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I need to understand a way to prove this statement

$f$ is uniformly continuous on $D$ $\iff$ (when $x,y \in D$ then $|x-y| \rightarrow 0 \implies |f(x)-f(y)| \rightarrow 0)$

I know $\delta$ can depend on $\epsilon$ and for all the examples I have done, $\delta$ always can be set to $\delta = k \epsilon$ which would allow me to prove the rightward implication quite easily, but how can I mathematically prove this?

I am also not at all sure how to prove the leftward implication

EDIT: I have been told I was vague. I will try to reformulate.

Let $D$ be any interval of $\mathbb{R}$. Prove that a continuous function $f:D \rightarrow \mathbb{R}$ is uniformly continuous on $D$ if and only if whenever $x_n, y_n \in D$ are such that if $|x_n - y_n| \rightarrow 0$, then $|f(x_n) - f(y_n)| \rightarrow 0$

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  • $\begingroup$ It might make things a bit clearer if you write explicitly (with $\epsilon$ and $\delta$) what the right-hand statement means. $\endgroup$ Commented Oct 31, 2023 at 14:26

2 Answers 2

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Your formulation is vague, and if interpreted a certain way, is false, as Lorenzo Vanni's answer shows. However, you can formulate it more precisely in a way that makes it true. Namely, $f$ is uniformly continuous if and only if for all sequences $x_n\in D$, $y_n\in D$ with $\lim_{n\to\infty}|x_n-y_n|= 0$ we have $\lim_{n\to\infty}|f(x_n)-f(y_n)|=0$.

As you observed, the rightward implication follows in a straightforward way from the $\epsilon$-$\delta$ definition of uniform continuity. [Update - I misread the question initially, see the update for an explanation of the rightward implication.]

To prove the leftward implication, note that if $f$ fails to be uniformly continuous, then there is some $\epsilon>0$ for which all $\delta$ fail to satisfy the definition of uniform continuity, meaning if we let $\delta=\frac{1}{n}$ we have some $x_n,y_n\in D$ with $|x_n-y_n|\leq \frac{1}{n}$ and $|f(x_n)-f(y_n)|> \epsilon$. Therefore letting $n\to \infty$ we have $|x_n-y_n|\to 0$, yet $|f(x_n)-f(y_n)|> \epsilon$ for all $n$, so we do not have $|f(x_n)-f(y_n)|\to 0$.

Update

Reading your question more closely, I'm not sure what you mean by $\delta=k\epsilon$. This requires something stronger than uniform continuity - it requires Lipschitz continuity.

Nonetheless, the usual $\epsilon$-$\delta$ definition of uniform continuity is enough to prove the rightward implication. Simply argue that given $\epsilon>0$, choose $\delta$ from the definition of uniform continuity, and then from the definition of convergence, eventually $|x_n-y_n|\leq \delta$ for large enough $n$, which implies from the definition of uniform continuity that $|f(x_n)-f(y_n)|\leq\epsilon$ for large enough $n$. Since $\epsilon$ was arbitrary, this in turn implies $|f(x_n)-f(y_n)|\to 0$.

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    $\begingroup$ I've edited my question to be more precise. You were correct that we were dealing with sequences $x_n, y_n \in D$ rather than simply singular points. I think you managed to work your way through the vagueness anyway and answer my question along with Lorenzo Vanni. $\endgroup$ Commented Oct 31, 2023 at 15:43
  • $\begingroup$ Also, I am still confused on the rightward implication. From uniform continuity we know that for any arbitrary $\epsilon$ there exists a corresponding $\delta$, but if we arbitrarily send this $\epsilon$ to 0, why would the $\delta$ also necessarily be sent to 0 as well? $\endgroup$ Commented Oct 31, 2023 at 15:52
  • $\begingroup$ We don't assume $\delta\to 0$ anywhere. We pick $\epsilon$, fix a $\delta$ depending on that $\epsilon$, and then eventually $|x_n-y_n|<\delta$. We are sending $|x_n-y_n|$ to $0$, not $\delta$. $\endgroup$ Commented Oct 31, 2023 at 15:55
  • $\begingroup$ Apologies as I am fairly new to analysis and don't grasp the nuances sometimes. As I understand it, the arbitrary nature of $\epsilon$ means that we can send it to 0, and thus $|f(x_n)-f(y_n)|$ will also be sent to 0, we can use the fact that $f$ is uniformly convergence to show that $\exists \delta$ based on this $\epsilon$ such that $|x_n-y_n| <\delta$, but how does this show that $|x_n-y_n| \rightarrow 0$ for large $n$ $\endgroup$ Commented Oct 31, 2023 at 16:07
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    $\begingroup$ Right. To prove that $A\implies (B\implies C)$ is the same as proving $(A\wedge B)\implies C$. So you can treat $A$ and $B$ as given. $\endgroup$ Commented Oct 31, 2023 at 16:17
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The question is a little bit unclear (you have to specify what limit you are taking when you say $|x-y|\to 0$ ), but I will try to answer anyway.

By definition, a univariate real function is uniformly continuous on a subset $A$ of the reals with respect to the usual metric if for every $\varepsilon > 0$ we can choose $\delta > 0$ such that

\begin{equation*} |f(x)-f(y)| < \varepsilon \end{equation*}

whenever $x,y \in A, |x-y| < \delta$.

Now if by $|x-y| \to 0$ you mean that $y$ tends to $x$, you are basically asking to prove that

\begin{equation*} \lim_{y \to x} |f(x)-f(y)| = 0 \end{equation*}

for the right arrow, that is, by the definition of limit

\begin{equation*} \forall \varepsilon > 0 \exists \delta > 0 \forall y |x-y| < \delta \implies |f(x)-f(y)| < \varepsilon \end{equation*}

that holds by the definition of uniform continuity. Note that $x$ is a constant in the preceeding equation.

The other arrow is not true, I will give a counterexample.

Let $f: (0,1) \to \mathbb{R}$, $f(x)=\frac{1}{x}$, $f$ is obviously not uniformly continuous, but it is indeed continuous, hence, for every $x \in (0,1)$

\begin{equation*} \lim_{y \to x} |f(x)-f(y)| = |\lim_{y \to x} f(x)-f(y)| = |f(x) - \lim_{y \to x} f(y)| = |f(x)-f(x)| = 0 \end{equation*} by the continuity of the absolute value and of $f$.

Note that you are basically trying to prove that uniform continuity is equivalent with ordinary continuity if every point of the set is an accumulation point (so that you can always take the limit at $x$) (in fact in the first part of the proof I have proved that uniform continuity implies continuity on such a set, because $\lim_{y \to x} |f(x)-f(y)| = 0 \Longleftrightarrow \lim_{y \to x} f(y) = f(x)$), and this is not true at all.

It is true anyway that if the set $D$ you are considering is compact, then continuity and uniform continuity are equivalent.

This is known as Heine-Borel theorem, and it applies even to general metric spaces.

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