I'm trying to prove the following elementary fact mentioned on page 238 of Folland's Real Analysis:
A function $f$ is called uniformly continuous if $\|\tau_y f - f \|_u \rightarrow 0$ as $y \to 0$. (The reader should pause to check that this is equivalent to the usual $\epsilon$-$\delta$ definition of uniform continuity.)
I am taking Folland's advice and trying to show the equivalence of these two definitions of uniform continuity. First, some notation and definitions:
- If $f$ is a function on $\mathbb{R}^d$ and $y \in \mathbb{R}^d$, then $\tau_y f(x) := f(x-y)$.
- $\|f \|_u := \sup\left\{|f(x)|: x \in \mathbb{R}^d \right\}$. ($\|\cdot \|_u$ is called the uniform norm.)
- $e_1 := (1,0,\ldots,0) \in \mathbb{R}^d$
- My working $\epsilon$-$\delta$ definition of uniform continuity:
A function $f: \mathbb{R}^d \to \mathbb{C}$ is uniformly continuous if, for each $\epsilon > 0$, there exists some $\delta > 0$ such that $$x,y \in \mathbb{R}^d \text{ and } |x-y| < \delta \implies |f(x) - f(y)| < \epsilon.$$
My proof: [Edit 3/21/22: This proof is wrong! See updated proof below.]
$\implies$ direction: Suppose $f$ is uniformly continuous in the $\epsilon$-$\delta$ sense. Then we may pick a sequence of positive real numbers $\{\delta_n\}_{n=1}^{\infty}$ such that
$$ x,y \in \mathbb{R}^d \text{ and } |x-y| \leq \delta_n \implies |f(x) - f(y)| < \frac{1}{n}$$
for each $n \in \mathbb{N}$. Then $|f(x - \delta_n e_1) - f(x)| < \frac{1}{n}$ for all $x \in \mathbb{R}^d$, for each $n \in \mathbb{N}$ (since $|(x - \delta_n e_1) - x| = |\delta_n e_1| = \delta_n \leq \delta_n$), and so $\sup\{|f(x-\delta_n e_1) - f(x)|: x \in \mathbb{R}^d \} \leq 1/n$ for each $n \in \mathbb{N}$. In other words, $\|\tau_{\delta_n e_1}f - f \|_u \leq 1/n$ for all $n$. Then since $\{\delta_n e_1 \}_{n=1}^{\infty}$ is a sequence in $\mathbb{R}^d$ and $\delta_n e_1 \rightarrow 0$ as $n \to \infty$ (with respect to $| \cdot |$, the standard metric on $\mathbb{R}^d$), we have
\begin{align*}
\lim_{y \to 0} \|\tau_{y}f - f \|_u &= \lim_{n \to \infty} \|\tau_{\delta_n e_1} f - f \|_u \leq \lim_{n \to \infty} \frac{1}{n} = 0.
\end{align*}
Thus, $\lim_{y \to 0} \|\tau_{y}f - f \|_u = 0$. (Just a note that $y \in \mathbb{R^d}$ here, so $y \to 0$ really means $y \to (0,\ldots,0)$.)
$\impliedby$ direction: Suppose $\|\tau_y f - f \|_u \to 0$ as $y \to 0$. Then for each $\epsilon > 0$, there exists a $\delta > 0$ such that
\begin{align*}
& y \in \mathbb{R}^d \text{ and } |y| < \delta \implies \|\tau_y f - f \|_u < \epsilon \\[4pt]
\implies \quad & y \in \mathbb{R}^d \text{ and } |y| < \delta \implies \sup \left\{|f(x-y) - f(x)|: x \in \mathbb{R}^d \right\} < \epsilon \\[4pt]
\implies \quad & y \in \mathbb{R}^d \text{ and } |y| < \delta \implies |f(x-y) - f(x)| < \epsilon \text{ for all } x \in \mathbb{R}^d \\[4pt]
\implies \quad & x,y \in \mathbb{R}^d \text{ and } |x - y| < \delta \implies |f(x) - f(y)| < \epsilon,
\end{align*}
and so $f$ is uniformly continuous in the $\epsilon$-$\delta$ sense. $\qquad \square$
Does this proof look ok? Any feedback or suggestions for improvement are welcomed.
Update 3/21/22: Thanks to @Mason for pointing out the issue with my "$\implies$" direction proof. I've attempted to write a corrected proof:
Corrected "$\implies$" proof:
Suppose $f$ is uniformly continuous in the $\epsilon$-$\delta$ sense. Then for each $\epsilon > 0$, there exists a $\delta > 0$ such that
\begin{align*}
& x,y \in \mathbb{R}^d \;\text{ and }\; |x-y| < \delta \implies |f(x)-f(y)| < \epsilon \\[3pt]
\implies \quad & x,y \in \mathbb{R}^d \;\text{ and }\; |(x-y) - x| < \delta \implies |f(x-y) - f(y)| < \epsilon \\[3pt]
\implies \quad & x,y \in \mathbb{R}^d \;\text{ and }\; |y| < \delta \implies |f(x-y) - f(x)| < \epsilon \\[3pt]
\implies \quad & y \in \mathbb{R}^d \;\text{ and }\; |y| < \delta \implies |f(x-y) - f(x)| < \epsilon \quad \text{for all } x \in \mathbb{R}^d \\[3pt]
\implies \quad & y \in \mathbb{R}^d \; \text{ and }\; |y| < \delta \implies \sup_{x \in \mathbb{R}^d} |f(x-y) - f(x)| \leq \epsilon \\[3pt]
\implies \quad & y \in \mathbb{R}^d \; \text{ and }\; |y| < \delta \implies \|\tau_y f - f \|_u \leq \epsilon.
\end{align*}
Does this look okay now?
The $\implies$ direction proof is wrong. You have only shown $\|\tau_{y_n}f - f\|_{u} \to 0$ for the particular sequence $y_n = \delta_ne_1$. But to show that $\|\tau_{y}f - f\| \to 0$ as $y \to 0$ with a sequence approach, you have to show that $\|\tau_{y_n}f - f\| \to 0$ for all sequences $y_n \to 0$ with $y_n \neq 0$ for all $n$. Sequences are unnecessary here, and the proof is much simpler using the $\varepsilon$-$\delta$ definition of $\|\tau_{y}f - f\| \to 0$ as $y \to 0$.
The $\impliedby$ direction looks good.