# uniform convergence preserve sequences

This exercise is taken from Terence Tao's book. Let $$(f^n)_{n = 1}^{\infty}$$ be a sequence of continuous functions from one metric space $$(X,d_X)$$ to another $$(Y,d_Y)$$, and suppose that this sequence converges uniformly to another function $$f : X \rightarrow Y$$. Let $$(x^n)$$ be a sequence of points in X which converge to some limit x. Then $$f^n(x^n)$$ converges (in Y) to $$f(x)$$.

Attempt:

We know that the norms are continuous:

$$d_Y(f^n(x^n) , f(x) ) \leq d_Y( f^n(x^n), f(x^n)) + d_Y( f(x^n), f(x)) \leq lim_{n \rightarrow \infty} d_Y( f^n(x^n), f(x^n) ) + d_Y( f(x^n), f(x) ) \leq d_Y( f(x), f(x)) + d_Y( f(x^n), f(x)) \rightarrow 0$$.

## 1 Answer

After the first inequality your proof is not correct. The correct way to write the proof is as follows: $$d_Y(f^{n}(x^{n}),f(x^{n})) \leq \sup_{u \in X} d_Y(f^{n}(u),f(u))$$. By uniform convergence, given $$\epsilon >0$$ we can find $$n_0$$ such that $$\sup_{u \in X} d_Y(f^{n}(u),f(u))<\epsilon$$ for $$n \geq n_0$$. Hence $$d_Y(f^{n}(x^{n}),f(x^{n})) <\epsilon$$ for $$n \geq n_0$$. Also uniform convergence implies that $$f$$ is continuous. [ I assume that you have seen this earlier]. Hence there exists $$n_1$$ such that $$d_Y(f(x^{n}),f(x)) <\epsilon$$ for $$n \geq n_1$$. Now let $$n_2$$ be the maximum of $$n_0$$ and $$n_1$$ and conclude that $$d_Y(f^{n}(x^{n}),f(x))<2\epsilon$$ for $$n \geq n_2$$.