Exercise 3.3.2 (Interchange of limits and uniform limits)

Question

Prove Proposition 3.3.3.\par 

	extbf{Proposition 3.3.3.} Let $\left(X, d_{X}ight)$ and $\left(Y, d_{Y}ight)$ be metric spaces, with $Y$ complete, and let $E$ be a subset of $X$. Let $\left(f^{(n)}ight)_{n=1}^{\infty}$ be a sequence of functions from $E$ to $Y$, and suppose that this sequence converges uniformly in $E$ to some function $f: E ightarrow Y$. Let $x_{0} \in X$ be an adherent point of $E$, and suppose that for each $n$ the limit $\lim _{x ightarrow x_{0} ; x \in E} f^{(n)}(x)$ exists. Then the limit $\lim _{x ightarrow x_{0} ; x \in E} f(x)$ also exists, and is equal to the limit of the sequence $\left(\lim _{x ightarrow x_{0} ; x \in E} f^{(n)}(x)ight)_{n=1}^{\infty}$; in other words we have the interchange of limits

$$
\lim _{n ightarrow \infty} \lim _{x ightarrow x_{0} ; x \in E} f^{(n)}(x)=\lim _{x ightarrow x_{0} ; x \in E} \lim _{n ightarrow \infty} f^{(n)}(x) .
$$

Prakash Anand · Accepted Answer

\begin{proof} Suppose $f^{(n)}$ converges uniformly to $f$. That is, given $\epsilon_{1}>0$ there exists $N_{1}>0$ such that $d_{Y}\left(f^{(n)}(x), f(x) ight)<\epsilon_{1}$ for all $n>N_{1}$ and $x \in X$. Let $x_{n} ightarrow x_{0}$. And so suppose for each $n$ that $\lim _{k ightarrow \infty} f^{(n)}\left(x_{k} ight)=L_{n}$. So we also have given $\epsilon_{2}>0$ there exists $N_{2}>0$ such that $d_{Y}\left(f^{(n)}\left(x_{k} ight), L_{n} ight)<\epsilon_{2}$ for all $n>N_{2}$ and $x \in X$. Now let us show that $\left(L_{n} ight)_{n=1}^{\infty}$ is a Cauchy sequence. Given $\epsilon>0$ take $\epsilon_{1}=\epsilon_{2}=\epsilon / 4$ and $N=\min \left(N_{1}, N_{2} ight)$. Then $$ \begin{aligned} &d_{Y}\left(L_{n}, L_{m} ight)\ & \leq d_{Y}\left(L_{n}, f^{(n)}\left(x_{k} ight) ight)+d_{Y}\left(f^{(n)}\left(x_{k} ight), f\left(x_{k} ight) ight)+d_{Y}\left(f\left(x_{k} ight), f^{(m)}\left(x_{k} ight) ight)+d_{Y}\left(f^{(m)}\left(x_{k} ight), L_{m} ight) \ &<\epsilon / 4+\epsilon / 4+\epsilon / 4+\epsilon / 4 \ &=\epsilon \end{aligned} $$ So $\left(L_{n} ight)_{n=1}^{\infty}$ is a Cauchy sequence, and since $Y$ is complete we know that this sequence converges to, say, $L_{0}$. That is, given $\epsilon_{3}>0$ there exists $N_{3}>0$ such that $d_{Y}\left(L_{n}, L_{0} ight)<\epsilon$ for all $n>N_{3}$. Now we will show that $$ \lim _{k ightarrow \infty} f\left(x_{k} ight)=L_{0}=\lim _{n ightarrow \infty} \lim _{k ightarrow \infty} f^{(n)}\left(x_{k} ight) $$ Given $\epsilon>0$ take $\epsilon_{1}=\epsilon_{2}=\epsilon_{3}=\epsilon / 3$ and $N=\min \left(N_{1}, N_{2}, N_{3} ight)$. Then we get that $$ \begin{aligned} d_{Y}\left(f\left(x_{k} ight), L_{0} ight) & \leq d_{Y}\left(f\left(x_{k} ight), f^{(n)}\left(x_{k} ight) ight)+d_{Y}\left(f^{(n)}\left(x_{k} ight), L_{n} ight)+d_{Y}\left(L_{n}, L_{0} ight) \ &<\epsilon / 3+\epsilon / 3+\epsilon / 3 \ &=\epsilon \end{aligned} $$ As desired. \end{proof}

Exercise 3.3.2 (Interchange of limits and uniform limits)

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Exercise 3.3.2 (Interchange of limits and uniform limits)

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