Continuity of a Limit Function of a Uniformly Convergent Seq. of Functs.

# Continuity of a Limit Function of a Uniformly Convergent Sequence of Functions

Recall from the Uniform Convergence of Sequences of Functions page that a sequence of real-valued functions $(f_n(x))_{n=1}^{\infty}$ with common domain $X$ is said to be uniformly convergent to the limit function $f(x)$ if for all $\epsilon > 0$ there exists an $N \in \mathbb{N}$ such that if $n \geq N$ we have that for all $x \in X$ that:

(1)\begin{align} \quad \mid f_n(x) - f(x) \mid < \epsilon \end{align}

Now consider a sequence of continuous real-valued functions $(f_n(x))_{n=1}^{\infty}$. What can be said about the limit function $f$ if this sequence uniformly converges? The following theorem answers that question.

Theorem 1: Let $(f_n(x))_{n=1}^{\infty}$ be a sequence of real-valued functions with common domain $X$, and suppose that this sequence of functions uniformly converges to the limit function $f$. If each function $f_n : X \to \mathbb{R}$ is continuous at $c \in X$ then $f$ is continuous at $c$. |

**Proof:**Let $\epsilon > 0$ be given. If $c$ is an isolated point of each $f_n$ then $c$ will also be an isolated point of $f$ and be continuous at $c$. Assume that $c$ is instead an accumulation point.

- Since $(f_n(x))_{n=1}^{\infty}$ is uniformly convergent to the limit function $f$ we have that for $\epsilon_1 = \frac{\epsilon}{3} > 0$ that there exists an $N \in \mathbb{N}$ such that if $n \geq N$ then for all $x \in X$ we have that:

\begin{align} \quad \mid f_n(x) - f(x) \mid < \epsilon_1 = \frac{\epsilon}{3} \quad (*) \end{align}

- Now since each function in the sequence $(f_n(x))_{n=1}^{\infty}$ is continuous at $c$ we have in particular that the function $f_N(x)$ is continuous at $c$. So for $\epsilon_2 = \frac{\epsilon}{3} > 0$ there exists a $\delta^* > 0$ such that if $x \in X$ and $\mid x - c \mid < \delta^*$ then:

\begin{align} \quad \mid f_N(x) - f_N(c) \mid < \epsilon_2 = \frac{\epsilon}{3} \quad (**) \end{align}

- Therefore if $\delta = \delta^*$ we have that $(*)$ and $(**)$ will both hold and so:

\begin{align} \quad \mid f(x) - f(c) \mid &= \mid f(x) - f_N(x) + f_N(x) - f(c) \mid \\ \quad & \leq \mid f(x) - f_N(x) \mid + \mid f_N(x) - f(c) \mid \\ \quad & \leq \mid f(x) - f_N(x) \mid + \mid f_N(x) - f_N(c) + f_N(c) - f(c) \mid \\ \quad & \leq \mid f(x) - f_N(x) \mid + \mid f_N(x) - f_N(c) \mid + \mid f_N(c) - f(c) \mid \\ \quad & < \epsilon_1 + \epsilon_2 + \epsilon_1 = \frac{\epsilon}{3} + \frac{\epsilon}{3} + \frac{\epsilon}{3} = \epsilon \end{align}