Exercise 3.2.1

Question

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a function. For any $a \in \mathbb{R}$, let $f_{a}: \mathbb{R} \rightarrow \mathbb{R}$ be the shifted function $f_{a}(x):=f(x-a)$.
\begin{enumerate}[label=(\alph*)]
\item Show that $f$ is continuous if and only if, whenever $\left(a_{n}\right)_{n=0}^{\infty}$ is a sequence of real numbers which converges to zero, the shifted functions $f_{a_{n}}$ converge pointwise to $f$.

\item Show that $f$ is uniformly continuous if and only if, whenever $\left(a_{n}\right)_{n=0}^{\infty}$ is a sequence of real numbers which converges to zero, the shifted functions $f_{a_{n}}$ converge uniformly to $f$.
\end{enumerate}

Prakash Anand · Accepted Answer

\begin{proof} \begin{description} \item[$(\Rightarrow)$] Let $x_{0} \in \mathbb{R}$. Suppose $f$ is continuous at $x_{0}$. That is, given $\epsilon>0$ there exists $\delta>0$ such that if $\left|x-x_{0}\right|<\delta$ then $\left|f(x)-f\left(x_{0}\right)\right|<\epsilon$. Assume $a_{n} \rightarrow 0$ as $n \rightarrow \infty$ and let $x_{n}=x_{0}-a_{n}$. So given $\delta>0$ there exists $N^{\prime}>0$ such that $\left|x_{n}-x_{0}\right|<\delta$ for all $n>N^{\prime}$. Given $\epsilon^{\prime}>0$ take $\epsilon=\epsilon^{\prime}$, and take $N=N^{\prime}$. So by the continuity of $f$ we get that $\left|x_{n}-x_{0}\right|<\delta$ for all $n>N^{\prime}=N$ Thus, $\left|f\left(x_{n}\right)-f\left(x_{0}\right)\right|<\epsilon=\epsilon^{\prime}$. But, $f\left(x_{n}\right)=f\left(x_{0}-a_{n}\right)=f_{a_{n}}\left(x_{0}\right)$ for all $n>N$ So we have $\left|f_{a_{n}}\left(x_{0}\right)-f\left(x_{0}\right)\right|<\epsilon$ for all $n>N$. Thus $f_{a_{n}}$ converges pointwise to $f$. \item[$(\Leftarrow)$] Suppose given $a_{n} \rightarrow 0$ that $f_{a_{n}}$ converges pointwise to $f$. Given $x_{0}$ take a sequence $x_{n}$ which converges to $x_{0}$. That is, $a_{n}=x_{0}-x_{n} \rightarrow 0$ Since $f_{a_{n}}$ converges pointwise to $f$, given $\epsilon>0$ there is some $N>0$ such that $\mid f_{a_{n}}\left(x_{0}\right)-$ $f\left(x_{0}\right) \mid<\epsilon$ for all $n>N$. But, $\left|f_{a_{n}}\left(x_{0}\right)-f\left(x_{0}\right)\right|=\left|f\left(x_{0}-a_{n}\right)-f\left(x_{0}\right)\right|=\left|f\left(x_{n}\right)-f\left(x_{0}\right)\right|$ Hence, $\left|f\left(x_{n}\right)-f\left(x_{0}\right)\right|<\epsilon$ for all $n>N$. That is, $f\left(x_{n}\right)$ converges to $f\left(x_{0}\right)$. Thus, $f$ is continuous. \end{description} \end{proof} For part (b) the argument is the same, except that now $\delta, N$ are independent of the base point $x_{0}$.

Exercise 3.2.1

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Exercise 3.2.1

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