Suppose we have continuous $f_{n} : D \rightarrow \mathbb{R}$ converging to $f : D \rightarrow \mathbb{R}$ uniformly on $D$. Then, is $f$ continuous?

Question

Suppose we have continuous $f_{n} : D \rightarrow \mathbb{R}$ converging to $f : D \rightarrow \mathbb{R}$ uniformly on $D$. Then, is $f$ continuous?

I know that if we have $f_{n}$ converging to $f$ and $f_{n}$ are all continuous then it is not necessary for $f$ to be continuous. This is shown by

$$x^{n} \hspace{1cm} 0 < x \leq 1 $$

for $n \geq 0$. But, it is the uniformly part that gets me here. Is this statement true?

Answer 1

Fix $d \in D$. Now, it is always true that

$$ \begin{align} |f(d) - f(y)| &\leq |f(d) - f_n(d)| + |f_n(d)-f_n(y)| + |f_n(y)-f(y)| \\ &\leq 2d(f_n,f) + |f_n(d)-f_n(y)|. \tag{1} \end{align} $$

You also know that $f_n \to f$, that is $d(f_n,f) \to 0$, and that each function $f_n$ is continuous. A solution follows, but I would encourage you to think about this for a while before reading it.

Let $\varepsilon > 0$. Now, there exists $m \gg 1$ such that $2d(f_n,f) < \varepsilon/2$ when $n \geq m$. Since $f_m$ is continuous, there exists $\delta > 0$ such that $|d-y| < \delta $ implies $|f_m(d)-f_m(y)| < \varepsilon /2$. Thus, if $|d-y| < \delta$, by $(1)$ when $n = m$, we have that $|f(d)-f(y)| < \varepsilon$, and so $f$ is continuous.