Thus

k|f|

p−1

(w)

|f(x)|

(p−1)q

w(x) dx

1/q

|f(x)|

w(x) dx

1/q

= kf k

p/q

(w)

= kf k

p−1

(w)

We use this equality in the last line of the next calculation, which estimates the

norm of f + g in L

(w),

kf + gk

(w)

|f(x) + g(x)|

p−1

|f(x) + g(x)|w(x) dx

≤

|f(x)+g(x)|

p−1

|f(x)|w(x) dx

|f(x)+g(x)|

p−1

|g(x)|w(x) dx

≤ k|f + g|

p−1

(w)

kfk

(w)

+ kgk

(w)

= kf + gk

p−1

(w)

kfk

(w)

+ kgk

(w)

Now divide by kf + gk

p−1

(w)

to obtain

kf + gk

(w)

≤ kfk

(w)

+ kgk

(w)

212 Normed Vector Spaces

Exercises for Section 7.7

A. Rework the H

older inequality to show: If a

≥ 0 and b

≥ 0 and 0 < t < 1, then

∞

n=1

1−t

≤

∞

n=1

∞

n=1

1−t

B. Show that the H

older inequality is sharp in the sense that for every f ∈ L

(w), there

is a nonzero g ∈ L

(w) so that

f(x)g(x)w(x) dx

= kfk

(w)

kgk

(w)

HINT: For nonzero f, set g(x) = |f(x)|

/f(x) if f(x) 6= 0 and g(x) = 0 if f(x) = 0.

C. Let f(t) be a continuous, strictly increasing function on [0, ∞) with f(0) = 0. Recall

that f has an inverse function g with the same properties by Theorem 5.7.6. Deﬁne

F (x) =

f(t) dt and G(x) =

g(t) dt.

(a) Prove Young’s Inequality: xy ≤ F (x) + G(y) for all x, y ≥ 0, with equality if

and only if f(x) = y.

HINT: Sketch f and ﬁnd two regions with areas F (x) and G(y).

(b) Take f(x) = x

p−1

for 1 < p < ∞. What inequality do you get?

D. Let f

= n

[

]

be deﬁned on [0, 1] for n ≥ 2.

(a) Show that f

converges pointwise to 0.

(b) Show that f

does not converge in L

[0, 1] for any 1 ≤ p < ∞.

E. (a) If f ∈ C[0, 1] and 1 ≤ r ≤ s < ∞, show that kfk

≤ kfk

∞

HINT: Let p =

and think of

|f|

· 1dx as the product of an L

function and

an L

function.

(b) Hence show that if f

∈ C[0, 1] converge uniformly to f, then they also converge

in the L

norm for all 1 ≤ p < ∞.

F. Let f

[0,n

]

be deﬁned on [0, ∞) for n ≥ 1.

(a) Show that f

converges uniformly to 0.

(b) Show that f

does not converge in L

[0, 1] for any 1 ≤ p < ∞.

G. (a) If a = (a

) ∈ `

and 1 ≤ r ≤ s < ∞, show that kak

∞

≤ kak

HINT: Show that for p ≥ 1,

k=1

≤

k=1

(b) Hence show that `

⊂ `

∞

H. Find continuous functions f and g on [0, ∞) so that f is in L

[0, ∞) but not in

[0, ∞), and g is in L

[0, ∞) but not in L

[0, ∞).