Exercise 3.2.2

Question

\begin{enumerate}[label=(\alph*)]
\item Let $\left(f^{(n)}\right)_{n=1}^{\infty}$ be a sequence of functions from one metric space $\left(X, d_{X}\right)$ to another metric space $\left(Y, d_{Y}\right)$, and let $f: X \rightarrow Y$ be another function from $X$ to $Y$. Show that if $f^{(n)}$ converges uniformly to $f$, then $f^{(n)}$ converges pointwise to $f$.

\item For each integer $n \geq 1$, let $f^{(n)}:(-1,1) \rightarrow \mathbb{R}$ be the function $f^{(n)}(x):=x^{n} .$ Prove that $f^{(n)}$ converges pointwise to the zero function, but does not converge uniformly to any function $f: \mathbb{R} \rightarrow \mathbb{R}$.

\item Let $g:(-1,1) \rightarrow \mathbb{R}$ be the function $g(x):=x /(1-x)$. With the notation as in (b), show that the partial sums $\sum_{n=1}^{N} f^{(n)}$ converges pointwise as $N \rightarrow \infty$ to $g$, but does not conform uniformly to $g$ on the open interval $(-1,1)$. What would happen if we replaced the open interval $(-1,1)$ with the closed interval $[-1,1]$?
\end{enumerate}

Prakash Anand · Accepted Answer

\begin{enumerate}[label=(\alph*)] \item This is obvious from the definitions. \item We have $$\lim _{n \rightarrow \infty} f^{(n)}(x)=\lim _{n \rightarrow \infty} x^{n}=0$$ for all $x \in(-1,1)$ (from previous work). Thus $f^{(n)}$ converges pointwise to 0. To show that $f^{(n)}$ does not converge uniformly to any function it is enough to show that it does not converge uniformly to 0 (because of the result in part (a)). So we need to show that there exists $\epsilon>0$ such that for all $N>0$ there is an $n>N$ and $x \in(-1,1)$ such that $\left|x^{n}\right| \geq \epsilon$. Consider, $0<\epsilon<1, n=N+1, x=\epsilon^{1 / n} \in(-1,1)$ Then we get $$\left|x^{n}\right|=\left|\left(\epsilon^{1 / n}\right)^{n}\right|=\epsilon.$$ So $\left|x^{n}\right| \geq \epsilon$. And hence $f^{(n)}$ does not conver uniformly to any function. \end{enumerate}

Exercise 3.2.2

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Exercise 3.2.2

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