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