📂Analysis

Necessary and Sufficient Condition for Uniform Convergence

Theorem 1

Let us suppose that a sequence of functions $\left\{ f_{n} \right\}$ defined on the metric space $E$ is given. The following two conditions are equivalent.

• $\left\{ f_{n} \right\}$ converges uniformly on $E$.
• For all $\varepsilon>0$, there exists a natural number $N$ such that the following equation holds. $$ \begin{equation} \quad m,n\ge N,\ x\in E \implies \left| f_{n}(x)-f_{m}(x) \right| \le \varepsilon \end{equation} $$

Explanation

In other words, for all $x \in E$, $\left\{ f_{n}(x) \right\}$ being a Cauchy sequence is equivalent to $\left\{ f_{n} \right\}$ converging uniformly on $E$.

Proof

• $(\implies)$

  Assume that $\left\{ f_{n} \right\}$ converges uniformly to $f$. Then, by definition, there exists a natural number $N$ such that the following equation holds.

  $$ n \le N, x\in E \implies \left| f_{n}(x)- f(x) \right| \le \frac{\varepsilon}{2} $$

  Therefore, for $n,m\ge N$, $x\in E$, the following equation holds.

  $$ \begin{align*} \left| f_{n}(x)-f_{m}(x) \right| &= \left| f_{n}(x)-f(x)+f(x)-f_{m}(x) \right| \\ &\le \left| f_{n}(x)-f(x) \right| + \left|f(x)-f_{m}(x) \right| \\ &\le \frac{\varepsilon}{2}+\frac{\varepsilon}{2} = \varepsilon \end{align*} $$

• $(\impliedby)$

  By assumption, $\left\{ f_{n}(x) \right\}$ is a Cauchy sequence and thus convergent. Let this limit be $f(x)$. Then $\left\{ f_{n} \right\}$ converges pointwise to $f$ on $E$.

  Now, to show that this convergence is uniform will complete the proof. Suppose $\varepsilon >0$ is given. Choose $N$ such that $(1)$ holds. Taking the limit $m \to \infty$ for the fixed $n$ in $(1)$, we get the following.

  $$ \lim \limits_{m\to \infty}f_{m}(x)=f(x) $$

  Therefore, for all $n\ge N$ and $x\in E$, the following holds.

  $$ \lim \limits_{m\to \infty}\left| f_{n}(x)-f_{m}(x) \right| =\left| f_{n}(x)-f(x) \right| \le \varepsilon $$

  Hence, $\left\{ f_{n} \right\}$ converges uniformly to $f$.

  ■

Theorem 2

Let us assume that the following holds for the metric spaces $E$ and $x \in E$.

$$ \lim \limits_{n\to \infty} f_{n}(x) =f(x) $$

Let’s set $M_{n}$ as follows.

$$ M_{n}=\sup \limits_{x\in E}\left| f_{n}(x)-f(x) \right| $$

Then, the following two conditions are equivalent:

• $\left\{ f_{n} \right\}$ converges uniformly to $f$ on $E$.

• $\lim \limits_{n \to \infty}M_{n}=0$

Explanation

Considering the definition of uniform convergence, it is just another way of describing the same concept.