Define $f_n \colon [0,1] \to \R$ as \begin{equation*} f_n(x) \coloneqq \begin{cases} 1-nx & \text{if } x \lt \dfrac{1}{n},\\ 0 & \text{if } x \geq \dfrac{1}{n}. \end{cases} \end{equation*}

Note that $f_n$ is continuous. Consider a fix $x \in (0,1].$
If $n \geq \dfrac{1}{x},$ then $x \geq \dfrac{1}{n}.$ Thus for $n \geq \dfrac{1}{x},$ we have that $f_n(x) = 0,$ and so \begin{equation*} \lim_{n \to \infty} f_n(x) = 0. \end{equation*}

Now if $x=0,$ then $\ds \lim_{n \to \infty} f_n(0)$ $\ds = \lim_{n \to \infty} 1 $ $ = 1.$

Thus $f_n$ converges pointwise to the function $f \colon [0,1] \to \R$ defined by \begin{equation*} f(x) \coloneqq \begin{cases} 1 & \text{if } x = 0,\\ 0 & \text{if } x > 0. \end{cases} \end{equation*} Observe that the function $f$ is not continuous at 0.

Theorem 9.4.1. Let $\{ f_n \}$ be a sequence of continuous functions $f_n \colon S \to \R$ converging uniformly to $f \colon S \to \R$. Then $f$ is continuous.

Theorem 9.4.1. Let $\{ f_n \}$ be a sequence of continuous functions $f_n \colon S \to \R$ converging uniformly to $f \colon S \to \R$. Then $f$ is continuous.

Proof. Fix $c\in S$ and let $\epsilon\gt 0.$ Choose $N$ so that

Since $f_N$ is continuous, there exists a $\delta \gt 0 $ such that

Theorem 9.4.1. Let $\{ f_n \}$ be a sequence of continuous functions $f_n \colon S \to \R$ converging uniformly to $f \colon S \to \R$. Then $f$ is continuous.

Proof.

$\Ra \abs{f(x)-f(c)} $

$ \ds\,\quad \lt \frac{\epsilon}{3} + \frac{\epsilon}{3} + \frac{\epsilon}{3} $ $ \ds = \epsilon. \;$ Hence, $f$ is continuous. $ \;\bs$

Define $f_n \colon [0,1] \to \R$ as \begin{equation*} f_n(x) \coloneqq \begin{cases} n^2x & \text{if } 0\leq x\leq \dfrac{1}{n},\\ -n^2\left(x-\dfrac{2}{n}\right) & \text{if } \dfrac{1}{n} \leq x \leq \dfrac{2}{n},\\ 0 & \text{if } \dfrac{2}{n}\leq x \leq 1. \end{cases} \end{equation*}

Since each of the functions $f_n$ is continuous on $[0,1]$, then $f$ Riemann integrable. For $n\geq 2$, using the FTC,

Now to compute $\lim f_n$ fix an $x\in[0,1].$ For $n\geq \dfrac{2}{x}$, we have $x \geq \dfrac{2}{n}$ and so $f_n(x) = 0$. Therefore, $\lim f_n = 0.$

Thus we have

Since $f_n = F'_n$ where $F_n(x):= -e^{-nx^2} ,$ then

We also have that $f(x)= \ds \lim_{n\to \infty} f_n = 0$ for all $x\in[0,1]$. Hence

Theorem 9.4.2. Let $\{ f_n \}$ be a sequence of Riemann integrable functions $f_n \colon [a,b] \to \R$ converging uniformly to $f \colon [a,b] \to \R.$ Then $f$ is Riemann integrable, and \begin{equation*} \int_a^b f = \lim_{n\to\infty} \int_a^b f_n . \end{equation*}

Suppose we wish to compute \begin{equation*} \lim_{n\to\infty} \int_0^1 \frac{nx+ \sin(nx^2)}{n} \,dx . \end{equation*}

Using Theorem 9.4.2, we have

If convergence is only pointwise, the limit need not even be Riemann integrable. On $[0,1]$ define \begin{equation*} f_n(x) \coloneqq \begin{cases} 1 & \text{if } x = \dfrac{p}{q} \text{ in lowest terms and } q \leq n, \\ 0 & \text{otherwise.} \end{cases} \end{equation*}

If the convergence is only pointwise, the limit of bounded functions is not even necessarily bounded. Define $f_n \colon [0,1] \to \R$ by \begin{equation*} f_n(x) \coloneqq \begin{cases} 0 & \text{if } x \lt \dfrac{1}{n},\\ \dfrac{1}{x} & \text{else.} \end{cases} \end{equation*}

Now we are going to consider derivatives of sequences of functions.

Let $\ds f_n(x) \coloneqq \frac{\sin(nx)}{n}$. Then $f_n$ converges uniformly to $0.$

The derivative of the limit is 0. But $f_n'(x) = \cos(nx)$, which does not converge even pointwise, for example consider $$f_n'(\pi) = {(-1)}^n.$$

Also note that $f_n'(0) = 1$ for all $n$, which converges, but not to $0$.

Let $\ds f_n(x) \coloneqq \dfrac{1}{1+nx^2}$. If $x \not= 0$, then $$\lim_{n \to \infty} f_n(x) = 0,$$ but $\lim_{n \to \infty} f_n(0) = 1$.

Thus, $\{ f_n \}_{n=1}^\infty$ converges pointwise to a function that is not continuous at $0$.

Let $\ds f_n(x) \coloneqq \dfrac{1}{1+nx^2}$. If $x \not= 0$, then $\lim_{n \to \infty} f_n(x) = 0,$ but $\lim_{n \to \infty} f_n(0) = 1$.

Thus, $\{ f_n \}_{n=1}^\infty$ converges pointwise to a function that is not continuous at $0$.

In this case we have that $\ds f_n'(x)$ $\ds = \frac{-2 n x}{\left(1+ n x^2\right)^2} .$

For every $x,$ $\lim_{n\to\infty} f_n'(x) = 0,$ then the derivatives converge pointwise to $0.$ However, the convergence is not uniform on any interval containing $0$.

The limit of $f_n$ is not differentiable at $0$. In fact, it is not even continuous at $0$.

1. If $x \not= 0$, then $\lim_{n \to \infty} f_n(x) = 0,$ but $\lim_{n \to \infty} f_n(0) = 1$.

2. For every $x$, $\lim_{n\to\infty} f_n'(x) = 0$. The convergence is not uniform on any interval containing $0$.

Thus we have that

The Fourier series of a real function $f$ defined on the interval $(-p,p)$ is given by

where

The Fourier series is named in honor of Jean Baptiste Joseph Fourier (1768-1830).

Example: Consider the function defined by $f(x)=x+\pi$ for $-\pi < x < \pi $ and $f(x)=f(x+2\pi)$ for $-\infty < x < \infty$. Its Fourier series expansion is \[ f(x)=\pi+2\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n+1}}{n}\sin (n\,x). \]