0
$\begingroup$

I have a sequence of continuous functions $f_n : I^k \rightarrow I^k$ converging uniformly to a continuous function $f$. Then for each $n$ I choose a point $x_n$ and since they're chosen in $I^n$ which is sequence-compact (not sure this is the right terminology) there will be a converging subsequence (which for simplicity we will call $x_n$ again) to some $x$. Will I then have that the sequence $f_n (x_n)$ converges to $f(x)$??

$\endgroup$
0
$\begingroup$

Yes, because the convergence is uniform. So, if $\varepsilon>0$, take $N\in\mathbb N$ such than $n\geqslant N$ and $y\in I^n$ implies that $\bigl\lvert f(y)-f_n(y)\bigr\rvert<\frac\varepsilon2$. Since the convergence is uniform, $f$ is continuous and so there is some $M\in\mathbb N$ such that $n\geqslant M\implies\bigl\lvert f(x)-f(x_n)\bigr\rvert<\frac\varepsilon2$. So, if $n\geqslant\max\{N,M\}$,$$\bigl\lvert f(x)-f_n(x_n)\bigr\rvert\leqslant\bigl\lvert f(x)-f(x_n)\bigr\rvert+\bigl\lvert f(x_n)-f_n(x_n)\bigr\rvert<\varepsilon.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.