Uniform convergence of a sequence of functions?

Question

let $f_n: R \rightarrow R$ be a sequence of continuous functions which converge uniformly to $f: R \rightarrow R$. Let $(x_n)$ be a sequence of real numbers which converges to $x \in R$. Show that $f_n(x_n) \rightarrow f(x)$.

So far this is my attempt at a solution but I wonder whether I can provide a shorter or more insightful proof:

1) $|f_n(x_n)-f(x)| \leq |f_n(x_n)-f_n(x)| + |f_n(x)-f(x)|$ and I argue that we can show that the expression on the right approaches zero as $n \rightarrow \infty$.

It's given that $\lim_{n \to +\infty}|f_n(x)-f(x)| = 0$ since $f_n \rightarrow f$ uniformly.

2) $\forall n, f_n(x)$ is continuous so $\forall \epsilon > 0 \exists \delta > 0, |f_n(x_n)-f_n(x)|<\epsilon \text{ when } |x_n-x|<\delta$.

Clearly, $\lim_{n \to +\infty}|f_n(x)-f_n(x)| = 0$ since $\lim_{n \to +\infty}|x_n-x| = 0$.

Q.E.D.

Answer 1

In step 2, your choice of $\delta$ will depend on $n$ since it depends on the function. You would need some form of "equicontinuity" to make the argument valid. It is easily corrected, though. Use the triangle inequality to write $$|f_n(x_n) - f(x)| \le |f_n(x_n) - f(x_n)| + |f(x_n) - f(x)|.$$ Uniform convergence will take care of the first term, continuity the second term.

Answer 2

Let $\varepsilon>0$. We want to show that for any such $\varepsilon$, there is an $N$ so that for every $n>N$, $\lvert f_n(x_n)-f(x) \rvert < \varepsilon$.

The triangle inequality gives $$ \lvert f_n(x_n)-f(x) \rvert \leqslant \lvert f_n(x_n)-f(x_n) \rvert + \lvert f(x_n)-f(x) \rvert $$

For the first term, you have to say that because the convergence is uniform, you can choose $N$ independent of $x_n$ so that $$\lvert f_n(x_n)-f(x_n) \rvert < \varepsilon/2 \tag{1}$$ for every $n>N$.

For the second term, $f$ is a limit of a uniformly convergent set of functions, so it is continuous. Therefore there is a $\delta>0$ so that $\lvert f(y)-f(x) \rvert< \varepsilon/2$ whenever $\lvert y-x \rvert < \delta$. Next, the sequence $x_n$ converges to $x$, so there is an $M>0$ such that $\lvert x_n-x \rvert< \delta$ for every $n>M$. Hence for every $n>M$, $$\lvert f(x_n)-f(x) \rvert < \varepsilon/2. \tag{2} $$

Now let $K=\max{\{N,M\}}$. For any $n>K$, both (1) and (2) hold, and so $$ \lvert f_n(x_n)-f(x) \rvert < \varepsilon/2+\varepsilon/2=\varepsilon. $$

Answer 3

given $f_n \to f$ uniformly so, for $\epsilon > 0,$ $ \exists n_1 \in \mathbb{N}$ such that $\forall n > n_1$, $|f_n(x)-f(x)|<\epsilon , \forall x$ in particular for $x_n$ again uniform convergence implies $f$ is continuous so $f(x_n) \to f(x) \implies \exists n_2\in \mathbb{N} | \forall n>n_2 ,|f(x_n)-f(x)|<\epsilon$ now the conclusion follows using triangle inequality