Question
Given a sequence of continuous functions $(f_n)_{n \in \mathbb N}$ and define
$$
f : X \rightarrow Y, \quad f(x) = \lim_{n \rightarrow \infty} f_n(x)
$$
where $X$ and $Y$ are metric spaces.
If $f$ is continuous, is it true that $(f_n)$ converges uniformly to $f$?
Are there any restrictions on $(f_n)$, $f$, $X$ and/or $Y$ for this to be true?
Thoughts
This seems like a very useful result, yet it does not seem to be mentioned anywhere which leads me to think it is either completely obvious, or false. This site usually has a question or two pertaining to 'obvious' results, and my attempts to find them have been unsuccessful, though I have found Dini's theorem which relaxes the condition of continuity but requires the sequence to be monotonic and $X$ to be compact.
I know this statement is false if $(f_n)$ are discontinuous functions. Take as counter-example
$$
f_n : \mathbb R \rightarrow \mathbb R, \quad f_n(x) = \begin{cases} x^n & x \in [0, 1) \\ 0 & \text{Else} \end{cases}
$$
then $f \equiv 0$ is continuous, but $(f_n)$ does not converge uniformly to it.
I have attempted a proof below, and if nothing is wrong, then we require that $X$ is compact.
Attempted Proof
We wish to show that given $\varepsilon > 0$, there exists an $N$ such that for $n > N$, then
$$
\forall x \in X, \quad d(f_n(x), f(x)) < \varepsilon.
$$
Since $f_n$ are continuous, then for $x \in X$, there exists a neighbourhood $N_x \subseteq X$ such that
$$
\forall y \in N_x, \quad d(f_n(x), f_n(y)) < \varepsilon
$$
and similarly with $f$.
Additionally, we know that for $x \in X$, there exists $N(x) \in \mathbb N$ such that
$$
\forall n > N(x), \quad d(f_n(x), f(x)) < \varepsilon
$$
We now set $\varepsilon > 0$, and suppose there does not exist an $N$ such that for $n > N$, then $\sup_{x \in X} \{d(f_n(x), f(x))\} < \varepsilon$. This means we can always pick some $x_0 \in X$ such that $d(f_n(x_0), f(x_0)) \geq \varepsilon$.
By the triangle inequality, we have that $\forall x \in X$
$$
d(f_n(x_0), f(x_0)) \leq d(f_n(x_0), f_n(x)) + d(f_n(x), f(x)) + d(f(x), f(x_0))
$$
As $f_n$ and $f$ are continuous, then we can pick $x$ in a neighbourhood such that the first and third term are less than any $\varepsilon > 0$. In other words, we have that
$$
d(f_n(x_0), f(x_0)) < d(f_n(x), f(x))
$$
Thus in the neighbourhood of any 'badly behaving' point, there exists another badly behaving point.
If we require $X$ to be compact, then there exists a point $y \in X$ such that $d(f_n(y), f(y)) = \sup_{x \in X}\{ d(f_n(x), f(x)) \}$ (since $f_n$, $f$ and $d$ are all continuous, thus the image of the distance function must be compact too), in which case the above result would lead to a contradiction.
—
Some of the answer have shown the proof to be false. Why?
Best Answer
That the claim is false even on $[0,1]$ can be seen by considering a function $f_n$ which is zero everywhere except for a peak between $(n-1)/n$ and $1$, with a maximum of $1$ and $f_n(1) = 0$.