## Section 11.2 Swapping limits

*Note: 2 lectures*

### Subsection 11.2.1 Continuity

Let us get back to swapping limits and expand on Chapter 6. Let \(\{ f_n \}\) be a sequence of functions \(f_n \colon X \to Y\) for a set \(X\) and a metric space \(Y\text{.}\) Let \(f \colon X \to Y\) be a function and for every \(x \in X\) suppose that

We say the sequence \(\{ f_n \}\) *converges pointwise* to \(f\text{.}\)

For \(Y=\C\text{,}\) a series *converges pointwise* if for every \(x \in X\text{,}\) we have

The question is: If \(f_n\) are all continuous, is \(f\) continuous? Differentiable? Integrable? What are the derivatives or integrals of \(f\text{?}\)

For example, for continuity of the pointwise limit of a sequence \(\{ f_n \}\text{,}\) we are asking if

We don't even a priory know if both sides exist, let alone if they are equal each other.

#### Example 11.2.1.

The functions \(f_n \colon \R \to \R\text{,}\)

are continuous and converge pointwise to the discontinuous function

So pointwise convergence is not enough to preserve continuity (nor even boundedness). For that, we need uniform convergence.

Let \(f_n \colon X \to Y\) be functions. Then \(\{f_n\}\) *converges uniformly* to \(f\) if for every \(\epsilon > 0\text{,}\) there exists an \(M\) such that for all \(n \geq M\) and all \(x \in X\text{,}\) we have

A series \(\sum f_n\) of complex-valued functions converges uniformly if the sequence of partial sums converges uniformly, that is for every \(\epsilon > 0\) there exists an \(M\) such that for all \(n \geq M\) and all \(x \in X\)

The simplest property preserved by uniform convergence is boundedness. We leave the proof of the following proposition as an exercise. It is almost identical to the proof for real-valued functions.

#### Proposition 11.2.2.

Let \(X\) be a set and \((Y,d)\) a metric space. If \(f_n \colon X \to Y\) are bounded functions and converge uniformly to \(f \colon X \to Y\text{,}\) then \(f\) is bounded.

If \(X\) is a set and \((Y,d)\) is a metric space, then a sequence \(f_n \colon X
\to Y\) is *uniformly Cauchy* if for every \(\epsilon > 0\text{,}\) there is an \(M\) such that for all \(n, m \geq M\) and all \(x \in X\text{,}\) we have

The notion is the same as for real-valued functions. The proof of the following proposition is again essentially the same as in that setting and is left as an exercise.

#### Proposition 11.2.3.

Let \(X\) be a set, \((Y,d)\) be a metric space, and \(f_n \colon X \to Y\) be functions. If \(\{ f_n \}\) converges uniformly, then \(\{f_n\}\) is uniformly Cauchy. Conversely, if \(\{f_n\}\) is uniformly Cauchy and \((Y,d)\) is Cauchy-complete, then \(\{f_n\}\) converges uniformly.

For \(f \colon X \to \C\text{,}\) we write

We call \(\snorm{\cdot}_u\) the *supremum norm* or *uniform norm*. Then a sequence of functions \(f_n \colon X \to \C\) converges uniformly to \(f \colon X \to \C\) if and only if

The supremum norm satisfies the triangle inequality: For every \(x \in X\text{,}\)

Take a supremum on the left to get

For a compact metric space \(X\text{,}\) the uniform norm is a norm on the vector space \(C(X,\C)\text{.}\) We leave it as an exercise. While we will not need it, \(C(X,\C)\) is in fact a complex vector space, that is, in the definition of a vector space we can replace \(\R\) with \(\C\text{.}\) Convergence in the metric space \(C(X,\C)\) is uniform convergence.

We will study a couple of types of series of functions, and a useful test for uniform convergence of a series is the so-called *Weierstrass \(M\)-test*.

#### Theorem 11.2.4. Weierstrass \(M\)-test.

Let \(X\) be a set. Suppose \(f_n \colon X \to \C\) are functions and \(M_n > 0\) numbers such that

Then

Another way to state the theorem is to say that if \(\sum \snorm{f_n}_u\) converges, then \(\sum f_n\) converges uniformly. Note that the converse of this theorem is not true. Also note that applying the theorem to \(\sum \sabs{f_n(x)}\) gives that a series satisfying the \(M\)-test also converges uniformly, so the series converges both absolutely and uniformly.

#### Proof.

Suppose \(\sum M_n\) converges. Given \(\epsilon > 0\text{,}\) we have that the partial sums of \(\sum M_n\) are Cauchy so there is an \(N\) such that for all \(m, n \geq N\) with \(m \geq n\text{,}\) we have

We estimate a Cauchy difference of the partial sums of the functions

We are done by Proposition 11.1.4.

#### Example 11.2.5.

The series

converges uniformly on \(\R\text{.}\) See Figure 11.2. This is a Fourier series, we will see more of these in a later section. Proof: The series converges uniformly because \(\sum_{n=1}^\infty \frac{1}{n^2}\) converges and

#### Example 11.2.6.

The series

converges uniformly on every bounded interval. This series is a power series that we will study shortly. Proof: Take the interval \([-r,r] \subset \R\) (every bounded interval is contained in some \([-r,r]\)). The series \(\sum_{n=0}^\infty \frac{r^n}{n!}\) converges by the ratio test, so \(\sum_{n=0}^\infty \frac{x^n}{n!}\) converges uniformly on \([-r,r]\) as

Now we would love to say something about the limit. For example, is it continuous?

#### Proposition 11.2.7.

Let \((X,d_X)\) and \((Y,d_Y)\) be metric spaces. Suppose \(f_n \colon X \to Y\) converge uniformly to \(f \colon X \to Y\text{.}\) Let \(\{ x_k \}\) be a sequence in \(X\) and \(x := \lim \, x_k\text{.}\) Suppose that

exists for all \(n\text{.}\) Then \(\{a_n\}\) converges and

In other words,

#### Proof.

First we show that \(\{ a_n \}\) converges. As \(\{ f_n \}\) converges uniformly it is uniformly Cauchy. Let \(\epsilon > 0\) be given. There is an \(M\) such that for all \(m,n \geq M\text{,}\) we have

Note that \(d_Y(a_n,a_m) \leq d_Y\bigl(a_n,f_n(x_k)\bigr) + d_Y\bigl(f_n(x_k),f_m(x_k)\bigr) + d_Y\bigl(f_m(x_k),a_m\bigr)\) and take the limit as \(k \to \infty\) to find

Hence \(\{a_n\}\) is Cauchy and converges since \(Y\) is complete. Write \(a := \lim \, a_n\text{.}\)

Find a \(k \in \N\) such that

for all \(p \in X\text{.}\) Assume \(k\) is large enough so that

Find an \(N \in \N\) such that for \(m \geq N\text{,}\)

Then for \(m \geq N\text{,}\)

We obtain an immediate corollary about continuity.

#### Corollary 11.2.8.

Let \(X\) and \(Y\) be metric spaces. If \(f_n \colon X \to Y\) are continuous functions such that \(\{ f_n \}\) converges uniformly to \(f \colon X \to Y\text{,}\) then \(f\) is continuous.

The converse is not true. Just because the limit is continuous doesn't mean that the convergence is uniform. For example: \(f_n \colon (0,1) \to \R\) defined by \(f_n(x) := x^n\) converge to the zero function, but not uniformly. However, if we add extra conditions on the sequence, we can obtain a partial converse such as Dini's theorem, see Exercise 6.2.10.

In Exercise 11.2.3 the reader is asked to prove that for a compact \(X\text{,}\) \(C(X,\C)\) is a normed vector space with the uniform norm, and hence a metric space. We have just shown that \(C(X,\C)\) is Cauchy-complete: Proposition 11.2.3 says that a Cauchy sequence in \(C(X,\C)\) converges uniformly to some function, and Corollary 11.2.8 shows that the limit is continuous and hence in \(C(X,\C)\text{.}\)

#### Corollary 11.2.9.

Let \((X,d)\) be a compact metric space. Then \(C(X,\C)\) is a Cauchy-complete metric space.

#### Example 11.2.10.

By Example 11.2.5 the Fourier series

converges uniformly and hence is continuous by Corollary 11.2.8 (as is visible in Figure 11.2).

### Subsection 11.2.2 Integration

#### Proposition 11.2.11.

Suppose \(f_n \colon [a,b] \to \C\) are Riemann integrable and suppose that \(\{ f_n \}\) converges uniformly to \(f \colon [a,b] \to \C\text{.}\) Then \(f\) is Riemann integrable and

Since the integral of a complex-valued function is just the integral of the real and imaginary parts separately, the proof follows directly by the results of Chapter 6. We leave the details as an exercise.

#### Corollary 11.2.12.

Suppose \(f_n \colon [a,b] \to \C\) are Riemann integrable and suppose that

converges uniformly. Then the series is Riemann integrable on \([a,b]\) and

#### Example 11.2.13.

Let us show how to integrate a Fourier series.

The swapping of integral and sum is possible because of uniform convergence, which we have proved before using the Weierstrass \(M\)-test (Theorem 11.2.4).

We remark that we can swap integrals and limits under far less stringent hypotheses, but for that we would need a stronger integral than the Riemann integral. E.g. the Lebesgue integral.

### Subsection 11.2.3 Differentiation

Recall that a complex-valued function \(f \colon [a,b] \to \C\text{,}\) where \(f(x) = u(x)+i\,v(x)\text{,}\) is differentiable, if \(u\) and \(v\) are differentiable and the derivative is

The proof of the following theorem is to apply the corresponding theorem for real functions to \(u\) and \(v\text{,}\) and is left as an exercise.

#### Theorem 11.2.14.

Let \(I \subset \R\) be a bounded interval and let \(f_n \colon I \to \C\) be continuously differentiable functions. Suppose \(\{ f_n' \}\) converges uniformly to \(g \colon I \to \C\text{,}\) and suppose \(\{ f_n(c) \}_{n=1}^\infty\) is a convergent sequence for some \(c \in I\text{.}\) Then \(\{ f_n \}\) converges uniformly to a continuously differentiable function \(f \colon I \to \C\text{,}\) and \(f' = g\text{.}\)

Uniform limits of the functions themselves are not enough, and can make matters even worse. In Section 11.7 we will prove that continuous functions are uniform limits of polynomials, yet as the following example demonstrates, a continuous function need not be differentiable anywhere.

#### Example 11.2.15.

There exist continuous nowhere differentiable functions. Such functions are often called *Weierstrass functions*, although this particular one, essentially due to Takagi^{ 1 }, is a different example than what Weierstrass gave.

Define

Extend the definition of \(\varphi\) to all of \(\R\) by making it 2-periodic: Decree that \(\varphi(x) = \varphi(x+2)\text{.}\) The function \(\varphi \colon \R \to \R\) is continuous, in fact \(\sabs{\varphi(x)-\varphi(y)} \leq \sabs{x-y}\) (why?). See Figure 11.3.

As \(\sum {\left(\frac{3}{4}\right)}^n\) converges and \(\sabs{\varphi(x)} \leq 1\) for all \(x\text{,}\) we have by the \(M\)-test (Theorem 11.2.4) that

converges uniformly and hence is continuous. See Figure 11.4.

We claim \(f \colon \R \to \R\) is nowhere differentiable. Fix \(x\text{,}\) and we will show \(f\) is not differentiable at \(x\text{.}\) Define

where the sign is chosen so that there is no integer between \(4^m x\) and \(4^m(x+\delta_m) = 4^m x \pm \frac{1}{2}\text{.}\)

We want to look at the difference quotient

Fix \(m\) for a moment. Consider the expression inside the series:

If \(n > m\text{,}\) then \(4^n\delta_m\) is an even integer. As \(\varphi\) is 2-periodic we get that \(\gamma_n = 0\text{.}\)

As there is no integer between \(4^m(x+\delta_m) = 4^m x\pm\nicefrac{1}{2}\) and \(4^m x\text{,}\) then on this interval \(\varphi(t) = \pm t + \ell\) for some integer \(\ell\text{.}\) In particular, \(\abs{\varphi\bigl(4^m(x+ \delta_m)\bigr)-\varphi(4^mx)} = \abs{4^mx\pm\nicefrac{1}{2}-4^mx} = \nicefrac{1}{2}\text{.}\) Therefore,

Similarly, suppose \(n < m\text{.}\) Since \(\sabs{\varphi(s) -\varphi(t)} \leq \sabs{s-t}\text{,}\)

And so

As \(m \to \infty\text{,}\) we have \(\delta_m \to 0\text{,}\) but \(\frac{3^m+1}{2}\) goes to infinity. Hence \(f\) cannot be differentiable at \(x\text{.}\)

### Subsection 11.2.4 Exercises

#### Exercise 11.2.1.

Prove Proposition 11.2.2.

#### Exercise 11.2.2.

Prove Proposition 11.2.3.

#### Exercise 11.2.3.

Suppose \((X,d)\) is a compact metric space. Prove that \(\snorm{\cdot}_u\) is a norm on the vector space of continuous complex-valued functions \(C(X,\C)\text{.}\)

#### Exercise 11.2.4.

Prove that \(f_n(x) := 2^{-n} \sin(2^n x)\) converge uniformly to zero, but there exists a dense set \(D \subset \R\) such that \(\lim_{n\to\infty} f_n'(x) = 1\) for all \(x \in D\text{.}\)

Prove that \(\sum_{n=1}^\infty 2^{-n} \sin(2^n x)\) converges uniformly to a continuous function, and there exists a dense set \(D \subset \R\) where the derivatives of the partial sums do not converge.

#### Exercise 11.2.5.

Suppose \((X,d)\) is a compact metric space. Prove that \(\snorm{f}_{C^1} := \snorm{f}_u+\snorm{f'}_u\) is a norm on the vector space of continuously differentiable complex-valued functions \(C^1(X,\C)\text{.}\)

#### Exercise 11.2.6.

Prove Theorem 11.2.14.

#### Exercise 11.2.7.

Prove Proposition 11.2.11 by reducing to the real result.

#### Exercise 11.2.8.

Work through the following counterexample to the converse of the Weierstrass \(M\)-test (Theorem 11.2.4). Define \(f_n \colon [0,1] \to \R\) by

Prove that \(\sum f_n\) converges uniformly, but \(\sum \snorm{f_n}_u\) does not converge.

#### Exercise 11.2.9.

Suppose \(f_n \colon [0,1] \to \R\) are monotone increasing functions and suppose that \(\sum f_n\) converges pointwise. Prove that \(\sum f_n\) converges uniformly.

#### Exercise 11.2.10.

Prove that

converges for all \(x > 0\) to a differentiable function.

^{ 2 }(1875–1960) was a Japanese mathematician.

`https://en.wikipedia.org/wiki/Teiji_Takagi`