Let $(X,d)$ be a metric space, and $f_n: X \to \mathbb{R}$ s.t $f_n \to f$ uniformly, f continuous. Show that if $x_n \to x$ then $f_n(x_n) \to f(x)$

Question

I cant see why I would need uniform convergence and not just pointwise for the result here. My attempt:

Let $\epsilon >0$. Since $x_n \to x$ then $\exists \ n_1 \ $ s.t $n>n_1 \implies d(x_n, x) <\delta \implies |f(x_n)-f(x)| < \dfrac{\epsilon}{2}$. Also, since $f_n \to f$, $\exists \ n_2 \ $ s.t $n>n_2 \implies |f_n(y)-f(y)| < \dfrac{\epsilon}{2}$.

Then, if $n > n_0 = \max\{n_1, n_2\} \implies |f_n(x_n)-f(x)| \leq |f_n(x_n) - f(x_n)| + |f(x_n)-f(x)| < \epsilon$.

Now, I don't get why pointwise convergence doesn't do it here, since i'm just using $f_n(x_n) \to f(x_n)$.

Answer 1

The flaw in your proof is that the statement $|f_n(y)-f(y)|<\epsilon/2$ is true for fixed $y$. So you cannot substitute $y=x_n$, which depends on $n$. The quantity $|f_n(x_n) - f(x_n)|$ behaves quite differently from the quantity $|f_n(y)-f(y)|$, since both the function $f_n$ and the point $x_n$ being observed are now varying with $n$. In particular, you cannot say that $f_n(x_n)$ is converging pointwise to $f(x_n)$, since $n$ is still present in $f(x_n)$.

Answer 2

Your proof does need uniform convergence. Pointwise convergence does not suffice.

Consider the metric space $[0,1]$ with the usual metric, and let $f_n(x)=nx(1-x)^n$. Then $f_n(x) \to 0$ as $n \to \infty$ for each point $x \in [0,1]$, yet $$f_n(1/n)=(1-1/n)^n \to e \quad \text{as} \quad n \to \infty \,.$$