Sequence of continous functions converges uniformly to a funtion then that function is continous

Question

I have to following exercise and I am not sure about its solution. Let $f_h\colon A\subset \mathbb{R}^n \rightarrow \mathbb{R}$ with $h=1,2,\dots$ a sequence of continous functions and let $f\colon A\rightarrow \mathbb{R}$ such that $f_h$ converges to $f$ uniformily on $A$. I have to prove that $f$ is continous.

Fist of all, $f_h$ converges to $f$ uniformily if and only if for any $\varepsilon>0$ exists $N\in \mathbb{N}$ such that for any $h> N$ such that \begin{equation} |f_h(x)-f(x)|\leq \varepsilon \end{equation} for any $x\in A$. Hence I can consider the following \begin{split} |f(x)-f(y)|= & |f(x)-f_h(x)+f_h(x)-f_h(y)+f_h(y)-f(y)| \\ \leq & |f(x)-f_h(x)| + |f_h(x)-f_h(y)| + |f_h(y)-f(y)| \end{split} now $|f(x)-f_h(x)|\leq \frac{\varepsilon}{3}$ for any $\varepsilon>0$ and $h> N$ because the uniform converge, $|f_h(x)-f_h(y)|\leq \frac{\varepsilon}{3}$ for any $\varepsilon>0$ beacause the continuity of $f_h$ and $|f_h(y)-f(y)|\leq \frac{\varepsilon}{3}$ for any $\varepsilon>0$ and $h> N$ because the uniform converge. hence we have \begin{equation} |f(x)-f(y)|\leq \frac{\varepsilon}{3}+\frac{\varepsilon}{3}+\frac{\varepsilon}{3}=\varepsilon \end{equation} for any $\varepsilon>0$ and from the the continuity of $f_{h}$ we have that for any $\varepsilon>0$ exists $\delta=\delta(\varepsilon)>0$ such that $|f_h(x)-f_h(y)|\leq \frac{\varepsilon}{3}$ provided that $|x-y|\leq \delta$. Consequently we have the continuity of the function $f$.

Do my solution is right?

Answer 1

Your general argument is in the right direction but your formulation is not precise. In fact, it lacks a crucial ingredient of the definition of uniform convergence. For example, instead of writing

First of all, $f_h$ converges to $f$ uniformily if and only if \begin{equation} |f_h(x)-f(x)|\leq \varepsilon \end{equation} for any $x\in A$ and for any $\varepsilon>0$

you should write

First of all, $f_h$ converges to $f$ uniformily if and only if for every $\varepsilon>0$ there exists some $N$ such that for every $h>N$ and every $x\in A$ \begin{equation} |f_h(x)-f(x)|\leq \varepsilon \end{equation}

and similarly in the continuation of your argument, where you keep claiming that some absolute value is smaller than $\varepsilon$ for every $\varepsilon>0$, which of course would imply that it is indentically zero!