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 $\lim_{n\to\infty}f_n(x_n)=f(x)$

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

1 Answer 1


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*}

  • $\begingroup$ 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. $\endgroup$
    – Daniil
    Mar 11, 2021 at 8:24
  • $\begingroup$ 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. $\endgroup$
    – Daniil
    Mar 11, 2021 at 8:53
  • 1
    $\begingroup$ @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 $|f_{n}(x_{n})-f_{n}(x)|<\varepsilon$. $\endgroup$ Mar 11, 2021 at 18:47

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