📂Analysis

Uniform Convergence and Continuity of Function Sequences

Theorem¹

Suppose that a sequence of functions $\left\{ f_{n} \right\}$ on a metric space $E$ converges uniformly to $f$.

$$ f_{n} \rightrightarrows f $$

For a limit point $x$ of $E$, let the sequence $A_{n}(x)$ be given as follows.

$$ A_{n}(x) = \lim\limits_{t \to x} f_{n}(t) $$

Then $\left\{ A_{n}(x) \right\}$ converges, and its limit is as follows.

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

In other words, the following holds.

$$ \lim \limits_{n\to \infty}\lim \limits_{t\to x}f_{n}(t) = \lim \limits_{t\to x}\lim \limits_{n\to \infty}f_{n}(t) \tag{1} $$

Corollary

If $f_{n}$ is continuous at $x \in E$ and $f_{n}$ converges uniformly to $f$, then $f$ is also continuous at $x$.

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

Explanation

The result $(1)$ of the theorem says that the value does not change even when the positions of the two limit symbols $\lim \limits_{n \to \infty}$ and $\lim\limits_{t \to x}$ are swapped. It can also be said that uniform convergence preserves continuity.

The reason we consider uniform convergence of function sequences in connection with continuity is that pointwise convergence does not preserve continuity.

Counterexample²

The fact that a sequence of functions $f_{n}$ consisting of continuous (differentiable) functions converges pointwise to $f$ does not guarantee that $f$ is continuous (differentiable).

Proof

The function $f_{n}(x) = x^{n}$ is continuous (differentiable) on $[0, 1]$. Now let us define the function $f$ as follows.

$$ f (x) = \begin{cases} 0 & \text{if } 0 \le x \lt 1 \\ 1 & \text{if } x = 1 \end{cases} $$

Then at every point $x \in [0, 1]$, $f_{n}(x)$ converges pointwise to $f(x)$. But clearly $f$ is not continuous (differentiable) at $x = 1$.

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Proof

Suppose a small positive number $\varepsilon >0$ is given. By assumption, $\left\{ f_{n} \right\}$ converges uniformly to $f$, and since this is equivalent to $\left\{ f_{n}(t) \right\}$ being a Cauchy sequence for every $t \in E$, there exists a natural number $N$ satisfying the following.

$$ n, m \ge N, t \in E \implies \left| f_{n}(t)-f_{m}(t) \right| \lt \varepsilon $$

Moreover, taking the limit $t \to x$ in the above expression, we obtain the following. For $n, m \ge N$,

$$ \left| A_{n} - A_{m} \right| \lt \varepsilon $$

Therefore $\left\{ A_{n}(x) \right\}$ is a Cauchy sequence and hence converges. Let its limit be $A$.

$$ A = \lim\limits_{n \to \infty} A_{n} $$

Now consider the following inequality.

$$ \left| f(t) - A \right| \le \left| f(t) - f_{n}(t) \right| + \left| f_{n}(t) - A_{n} \right| + \left| A_{n} - A \right| \tag{2} $$

Now since $f_{n} \to f$ and $A_{n} \to A$, let us choose an $n$ satisfying the following two expressions.

$$ \left| f(t) - f_{n}(t) \right| \le \frac{\varepsilon}{3} \quad \forall t \in E $$

$$ \left| A_{n} - A \right| \le \frac{\varepsilon}{3} $$

And for this chosen $n$, (since $A_{n}(x) = \lim\limits_{t \to x} f_{n}(t)$) we can choose a neighborhood $V$ of $x$ satisfying the following.

$$ \left| f_{n}(t) - A_{n}(x) \right| \le \frac{\varepsilon}{3} \qquad x \ne t \in V \cap E $$

Now from inequality $(2)$ and the above results, we obtain the following.

$$ \left| f(t) - A \right| \le \frac{\varepsilon}{3} + \frac{\varepsilon}{3} + \frac{\varepsilon}{3} = \varepsilon \qquad x \ne t \in V \cap E $$

Rewriting this, we get the following.

$$ \left| f(t) - \lim\limits_{n \to \infty} A_{n}(x) \right| \le \varepsilon \qquad x \ne t \in V \cap E $$

Therefore the following holds.

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

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Proof of the Corollary³

The proof below establishes the corollary without using the result of the main theorem.

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Suppose $\varepsilon >0$ is given. Since $\left\{ f_{n} \right\}$ converges uniformly to $f$, by definition there exists a natural number $N$ satisfying the following.

$$ \begin{equation} n\ge N,\ t\in E \implies \left| f_{n}(t)-f(t) \right| < \frac{\varepsilon}{3} \label{eq1} \end{equation} $$

Since we assumed that $f_{n}$ is continuous at $x$, there exists a $\delta >0$ satisfying the following.

$$ \begin{equation} \left| t-x \right|<\delta \implies \left| f_{n}(t)-f_{n}(x) \right|\le \frac{\varepsilon}{3} \label{eq2} \end{equation} $$

Therefore, by $(1), (2)$, when $\left| t-x \right| \lt \delta$ the following expression holds.

$$ \begin{align*} \left| f(t)-f(x) \right| &= \left| f(t)-f_{n}(t)+f_{n}(t)-f_{n}(x)+f_{n}(x)-f(x) \right| \\ &\le \left| f(t)-f_{n}(t)\right|+\left| f_{n}(t)-f_{n}(x) \right|+ \left| f_{n}(x)-f(x) \right| \\ &\le \frac{\varepsilon}{3}+ \frac{\varepsilon}{3} + \frac{\varepsilon}{3} \\ &= \varepsilon \end{align*} $$

Therefore $f$ is continuous at $x$.

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