# Let $(f_n)$ be a sequence of continuous functions such that $f_n\to f$ uniformly on $\mathbf{R}$. Show that $\lim_{n\to\infty}f_n(x_n)=f(x)$

Let $$(f_n)$$ be sequence of continuous functions such that $$f_n\to f$$ uniformly on $$\mathbf{R}$$. Show that $$\lim_{n\to\infty}f_n(x_n)=f(x)$$ for all sequence of points $$x_n \in \mathbf{R}$$ such that $$x_n\to x$$ as $$n\to\infty$$ for $$x\in \mathbf{R}$$. I would like to know if my proof holds and have a feedback, please.

As $$f_n\to f$$ uniformly, we have by definition that: $$\forall \epsilon>0 \ \exists N \ \forall n\ge N \ \forall x \in \mathbf{R}$$: $$|f_n(x)-f(x)|<\epsilon/3$$. As $$x_n\in\mathbf{R}$$, we have as well that $$|f_n(x_n)-f(x_n)|<\epsilon/3$$ $$\forall \epsilon>0$$.

Then, $$f_n$$ is a continuous sequence of functions. Let $$a \in \mathbf{R}$$. Thus, $$\forall \epsilon>0 \ \exists\delta>0 \ \forall x \in \mathbf{R}$$: $$|x-a|<\delta\implies |f_n(x)-f_n(a)|<\epsilon/3$$.

Moreover, $$x_n$$ is a converging seuquence. Therefore $$\forall \epsilon>0 \ \exists N' \ \forall n\ge N': |x_n-x|<\epsilon$$

We have then $$\forall \epsilon>0 \ \forall n\ge n_0:=\max(N,N')$$:

$$|f_n(x_n)-f(x)|\le|f_n(x_n)-f_n(a)|+|f_n(a)-f_n(x)|+|f_n(x)-f(x)|=\epsilon/3 + \epsilon/3 + \epsilon/3 = \epsilon$$. So we showed that $$\lim_{n\to\infty}f_n(x_n)=f(x)$$

• The proof is invalid because $\delta$ also depends on $n$. Mar 10, 2021 at 20:25
• also, what is $a$ in the last paragraph? Mar 10, 2021 at 20:26

Let $$\varepsilon>0$$ be given. Choose $$N_{1}\in\mathbb{N}$$ such that $$|f_{n}(y)-f(y)|<\varepsilon/2$$ whenever $$n\geq N_{1}$$ and $$y\in\mathbb{R}$$. Since $$f$$ is continuous at $$x$$ (recall that uniform limit of continuous functions is also continuous), we may choose $$\delta>0$$ such that $$|f(y)-f(x)|<\varepsilon/2$$ whenever $$|y-x|<\delta$$. Sine $$x_n\rightarrow x$$, we may choose $$N_{2}\in\mathbb{N}$$ such that $$|x_{n}-x|<\delta$$ whenever $$n\geq N_{2}$$. Let $$N=\max(N_{1},N_{2})$$. Let $$n\geq N$$ be arbitrary. We have that $$\begin{eqnarray*} & & |f_{n}(x_{n})-f(x)|\\ & \leq & |f_{n}(x_{n})-f(x_{n})|+|f(x_{n})-f(x)|\\ & < & \varepsilon. \end{eqnarray*}$$
• Thank you for solution. I just would like to as a question concerning this type of proofs. What is the intuition behind this kind of proofs? I mean, you said to me that my proof is invalid because $\delta$ depends on $n$. But how can I know that my proof doesn't hold? Looking at triangular inequality, everything works. When for example I have to prove things on series, sequences or functions(not sequences of functions) I can see if my proof holds or not looking some "basic" examples or look if my hypothesises are enough to conclude. If you could give an advice I would really appreciate it. Mar 11, 2021 at 8:24
• Oh, I think I see. We have to use the continuity of $f$ because we want to make sure that $f(x)$ exists. Is it right? But, why the continuity of $f_n$+uniforme convergence doesn't conclude the proof? Because these 2 hypothesis implie that $f$ is continuous. I still would like to have an answer to the previous question, if possible, please. Mar 11, 2021 at 8:53
• @Daniil The problem of your proof is: Since $f_{n}$ is continuous at $x$, there exists $\delta_{n}>0$ such that $|f_{n}(y)-f_{n}(x)|<\varepsilon$ whenever $|y-x|<\delta_{n}.$ (For different function, this $\delta$ of course may be different!) Although $x_{n}\rightarrow x$, we cannot guarantee that $|x_{n}-x|<\delta_{n}$ because $\delta_{n}$ may approach $0$ much faster than $|x_{n}-x|$ does. Since the condition $|x_{n}-x|<\delta_{n}$ may not be fulfilled, we cannot infer that $|f_{n}(x_{n})-f_{n}(x)|<\varepsilon$. Mar 11, 2021 at 18:47