Exercise 3.3.4

Question

Prove Proposition 3.3.4.\par 

	extbf{Proposition 3.3.4.} Let $\left(f^{(n)}ight)_{n=1}^{\infty}$ be a sequence of continuous functions from one metric space $\left(X, d_{X}ight)$ to another $\left(Y, d_{Y}ight)$, and suppose that this sequence converges uniformly to another function $f: X ightarrow Y$. Let $x^{(n)}$ be a sequence of points in $X$ which converge to some limit $x$. Then $f^{(n)}\left(x^{(n)}ight)$ converges to $f(x)$.

Prakash Anand · Accepted Answer

\begin{proof} Let $x^{(n)}$ be a sequence that converges to $x$. By the continuity of the $f^{(n)}$ 's we have that $f^{(n)}\left(x^{(m)} ight) ightarrow f^{(n)}(x)$ as $m ightarrow \infty$. That is, given $\epsilon_{1}>0$ there exists $N_{1}>0$ such that $d_{Y}\left(f^{(n)}\left(x^{(m)} ight), f^{(n)}(x) ight)<\epsilon_{1}$ for all $m>N_{1}$. But since $f^{(n)} ightarrow f$ uniformly we know given $\epsilon_{2}>0$ there exists $N_{2}>0$ such that $d_{Y}\left(f^{(n)}(x), f(x) ight)<\epsilon_{2}$ for all $n>N_{2}$ and $x \in X$. Given $\epsilon>0$ take $\epsilon_{1}=\epsilon_{2}=\epsilon / 3$ and $N=\max \left\{N_{1}, N_{2} ight\}$. Then we have that $d_{Y}\left(f^{(n)}\left(x^{(m)} ight), f^{(n)}(x) ight)<\epsilon / 3$ and $d_{Y}\left(f^{(n)}(x), f(x) ight)<\epsilon / 3$ for all $m, n>N$ and $x \in X$. In particular we have that $d_{Y}\left(f^{(n)}\left(x^{(m)} ight), f^{(n)}\left(x^{(n)} ight) ight)<\epsilon / 3$. But by the triangle inequality we have: $$ \begin{aligned} d_{Y}\left(f^{(n)}\left(x^{(n)} ight), f(x) ight) \leq d_{Y}(&\left.f^{(n)}\left(x^{(n)} ight), f^{(n)}\left(x^{(m)} ight) ight)+d_{Y}\left(f^{(n)}\left(x^{(m)} ight), f^{(n)}(x) ight) \ &+d_{Y}\left(f^{(n)}(x), f(x) ight)<\epsilon ext { for all } n>N ext { and } x \in X \end{aligned} $$ Thus $f^{(n)}\left(x^{(n)} ight)$ converges to $f(x)$. \end{proof}

Exercise 3.3.4

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Exercise 3.3.4

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