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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:

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

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1 Answer 1

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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 $n \geq n_2$.

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