# User:TakuyaMurata/Continuous functions on a compact space

In this section, we will undergo a throughout study of , the space of all real-valued or complex-valued functions on a compact space. The most important result in the line of this study is Ascoli's theorem and Stone-Weierstrass theorem. For simplicity, we assume functions in are real-valued. (A discussion will be given later as to why this does not diminish the generality.)

As usual, we topologizes by the norm . To say that is complete is precisely:

**2. Lemma** *The limit of a uniformly convergent sequence of continuous functions is continuous.*

Proof: Suppose is a sequence such that for some function defined on . For any , by the iterated limit theorem,

Hence, is a Banach space. (For the definition and basic results of Banach spaces, see Functional Analysis.)

**2 Theorem (Ascoli)** *Let . Then is relatively compact if and only if*

*(i) Given an and , we can find a neighborhood of such that**for every and every*

*(ii) for every*

Proof: First, assume that is relatively compact. (ii) is then obvious. For (i), let and be given. For each , by continuity, we can find a neighborhood of such that:

- for every .

Since is relatively compact, contains a finite subset such that is the union of the sets of the form

over . Let be the intersection of over . Then for every , there is with , and so:

- for any .

This proves (i). Next, suppose satisfies (i) and (ii). To show that is totally bounded, let be given. For each , by (i), we can find a neighborhood of such that:

- for every and .

Since is compact, we can find such that is the union of over . Let

- .

By (ii), is a bounded (thus totally bounded) subset of . That means that contains a finite subset such that:

It now follows: given , we can find such that:

- .

Then, for each , since for some ,

In other words, . Hence, is totally bounded, or equivalently, relatively compact.

**2 Corollary** *Let . is uniformly convergent if and only if it is pointwise convergent and equicontinuous.*

**2 Theorem** *Let converge pointwise to . If exists and converges uniformly to , then converges uniformly to . Moreover, is differentiable and its derivative is .* Proof: Let . is finite by uniform convergence. By the mean value theorem,

Thus, is equicontinuous and converges uniformly by Ascoli's thoerem (or one of its corollaries.)

**2 Theorem** *Let be an equicontinous set of real-valued functions on . If , then there exist an such that:*

Proof: Let be such that:

- for all and

( isn't a typo; it is meant to simplify the computation.) Let be fixed. Then, for any ,

- ,

and we estimate:

where is such that . Thus,

Since we can get the same estimate for , the proof is complete.

**2 Corollary (Dini's theorem)** *Let be a sequence such that for every . If is increasing, then .*

Proof: Set . Then is decreasing and thus satisfies the hypothesis of Ascoli's theorem. Hence, admits a convergent subsequence, which converges to 0 since the subsequence converges pointwise to 0. Since is decreasing, converges as well.

**2 Theorem** *Suppose is a metric space. Then is equicontinuous if and only if for every there exists such that*

for every and with . Proof: holds vacuously. For the converse, let be given. Then for each , we can find such that

- implies for every

By compactness, we find such that:

Let , and then suppose we are given with . It follows: there is a with . Since

- ,

we have:

for every .

The Stone-Weierstrass theorem states that polynomials are dense in C(K, *R*). It is however not the case that the space of polynomials in *z* is a dense in C(K, *C*). If it were, we have the equality in the below

But if K has nonempty interior.

**2 Theorem (intermediate value theorem)** *A function is continuous if and only if*

*(i) If , then*c*is in .**(ii) If is closed for every real*c*.*

Proof: () Obvious. () Suppose . Since the complement of , which contains *x*, is open, we have: in some interval *U* in containing *x*. We actually have: in *U*. In fact, if and , then , which implies *U* contains a point *z* such that , a contradiction. Hence, *f* is upper semicontinous at *x*. The same argument applied to shows that *f* is also lower semicontinous at *x*.

**2 Theorem** *Let be a real-valued continuous function on an open interval. Then the following are equivalent.*

*(i)*f*is injective.**(ii)*f*is strictly monotonic.**(iii)*f*is an open mapping.*

Proof: (ii) (i) is obvious. (iii) (ii): If (ii) is false, then we can assume there exists such that and . By continuity and compactness, *f* attains a maximum in some point *x* in but by hypothesis and so is a non-interior point of , falsifying (iii). If (iii) is false, then contains a *x* such that is not an interior point of . Since is an interval, we may assume that . It then follows from the intermediate value theorem that *f* is not injective.

**2 Theorem (mean value theorem)** *Suppose is differentiable on the open interval . Then*

*for some *

Proof: FIrst assume . By the theorem preceding this one, attains a maximum or minimum at ; say, a maximum. By definition, we can write:

- as

Then since *x* is a maximum,

- .

If ,

and letting gives that . If , then by the same argument, we find that . Thus, . For the general case, let

- where .

Then . Hence, applying the first part of the proof gives: for some *c*. Since , *c* is a solution of the equation.

**2 Corollary** *Let be a differentiable function. If , then* c *is in .*

Proof:Let . Then and . In other words, is increasing at *a* and decreasing at *b*. By the theorem above, *g* is not injective on ; i.e., for some in . It follows:

- for some .

and .

**2 Corollary** *A strictly monotonic continuous function with closed range is a homeomorphism.*

A real-valued function on is said to be homogeneous of degree ( could be any real number) if

for all and all .

**2 Theorem (Euler's relation)** *Let be a function differentiable on . Then is homogeneous of degree if and only if:*

for all .

Proof: () Differentiate with respect to both sides of and then put . () Note

- .

Thus, if we let

- ,

then the derivative of vanishes identically. Since , is identically zero.

The theorem permits a generalization. By definition, any (distribution) solution of the equation

is said to be *homogeneous of degree .*

A homogeneous distribution is tempered and its Fourier transform is homogeneous.

**2 Theorem (l'Hôpital's rule)** *Let . If as or if as , then*

provided the limit in the right-hand side exists.

Proof: First assume is finite. We may redefine (since the values of functions at are immaterial when we compute the limit.) Fix , and define

Since , by the mean value theorem, we can find a between and such that

Since when is close to we may assume never vanishes,

Since as , the proof of this case is complete. (TODO: handle other cases.)

The next theorem can be skipped without the loss of continuity, for more general results will later be obtained.

**2 Theorem (The Weierstrass approximation theorem)** *Let , and define*

- with (w:Bernstein polynomial).

Then uniformly on

Proof [1]: First note that

is a partition of unity, by the binomial theorem applied to . Moreover, a simple computation gives the identity:

It thus follows: for any

Since is uniformly continuous on by compactness, the theorem now follows.

**2 Corollary** *Any continuous function vanishing at infinity is uniformly approximated by Hermite function. (TODO: made the statement more precise and give a proof.)*

Example: Let . Let be the linear span of polynomials. Then is a dense subspace of by the above theorem since

**2 Theorem** *If is uniformly continuous and integrable, then .*

Proof: Define . That is integrable means that exists and is finite. Let be given. By uniform continuity, there is a such that

- whenever .

Then there is an such that

- whenever

Now, let be given. By the mean value theorem, we find such that and . Thus,

**Example** *No function is continuous only on rational points. To see this, let be a function, and let be the set of all points at which is continuous. It follows immediately from the definition of continuity that is ; i.e., it is an intersection of countably many open sets. On the other hand, is not .*

**2 Theorem** *Let be a nonempty open subset.*

Proof: Let and . We assume and are finite; otherwise the inequality is trivial. Given , we can find so that an interval . By Taylor's formula,

- (where )

and so:

Now, take .

**2 Theorem (Whitney extension theorem)** *Every real-valued L-Lipschitz function on a subset of is the restriction of a L-Lipschitz function on .*

Proof ([2] pg. 5): Let be a L-Lipschitz function on a subset . Define

It is clear that is L-Lipschitz continuous.

Exercise: Every closed set is a zero set of a function.