theorem of sequence of continuous functions

Question

If $f_n$ is a sequence of continuous functions on $D$ and $f_n \xrightarrow{u} f$ show that $f$ is a continuous function. I found it in many demonstrantions and I wonder how to prove it!

Answer 1

As currently written, the statement is false. That is, a sequence of continuous functions $f_n$ can converge pointwise to a function $f$ that is not continuous.

The classic counterexample is $f_n = x^n$ on $[0,1]$. Then clearly each $f_n$ is continuous. But the limit is $$ f(x) = \begin{cases} 0 & \text{ if } x < 1 \\ 1 & \text{ if } x = 1 \end{cases},$$ which is not continuous.

There is a related statement. If the continuous functions $f_n$ converge to $f$ uniformly, then $f$ is continuous. This is sometimes called the Uniform Limit Theorem. More can be found at this other question on this site.

Answer 2

As mentioned in the comments, the convergence must be uniform for that to hold. To see how this could fail when you only have pointwise convergence, consider the functions $f_n(x)=\tan^{-1}(nx)$ defined on the real line. If $x>0$, then $nx$ will get arbitrarily large, and hence $f_n(x)$ will get arbitrarily close to $\frac{\pi}{2}$. Similarly if $x<0$, $f_n(x)$ will approach $-\frac{\pi}{2}$. Finally, $f_n(0)$ will always remain $0$. So the function that $(f_n)$ converges to pointwise is discontinuous at $x=0$.