Suppose that each $f_n$ is continuous, it converges uniformly to $f$ and $x_n$ converges to $x$. Prove that $f_n(x_n)$ converges to $f(x)$.

Question

Let $f_{n}$ be a sequence of continuous functions from one metric space $(X,d_{X})$ to another $(Y,d_{Y})$, and suppose that this sequence converges uniformly to another function $f:X\to Y$. Let $x_{n}$ be a sequence of points in $X$ which converge to some limit $x$. Then $f_{n}(x_{n})$ converges (in $Y$) to $f(x)$.

MY ATTEMPT

I know that continuous functions between metric spaces map convergent sequences onto convergent sequences. I do also know that uniform convergence preserves continuity. However I am not able to apply such results in order to prove the desired result. I feel that we should apply the triangle inequality, but still I am not sure how to do it.

Can someone help me with this?

Answer 1

Use the classical decomposition: \begin{align} d_Y(f_n(x_n), f(x)) \leq d_Y(f_n(x_n), f(x_n)) + d_Y(f(x_n), f(x)) \end{align}

Analyze the first term using uniform convergence of $f_n$ to $f$. Analyze the second term using the continuity of $f$.

Answer 2

hint

Given an $ \; \epsilon>0$,

for $n $ great enough, $$d(f_n(x_n),f(x_n))\le$$ $$\sup_{x\in X}d(f_n(x),f(x))<\frac{\epsilon}{2},$$

$x_n\to x$ and $ f $ continuous at $x$ implies $$d(f(x_n),f(x))\le \frac{\epsilon}{2},$$

Now use the triangular inequality $$(\forall n\in \Bbb N)\;\;$$ $$ d(f_n(x_n),f(x))\le d(f_n(x_n),f(x_n))+d(f(x_n),f(x))$$