# $\lim$ in $X$ and $\lim_{n\rightarrow \infty} f_n = f$ in $C(X)$, how to follow that $\lim_{n\rightarrow \infty} f_n (x_n) = f(x)$

$X$ is a compact metric space and $( C(X), ||.||_\infty )$ the space of the continous functions on X with the maximum norm.

If it holds that every sequence converges in $X$ : $$\lim _{n\rightarrow \infty} x_n = x$$ in $X$ and every sequence of functions converges in $C(X)$: $$\lim_{n\rightarrow \infty} f_n = f$$ in $C(X)$

how can one conclude by choosing ($\frac{\epsilon}{3}$) as boundary for $||f_n-f||$ and $|f(x_n)-f(x)|$ :

$$|f_{n}(x_n)-f(x)| = |f_n(x_n)+f(x_n)-f(x_n)+f_n(x)-f_n(x) - f(x)| \le 2 || f_n-f|| + |f(x_n)-f(x)| = \epsilon$$

that $$\lim_{n\rightarrow \infty} f_n(x_n)=f(x)$$ ?

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Your "every"'s in the second sentence need to "a"'s... That is, $(x_n)$ and $(f_n)$ are given convergent sequences. –  David Mitra Jan 18 '13 at 14:58
How does one find these boundaries or rather what is the reasoning behind choosing $\epsilon / 3$ ? And would this method of choosing $\epsilon /3$ also work if X were not compact? –  bakabakabaka Jan 18 '13 at 15:45

First, it seems the statement you want to prove is "if $(x_n)$ is a sequence in $X$ with $\lim x_n=x$ and $(f_n)$ is a convergent sequence in $C(X)$ with limit $f$, then $\lim\limits_{n\rightarrow\infty}f_n(x_n)=f(x)$".

In your proof, it seems you've written $$|f_n(x_n)-f(x)| \le |f_n(x_n)-f(x_n)|+|f(x_n)-f_n(x) |+| f(x_n)-f(x)|.$$ You do know that $|f_n(x_n)-f(x_n)|\le\Vert f_n-f\Vert$, but how do you conclude the same for $|f(x_n)-f_n(x) |$? This is true for sufficiently large $n$, but would require an additional argument.

An alternative method would be to write, for any fixed $N$ and arbitrary $n$: \eqalign{ |f_n(x_n)-f(x)| &=| f_n(x_n)-f_N(x_n) +f_N(x_n)-f_N(x)+f_N(x) -f(x) | \cr &\le \color{maroon}{|f_n(x_n)-f_N(x_n)|} +\color{darkgreen}{|f_N(x_n)-f_N(x)|}+\color{darkblue}{|f_N(x)-f(x)|}.}

Then, given $\epsilon>0$, you can select $N$ so large that:

$\ \ \ 1)\ \ \color{darkblue}{|f_N(x)-f(x)|}<\epsilon/3$, since $\Vert f_n-f\Vert\rightarrow 0$.

$\ \ \ 2)\ \ \color{darkgreen}{|f_N(x_n)-f_N(x)|}<\epsilon/3$ for all $n\ge N$, since $f_N$ is continuous and since $x_n\rightarrow x$,

and

$\ \ \ 3)\ \ \color{maroon}{|f_n(x_n)-f_N(x_n)|}<\epsilon/3$ for all $n\ge N$, since $\Vert f_n-f\Vert\rightarrow 0$.

It would follow then that for every $\epsilon>0$, there is an $N$ so that for all $n\ge N$, $$|f_n(x_n)-f(x)|<\epsilon;$$ which shows that $\lim\limits_{n\rightarrow\infty}f_n(x_n)=f(x)$.

Incidentally, I don't think the compactness of $X$ is required.

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Compactness of $X$ is only needed to make $C(X)$ a normed space. But it won't hurt to use only the bounded continuous functions. –  Hagen von Eitzen Jan 18 '13 at 15:57
Thank you very much. –  bakabakabaka Jan 18 '13 at 15:58