3
$\begingroup$

Let us introduce the definitions with which I am working.

Definition 1 (Continuity) Let $f:A\to\mathbb{R}$. Then $f$ is continuous on $A$ if for all $y\in A$ we have that for all $\epsilon>0$ there exists $\delta>0$ such that for all $x\in A$ with $|x-y|<\delta$, we have $|f(x)-f(y)|<\epsilon$.

Definition 2 (Uniform Continuity) Let $f:A\to\mathbb{R}$. Then $f$ is uniformly continuous on $A$ if for all $\epsilon>0$ there exists $\delta>0$ such that for all $x,y\in A$ with $|x-y|<\delta$ we have $|f(x)-f(y)|<\epsilon$.

I notice that these definitions are very similar. However I must clarify I know that continuity does not imply uniform continuity.

My question is this: suppose I have in front of me some proof that some $f$ is continuous. Uniform continuity cannot allow the choice of $\delta$ to be dependent on the points $x,y$. I wondered whether if I could see that in the proof that $f$ is continuous, that the choice of $\delta$ has no dependence on $x,y$, that in fact I have uniform continuity as well.

In other words, if I have $\delta$ dependent only on $\epsilon$ in some continuity proof, do I automatically have that $f$ is also uniformly continuous, or does a separate proof need to be written?

$\endgroup$
  • $\begingroup$ I'd say, as a rule, that any assertion that is not the original statement of the theorem you are proving should need a separate proof. $\endgroup$ – Gae. S. Nov 16 '19 at 23:47
  • 1
    $\begingroup$ If you’re absolutely positive (and only then) that your $\delta$ depends only on $\epsilon$ and not in $y$ (including some possibly non-explicit ways for convoluted $A$ or $f$), you can just say (and it is rigorous): “we see in the above proof [of continuity] that the choice of $\delta$ depended only on the $\epsilon$, thus the previous argument actually gives uniform continuity.” I would suggest, for homework or similar, that you write a distinct proof at least once, though. $\endgroup$ – Mindlack Nov 16 '19 at 23:49
  • $\begingroup$ @Gae.S. I understand what you mean but my question was, more precisely, if $\delta$ depends only on $\epsilon$, and $f$ is continuous, is $f$ automatically uniformly continuous? $\endgroup$ – Benjamin Nov 16 '19 at 23:49
  • $\begingroup$ @Mindlack This is what I thought. In almost any circumstance I imagine I would prove these things distinctly anyway, however I wanted to know essentially, whether when you have $\delta$ not dependent on $y$, whether you are allowed to move the position of the "$\forall y$" in the definition of continuity. $\endgroup$ – Benjamin Nov 16 '19 at 23:53
2
$\begingroup$

If in your proof the value of $\delta$ doesn't depend on your choice of $x$, but uniquely on $\varepsilon$, then you could have fixed $x$ after fixing $\varepsilon$ and determining $\delta$, therefore proving absolute continuity of the function, hence you can basically use the same proof, except for fixing $x$ and $y$ after determining $\delta$.

| cite | improve this answer | |
$\endgroup$
2
$\begingroup$

No, you don't need another proof: uniform continuity is just continuity independent from the point, as you stated above. Written in first order logic symbols (exists, for all etc.) the difference is in fact the position of quantifier $\forall y$ and nothing more.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ So when $\delta$ does not need to depend on $y$, we are able to move it after the $\delta$? $\endgroup$ – Benjamin Nov 16 '19 at 23:51
  • 1
    $\begingroup$ Continuity: $\forall\varepsilon>0\forall y\in A\exists\delta\dots$ UC: $\forall\varepsilon>0\exists\delta\forall y\in A\dots$ and therefore you move it $\endgroup$ – Lorenzo Cecchi Nov 16 '19 at 23:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.