# How do we prove that we can exchange limits when $f_n\to f$ uniformly?

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*}

MY ATTEMPT

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 $$L\neq L'$$.

• 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 ? May 21, 2020 at 20:29
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 $$0 and $$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 and $$x \in E$$.
• Btw, $d_Y:Y \times Y \to \mathbb{R}$ is continuous. May 21, 2020 at 23:16