Problem: Let $A$ be a non-empty subset of $\mathbb{R}$ and let $f: A \rightarrow \mathbb{R}$ be a function. We have the following statements:
a) If $(x_n)$ is a Cauchy sequence in $A$, then $f(x_n)$ is a Cauchy sequence in $\mathbb{R}$.
b) For all $M > 0$ the restriction of $f$ to $A \cap [-M,M]$ is uniformly continuous.
Prove the implication $a \Rightarrow b$.
Attempt: I think I almost found it, but I'm not quite sure yet. Here is my attempt:
Suppose a) holds. Let $M > 0$. Let $ \epsilon > 0$. We need to a find a $\delta > 0$ such that for all $x,y \in A \cap [-M,M]$ with $|x-y| < \delta$ it holds that $|f(x) – f(y)| < \epsilon.$
Choose $\delta = \epsilon/3$. Let $x,y \in A \cap [-M,M]$ and suppose that $|x-y| < \delta$. Then $x,y \in [-M,M]$, and since $[-M,M]$ is closed, there are (Cauchy) sequences $(x_n), (y_n) \in [-M,M]$ such that $(x_n) \to x$ and $(y_n) \to y$. From a), we then also know that $f(x_n)$ and $f(y_n)$ are Cauchy sequences in $\mathbb{R}$. Now, we have that \begin{align*} |f(x) – f(y)| \leq |f(x) – f(x_n)| + |f(x_n) – f(y_n) | + |f(y_n) – f(y)|. \end{align*} Since $f(x_n)$ and $f(y_n)$ are Cauchy sequences, I know I can get the middle term smaller than $\epsilon/3$. But I don't know what to do about the other two terms? Can I assume that $f(x_n) \to f(x)$ and $f(y_n) \to f(y)$? If I do this, then I assume continuity of $f$, while this is not given?
This is the part where I'm stuck. Any help is appreciated!
Best Answer
By contradiction. Assume that there exists $M>0$ such that the restriction of $f$ to $A\cap[-M,M]$ is not uniformly continuous. Then there exist $\epsilon > 0$ and two sequences $(x_n), (y_n) \subset A$ such that $$ |x_n - y_n| \to 0, \qquad |f(x_n) - f(y_n)| \geq \epsilon, \quad \forall n\in\mathbb{N}. $$ Since $(x_n)$ and $(y_n)$ are bounded and $|x_n - y_n| \to 0$, we can extract a common subsequence (not relabeled) converging to the same point $x$. Let $(z_n)\subset A$ be the sequence defined by $z_{2n} = x_n$, $z_{2n+1} = y_n$. Then $(z_n)$ is a Cauchy sequence (converging to $x$) but $(f(z_n))$ is not a Cauchy sequence, since $|f(z_{2n}) - f(z_{2n+1})| \geq \epsilon$ for every $n$.