Uniform convergence of sequence of functions

Consider a sequence of functions $\{f_n(x)\}$ on interval $I$:

$$f_1(x),f_2(x),f_3(x),\cdots .$$

This is an indexed collection of functions on $I$. We can consider different types of convergence of such a sequence of functions: $f_n(x)\to f(x)$. If the convergence is ``uniform,'' then some properties of the functions $f_n(x)$, such as continuity, integrability, and differentiability, are inherited to the limiting function $f(x)$.

Given a sequence of functions $\{f_n(x)\}$ on $I$, we can define a sequence of real numbers $\{f_n(a)\}$ for each $a\in I$. If the sequence $\{f_n(a)\}$ converges for each $a\in I$, we can define its limit as $f(a)$ thereby defining a function $f(x)$ on $I$. In this case, we say the sequence of functions $\{f_n(x)\}$ converges to the function $f(x)$ and write

$$\underset{n\to \mathrm{\infty }}{lim}{f}_n(x)=f(x).$$

We refer to this type of convergence as point-wise convergence. 

Remark. In a logical form, the point-wise convergence is denoted as

$$\mathrm{\forall }a\in I\mathrm{\forall }\epsilon >0\mathrm{\exists }N\in \mathbb{N}\mathrm{\forall }n\in \mathbb{N}(n\ge N⟹|f(a)-f_n(a)|<\epsilon ).$$

Note in particular that the value of $N$ may depend on $a$ in this expression. □

On the contrary, we say the sequence of functions $\{f_n(x)\}$ uniformly converges to the function $f(x)$ if the following holds:

$$\mathrm{\forall }\epsilon >0\mathrm{\exists }N\in \mathbb{N}\mathrm{\forall }a\in I\mathrm{\forall }n\in \mathbb{N}(n\ge N⟹|f(a)-f_n(a)|<\epsilon ).$$

Note the difference of the position of the quantifier $\mathrm{\forall }a\in I$. In this expression, the value of $N$ does not depend on $a$ (but does depend on $\epsilon$)  That's why this is called "uniform" convergence. This corresponds to the condition (*) in the proof of the Theorem (Continuous power series).

See also: Calculus of power series

Thus, we have the following theorem.

Theorem (Uniform convergence and continuity)

Suppose that the sequence $\{f_n(x)\}$ of functions on interval $I$ uniformly converges to the function $f(x)$ on $I$. If each $f_n(x)$ is continuous on $I$, then so is $f(x)$.

Proof. Exercise. (See Step 2 of the proof of the Theorem (Continuous power series) in Calculus of power series.) ■

Example. For each $n\in \mathbb{N}$, let us define the function $f_n(x)$ on $\mathbb{R}$ as follows:

$$f_n(x)=\{\begin{array}{cc}0& (x\le 0),\\ nx& (0<x<\frac{1}{n}),\\ 1& (x\ge \frac{1}{n}).\end{array}$$

The sequence of function $\{f_n(x)\}$ converges to the following function $f(x)$:

$$f(x)=\{\begin{array}{cc}0& (x\le 0),\\ 1& (x>0).\end{array}$$

For each $n$, $f_n(x)$ is a continuous function. However, $f(x)$ is not continuous. In fact, $\underset{n\to \mathrm{\infty }}{lim}{f}_n(x)=f(x)$ is point-wise convergence, not uniform convergence. □

Exercise. (Try to) show directly that the sequence $\{f_n(x)\}$ in the above example does not converge uniformly to $f(x)$. □

Theorem (Uniform convergence and definite integral)

Suppose that the sequence $\{f_n(x)\}$ of functions on a bounded interval $I$ uniformly converges to the function $f(x)$ on $I$. For each $a\in I$, let $F_n(x)={\int }_a^x{f}_n(t)dt$ and $F(x)={\int }_a^{x}f(t)dt$. Then the sequence of functions $\{F_n(x)\}$ uniformly converges to $F(x)$ on $I$.

Proof. Suppose $I\subset [\alpha ,\beta ]$. For any $\epsilon >0$, we can find a natural number $N$ such that, for all $n\ge N$ and for any $x\in I$, $|f(x)-f_n(x)|<\frac{\epsilon }{\beta -\alpha }$ (uniform convergence). Then, for any $x\in I$, we have

$$|F(x)-F_n(x)|\le {\int }_a^x|f(t)-f_n(t)|dt<\epsilon \cdot \frac{|x-a|}{\beta -\alpha }\le \epsilon .$$

This means that $\{F_n(x)\}$ uniformly converges to $F(x)$ on $I$. ■

Theorem (Uniform convergence and derivative)

Let $\{f_n(x)\}$ be a sequence of $C^1$ functions that converges to $f(x)$ on $I$ (point-wise). Suppose that the sequence of derivatives $\{f_n^{\prime }(x)\}$ uniformly converges to the function $g(x)$ on $I$. Then, $f(x)$ is also of class $C^1$ and $f^{\prime }(x)=g(x)$.

Proof. On the one hand, by the above Theorem (Uniform convergence and definite integral), ${\int }_a^x{f}_n^{\prime }(t)dt$ uniformly converges to ${\int }_a^{x}g(t)dt$ on $I$ where $a\in I$ is arbitrary. On the other hand,

$${\int }_a^x{f}_n^{\prime }(t)dt=f_n(x)-f_n(a)\to f(x)-f(a)$$

as $n\to \mathrm{\infty }$. Thus, the convergence of $\{f_n(x)\}$ to $f(x)$ is uniform and

$$f(x)-f(a)={\int }_a^{x}g(t)dt.$$

By differentiating both sides with respect to $x$, we have $f^{\prime }(x)=g(x)$. ■