Let $(X,d_{X})$ and $(Y,d_{Y})$ be metric spaces, with $Y$ complete, and let $E$ be a subset of $X$. Let $f_{n}$ be a sequence of functions from $E$ to $Y$, and suppose that this sequence converges uniformly in $E$ to some function $f:E\to Y$. Let $x_{0}\in X$ be an adherent point of $E$, and suppose that for each $n$ the limit $\lim_{x\to x_{0};x\in E}f_{n}(x)$ exists. Then the limit $\lim_{x\to x_{0};x\in E}f(x)$ also exists and we have that \begin{align*} \lim_{n\rightarrow\infty}\lim_{x\rightarrow x_{0};x\in E}f_{n}(x) = \lim_{x\rightarrow x_{0};x\in E}\lim_{n\rightarrow\infty}f_{n}(x) \end{align*}


Let us reinforce the definitions involved.

Since $f_{n}\to f$ uniformly, let $\varepsilon > 0$. Then there corresponds a $N\geq 0$ such that for every $x\in E$ we have that \begin{align*} n\geq N \Rightarrow d_{Y}(f_{n}(x),f(x)) < \varepsilon \end{align*}

In accordance to the RHS, let $\varepsilon > 0$. Then there corresponds a $\delta > 0$ such that for every $x\in E$ we have that \begin{align*} d_{X}(x,x_{0}) < \delta \Rightarrow d_{Y}(f(x),L) < \varepsilon \end{align*}

Similarly, accoring to the LHS, let $\varepsilon > 0$. Then there corresponds a natural $M \geq 0$ such that \begin{align*} n\geq M \Rightarrow d_{Y}(L_{n},L') < \varepsilon \end{align*}

We have to prove that $L = L'$. Let us consider otherwise that $L\neq L'$.

But then I get stuck. Can someone please help me to prove it?

  • $\begingroup$ There something missing in the statement : "Then the limit $\lim_{x \to x_0,x \in E} $ also exists." What are you taking the limit of ? $\endgroup$
    – Digitallis
    May 21, 2020 at 20:29
  • $\begingroup$ Thanks for the comment. I have already fixed it. $\endgroup$
    – user0102
    May 21, 2020 at 20:31

1 Answer 1


Let $L_n:=\lim_{x \to x_0; x \in E} f_n(x)$. Since $d_Y(f_n(x),f(x))<\varepsilon$ whenever $n \geq N$, we have $$d_Y(f_n(x),f_m(x)) \leq d_Y(f_n(x),f(x))+d_Y(f(x),f_m(x)) < 2 \varepsilon$$ for all $n,m \geq N$ and all $x \in E$. And so $$\lim_{x \to x_0; x \in E}d_Y(f_n(x),f_m(x))=d_Y(L_n, L_m)\leq 2\varepsilon $$ for all $n, m \geq N$. Since $Y$ is complete and $\{L_n\}_{n=1}^\infty$ is a Cauchy sequence in $Y$, $\lim_{n \to \infty} L_n=L$ exists. So there is $N' \in \mathbb{N}$ such that $d_Y(L_n,L)<\varepsilon$ whenever $n \geq N'$. Let $n_0:=\max\{N,N'\}$. We may choose $\delta>0$ so that $d_Y(f_{n_0}(x),L_{n_0}) < \varepsilon$ whenever $0<d_X(x,x_0)<\delta$ and $x \in E$. We now combine to see that we have \begin{aligned} d_Y(f(x),L) &\leq d_Y(f(x),f_{n_0}(x)) + d_Y(f_{n_0}(x),L_{n_0})+d_Y(L_{n_0},L) \\& <\varepsilon+\varepsilon+\varepsilon=3\varepsilon \end{aligned} whenever $0<d_X(x,x_0)<\delta$ and $x \in E$.

  • $\begingroup$ Btw, $d_Y:Y \times Y \to \mathbb{R}$ is continuous. $\endgroup$
    – M A Pelto
    May 21, 2020 at 23:16

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