## Section6.1Pointwise and uniform convergence

Note: 1–1.5 lecture

Up till now, when we talked about limits of sequences we talked about sequences of numbers. A very useful concept in analysis is a sequence of functions. For example, a solution to some differential equation might be found by finding only approximate solutions. Then the actual solution is some sort of limit of those approximate solutions.

When talking about sequences of functions, the tricky part is that there are multiple notions of a limit. Let us describe two common notions of a limit of a sequence of functions.

### Subsection6.1.1Pointwise convergence

#### Definition6.1.1.

For every $$n \in \N\text{,}$$ let $$f_n \colon S \to \R$$ be a function. The sequence $$\{ f_n \}_{n=1}^\infty$$ converges pointwise to $$f \colon S \to \R$$ if for every $$x \in S\text{,}$$ we have

\begin{equation*} f(x) = \lim_{n\to\infty} f_n(x) . \end{equation*}

Limits of sequences of numbers are unique, and so if a sequence $$\{ f_n \}$$ converges pointwise, the limit function $$f$$ is unique. It is common to say that $$f_n \colon S \to \R$$ converges to $$f$$ on $$T \subset S$$ for some $$f \colon T \to \R\text{.}$$ In that case we mean $$f(x) = \lim\, f_n(x)$$ for every $$x \in T\text{.}$$ In other words, the restrictions of $$f_n$$ to $$T$$ converge pointwise to $$f\text{.}$$

#### Example6.1.2.

On $$[-1,1]\text{,}$$ the sequence of functions defined by $$f_n(x) := x^{2n}$$ converges pointwise to $$f \colon [-1,1] \to \R\text{,}$$ where

\begin{equation*} f(x) = \begin{cases} 1 & \text{if } x=-1 \text{ or } x=1, \\ 0 & \text{otherwise.} \end{cases} \end{equation*}

See Figure 6.1.

To see this is so, first take $$x \in (-1,1)\text{.}$$ Then $$0 \leq x^2 < 1\text{.}$$ We have seen before that

\begin{equation*} \abs{x^{2n} - 0} = {(x^2)}^n \to 0 \quad \text{as} \quad n \to \infty . \end{equation*}

Therefore, $$\lim\,f_n(x) = 0\text{.}$$

When $$x = 1$$ or $$x=-1\text{,}$$ then $$x^{2n} = 1$$ for all $$n$$ and hence $$\lim\,f_n(x) = 1\text{.}$$ For all other $$x\text{,}$$ the sequence $$\{ f_n(x) \}$$ does not converge.

Often, functions are given as a series. In this case, we use the notion of pointwise convergence to find the values of the function.

#### Example6.1.3.

We write

\begin{equation*} \sum_{k=0}^\infty x^k \end{equation*}

to denote the limit of the functions

\begin{equation*} f_n(x) := \sum_{k=0}^n x^k . \end{equation*}

When studying series, we saw that for $$(-1,1)$$ the $$f_n$$ converge pointwise to

\begin{equation*} \frac{1}{1-x} . \end{equation*}

The subtle point here is that while $$\frac{1}{1-x}$$ is defined for all $$x \not=1\text{,}$$ and $$f_n$$ are defined for all $$x$$ (even at $$x=1$$), convergence only happens on $$(-1,1)\text{.}$$ Therefore, when we write

\begin{equation*} f(x) := \sum_{k=0}^\infty x^k \end{equation*}

we mean that $$f$$ is defined on $$(-1,1)$$ and is the pointwise limit of the partial sums.

#### Example6.1.4.

Let $$f_n(x) := \sin(nx)\text{.}$$ Then $$f_n$$ does not converge pointwise to any function on any interval. It may converge at certain points, such as when $$x=0$$ or $$x=\pi\text{.}$$ It is left as an exercise that in any interval $$[a,b]\text{,}$$ there exists an $$x$$ such that $$\sin(xn)$$ does not have a limit as $$n$$ goes to infinity. See Figure 6.2.

Before we move to uniform convergence, let us reformulate pointwise convergence in a different way. We leave the proof to the reader—it is a simple application of the definition of convergence of a sequence of real numbers.

The key point here is that $$N$$ can depend on $$x\text{,}$$ not just on $$\epsilon\text{.}$$ That is, for each $$x$$ we can pick a different $$N\text{.}$$ If we could pick one $$N$$ for all $$x\text{,}$$ we would have what is called uniform convergence.

### Subsection6.1.2Uniform convergence

#### Definition6.1.6.

Let $$f_n \colon S \to \R$$ and $$f \colon S \to \R$$ be functions. The sequence $$\{ f_n \}$$ converges uniformly to $$f$$ if for every $$\epsilon > 0\text{,}$$ there exists an $$N \in \N$$ such that for all $$n \geq N\text{,}$$ we have

\begin{equation*} \abs{f_n(x) - f(x)} < \epsilon \qquad \text{for all } x \in S. \end{equation*}

In uniform convergence, $$N$$ cannot depend on $$x\text{.}$$ Given $$\epsilon > 0\text{,}$$ we must find an $$N$$ that works for all $$x \in S\text{.}$$ See Figure 6.3 for an illustration.

Uniform convergence implies pointwise convergence, and the proof follows by Proposition 6.1.5:

The converse does not hold.

#### Example6.1.8.

The functions $$f_n(x) := x^{2n}$$ do not converge uniformly on $$[-1,1]\text{,}$$ even though they converge pointwise. To see this, suppose for contradiction that the convergence is uniform. For $$\epsilon := \nicefrac{1}{2}\text{,}$$ there would have to exist an $$N$$ such that $$x^{2N} = \abs{x^{2N} - 0} < \nicefrac{1}{2}$$ for all $$x \in (-1,1)$$ (as $$f_n(x)$$ converges to 0 on $$(-1,1)$$). But that means that for every sequence $$\{ x_k \}$$ in $$(-1,1)$$ such that $$\lim\, x_k = 1\text{,}$$ we have $$x_k^{2N} < \nicefrac{1}{2}$$ for all $$k\text{.}$$ On the other hand, $$x^{2N}$$ is a continuous function of $$x$$ (it is a polynomial). Therefore, we obtain a contradiction

\begin{equation*} 1 = 1^{2N} = \lim_{k\to\infty} x_k^{2N} \leq \nicefrac{1}{2} . \end{equation*}

However, if we restrict our domain to $$[-a,a]$$ where $$0 < a < 1\text{,}$$ then $$\{ f_n \}$$ converges uniformly to 0 on $$[-a,a]\text{.}$$ Note that $$a^{2n} \to 0$$ as $$n \to \infty\text{.}$$ Given $$\epsilon > 0\text{,}$$ pick $$N \in \N$$ such that $$a^{2n} < \epsilon$$ for all $$n \geq N\text{.}$$ When $$x \in [-a,a]\text{,}$$ we have $$\abs{x} \leq a\text{.}$$ So for all $$n \geq N$$ and all $$x \in [-a,a]\text{,}$$

\begin{equation*} \abs{x^{2n}} = \abs{x}^{2n} \leq a^{2n} < \epsilon . \end{equation*}

### Subsection6.1.3Convergence in uniform norm

For bounded functions, there is another more abstract way to think of uniform convergence. To every bounded function we assign a certain nonnegative number (called the uniform norm). This number measures the “distance” of the function from 0. We can then “measure” how far two functions are from each other. We then translate a statement about uniform convergence into a statement about a certain sequence of real numbers converging to zero.

#### Definition6.1.9.

Let $$f \colon S \to \R$$ be a bounded function. Define

\begin{equation*} \norm{f}_u := \sup \bigl\{ \abs{f(x)} : x \in S \bigr\} . \end{equation*}

We call $$\norm{\cdot}_u$$ the uniform norm.

To use this notation 1  and this concept, the domain $$S$$ must be fixed. Some authors use the notation $$\norm{f}_S$$ to emphasize the dependence on $$S\text{.}$$

#### Proof.

First suppose $$\lim \norm{f_n - f}_u = 0\text{.}$$ Let $$\epsilon > 0$$ be given. Then there exists an $$N$$ such that for $$n \geq N\text{,}$$ we have $$\norm{f_n - f}_u < \epsilon\text{.}$$ As $$\norm{f_n-f}_u$$ is the supremum of $$\abs{f_n(x)-f(x)}\text{,}$$ we see that for all $$x \in S\text{,}$$ we have $$\abs{f_n(x)-f(x)} \leq \norm{f_n - f}_u < \epsilon\text{.}$$

On the other hand, suppose $$\{ f_n \}$$ converges uniformly to $$f\text{.}$$ Let $$\epsilon > 0$$ be given. Then find $$N$$ such that $$\abs{f_n(x)-f(x)} < \epsilon$$ for all $$x \in S\text{.}$$ Taking the supremum we see that $$\norm{f_n - f}_u \leq \epsilon\text{.}$$ Hence $$\lim \norm{f_n-f}_u = 0\text{.}$$

Sometimes it is said that $$\{ f_n \}$$ converges to $$f$$ in uniform norm instead of converges uniformly if $$\snorm{f_n-f} \to 0\text{.}$$ The proposition says that the two notions are the same thing.

#### Example6.1.11.

Let $$f_n \colon [0,1] \to \R$$ be defined by $$f_n(x) := \frac{nx+ \sin(nx^2)}{n}\text{.}$$ We claim $$\{ f_n \}$$ converges uniformly to $$f(x) := x\text{.}$$ Let us compute:

\begin{equation*} \begin{split} \norm{f_n-f}_u & = \sup \left\{ \abs{\frac{nx+ \sin(nx^2)}{n} - x} : x \in [0,1] \right\} \\ & = \sup \left\{ \frac{\abs{\sin(nx^2)}}{n} : x \in [0,1] \right\} \\ & \leq \sup \bigl\{ \nicefrac{1}{n} : x \in [0,1] \bigr\} \\ & = \nicefrac{1}{n}. \end{split} \end{equation*}

Using uniform norm, we define Cauchy sequences in a similar way as we define Cauchy sequences of real numbers.

#### Definition6.1.12.

Let $$f_n \colon S \to \R$$ be bounded functions. The sequence is Cauchy in the uniform norm or uniformly Cauchy if for every $$\epsilon > 0\text{,}$$ there exists an $$N \in \N$$ such that for all $$m,k \geq N\text{,}$$

\begin{equation*} \norm{f_m-f_k}_u < \epsilon . \end{equation*}

#### Proof.

Let us first suppose $$\{ f_n \}$$ is Cauchy in the uniform norm. Let us define $$f\text{.}$$ Fix $$x\text{,}$$ then the sequence $$\{ f_n(x) \}$$ is Cauchy because

\begin{equation*} \abs{f_m(x)-f_k(x)} \leq \norm{f_m-f_k}_u . \end{equation*}

Thus $$\{ f_n(x) \}$$ converges to some real number. Define $$f \colon S \to \R$$ by

\begin{equation*} f(x) := \lim_{n \to \infty} f_n(x) . \end{equation*}

The sequence $$\{ f_n \}$$ converges pointwise to $$f\text{.}$$ To show that the convergence is uniform, let $$\epsilon > 0$$ be given. Find an $$N$$ such that for all $$m, k \geq N\text{,}$$ we have $$\norm{f_m-f_k}_u < \nicefrac{\epsilon}{2}\text{.}$$ In other words, for all $$x\text{,}$$ we have $$\abs{f_m(x)-f_k(x)} < \nicefrac{\epsilon}{2}\text{.}$$ For any fixed $$x\text{,}$$ take the limit as $$k$$ goes to infinity. Then $$\abs{f_m(x)-f_k(x)}$$ goes to $$\abs{f_m(x)-f(x)}\text{.}$$ Consequently for all $$x\text{,}$$

\begin{equation*} \abs{f_m(x)-f(x)} \leq \nicefrac{\epsilon}{2} < \epsilon . \end{equation*}

And hence $$\{ f_n \}$$ converges uniformly.

Next, we prove the other direction. Suppose $$\{ f_n \}$$ converges uniformly to $$f\text{.}$$ Given $$\epsilon > 0\text{,}$$ find $$N$$ such that for all $$n \geq N\text{,}$$ we have $$\abs{f_n(x)-f(x)} < \nicefrac{\epsilon}{4}$$ for all $$x \in S\text{.}$$ Therefore, for all $$m, k \geq N\text{,}$$

\begin{equation*} \abs{f_m(x)-f_k(x)} = \abs{f_m(x)-f(x)+f(x)-f_k(x)} \leq \abs{f_m(x)-f(x)}+\abs{f(x)-f_k(x)} < \nicefrac{\epsilon}{4} + \nicefrac{\epsilon}{4} . \end{equation*}

Take supremum over all $$x$$ to obtain

\begin{equation*} \norm{f_m-f_k}_u \leq \nicefrac{\epsilon}{2} < \epsilon . \qedhere \end{equation*}

### Subsection6.1.4Exercises

#### Exercise6.1.1.

Let $$f$$ and $$g$$ be bounded functions on $$[a,b]\text{.}$$ Prove

\begin{equation*} \norm{f+g}_u \leq \norm{f}_u + \norm{g}_u . \end{equation*}

#### Exercise6.1.2.

1. Find the pointwise limit $$\dfrac{e^{x/n}}{n}$$ for $$x \in \R\text{.}$$

2. Is the limit uniform on $$\R\text{?}$$

3. Is the limit uniform on $$[0,1]\text{?}$$

#### Exercise6.1.3.

Suppose $$f_n \colon S \to \R$$ are functions that converge uniformly to $$f \colon S \to \R\text{.}$$ Suppose $$A \subset S\text{.}$$ Show that the sequence of restrictions $$\{ f_n|_A \}$$ converges uniformly to $$f|_A\text{.}$$

#### Exercise6.1.4.

Suppose $$\{ f_n \}$$ and $$\{ g_n \}$$ defined on some set $$A$$ converge to $$f$$ and $$g$$ respectively pointwise. Show that $$\{ f_n+g_n \}$$ converges pointwise to $$f+g\text{.}$$

#### Exercise6.1.5.

Suppose $$\{ f_n \}$$ and $$\{ g_n \}$$ defined on some set $$A$$ converge to $$f$$ and $$g$$ respectively uniformly on $$A\text{.}$$ Show that $$\{ f_n+g_n \}$$ converges uniformly to $$f+g$$ on $$A\text{.}$$

#### Exercise6.1.6.

Find an example of a sequence of functions $$\{ f_n \}$$ and $$\{ g_n \}$$ that converge uniformly to some $$f$$ and $$g$$ on some set $$A\text{,}$$ but such that $$\{ f_ng_n \}$$ (the multiple) does not converge uniformly to $$fg$$ on $$A\text{.}$$ Hint: Let $$A := \R\text{,}$$ let $$f(x):=g(x) := x\text{.}$$ You can even pick $$f_n = g_n\text{.}$$

#### Exercise6.1.7.

Suppose there exists a sequence of functions $$\{ g_n \}$$ uniformly converging to $$0$$ on $$A\text{.}$$ Now suppose we have a sequence of functions $$\{ f_n \}$$ and a function $$f$$ on $$A$$ such that

\begin{equation*} \abs{f_n(x) - f(x)} \leq g_n(x) \end{equation*}

for all $$x \in A\text{.}$$ Show that $$\{ f_n \}$$ converges uniformly to $$f$$ on $$A\text{.}$$

#### Exercise6.1.8.

Let $$\{ f_n \}\text{,}$$ $$\{ g_n \}$$ and $$\{ h_n \}$$ be sequences of functions on $$[a,b]\text{.}$$ Suppose $$\{ f_n \}$$ and $$\{ h_n \}$$ converge uniformly to some function $$f \colon [a,b] \to \R$$ and suppose $$f_n(x) \leq g_n(x) \leq h_n(x)$$ for all $$x \in [a,b]\text{.}$$ Show that $$\{ g_n \}$$ converges uniformly to $$f\text{.}$$

#### Exercise6.1.9.

Let $$f_n \colon [0,1] \to \R$$ be a sequence of increasing functions (that is, $$f_n(x) \geq f_n(y)$$ whenever $$x \geq y$$). Suppose $$f_n(0) = 0$$ and $$\lim\limits_{n \to \infty} f_n(1) = 0\text{.}$$ Show that $$\{ f_n \}$$ converges uniformly to $$0\text{.}$$

#### Exercise6.1.10.

Let $$\{f_n\}$$ be a sequence of functions defined on $$[0,1]\text{.}$$ Suppose there exists a sequence of distinct numbers $$x_n \in [0,1]$$ such that

\begin{equation*} f_n(x_n) = 1 . \end{equation*}

Prove or disprove the following statements:

1. True or false: There exists $$\{ f_n \}$$ as above that converges to $$0$$ pointwise.

2. True or false: There exists $$\{ f_n \}$$ as above that converges to $$0$$ uniformly on $$[0,1]\text{.}$$

#### Exercise6.1.11.

Fix a continuous $$h \colon [a,b] \to \R\text{.}$$ Let $$f(x) := h(x)$$ for $$x \in [a,b]\text{,}$$ $$f(x) := h(a)$$ for $$x < a$$ and $$f(x) := h(b)$$ for all $$x > b\text{.}$$ First show that $$f \colon \R \to \R$$ is continuous. Now let $$f_n$$ be the function $$g$$ from Exercise 5.3.7 with $$\epsilon = \nicefrac{1}{n}\text{,}$$ defined on the interval $$[a,b]\text{.}$$ That is,

\begin{equation*} f_n(x) := \frac{n}{2} \int_{x-1/n}^{x+1/n} f . \end{equation*}

Show that $$\{ f_n \}$$ converges uniformly to $$h$$ on $$[a,b]\text{.}$$

#### Exercise6.1.12.

Prove that if a sequence of functions $$f_n \colon S \to \R$$ converge uniformly to a bounded function $$f \colon S \to \R\text{,}$$ then there exists an $$N$$ such that for all $$n \geq N\text{,}$$ the $$f_n$$ are bounded.

#### Exercise6.1.13.

Suppose there is a single constant $$B$$ and a sequence of functions $$f_n \colon S \to \R$$ that are bounded by $$B\text{,}$$ that is $$\abs{f_n(x)} \leq B$$ for all $$x \in S\text{.}$$ Suppose that $$\{ f_n \}$$ converges pointwise to $$f \colon S \to \R\text{.}$$ Prove that $$f$$ is bounded.

#### Exercise6.1.14.

(requires Section 2.6)   In Example 6.1.3 we saw $$\sum_{k=0}^\infty x^k$$ converges pointwise to $$\frac{1}{1-x}$$ on $$(-1,1)\text{.}$$

1. Show that whenever $$0 \leq c < 1\text{,}$$ the series $$\sum_{k=0}^\infty x^k$$ converges uniformly on $$[-c,c]\text{.}$$

2. Show that the series $$\sum_{k=0}^\infty x^k$$ does not converge uniformly on $$(-1,1)\text{.}$$

The notation nor terminology is not completely standardized. The norm is also called the sup norm or infinity norm, and in addition to $$\norm{f}_u$$ and $$\norm{f}_S$$ it is sometimes written as $$\norm{f}_{\infty}$$ or $$\norm{f}_{\infty,S}\text{.}$$
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