4
$\begingroup$

Recently I was working on a proof where I eventually wanted to show that for a continuous function $f$ we have the following property at $x$:

For all $\delta > 0$ there exists some $\epsilon > 0$ such that for every $x' \in N_\delta(x)$:

$$|f(x') - f(x)| \leq \epsilon$$

This statement looks superficially like a statement about the continuity of $f$, but it seems to reduce to a claim about $f$ being bounded on a neighborhood at $x$:

$$\sup_{x' \in N_\delta(x)} f(x') < \infty$$

Is there more to this "opposite" definition of continuity? Does it have another name? Is it really just that $f$ is bounded on a neighborhood?

$\endgroup$
2
  • $\begingroup$ This has nothing to deal with coninuity. Functions satisfying this condition are simply locally bounded functions. $\endgroup$ – Crostul Dec 5 '20 at 22:15
  • 1
    $\begingroup$ The proposed property need not hold for any continuous function. It is in general much stronger than local boundedness. For example, $f:(0,\infty)\to\mathbb{R}$ with $f(x)=1/x$ is continuous (and hence locally bounded), but it does not satisfy OP's property. Another line of example comes when the domain is a bounded metric space, in which OP's property is the same as boundedness of $f$. $\endgroup$ – Sangchul Lee Dec 5 '20 at 22:16
2
$\begingroup$

Hölder continuity is defined as :

$$\forall x,y \in \mathbb R^d , |f(x)-f(y)| \leq C|x-y|^\alpha$$

Maybe if you set $\alpha=0$ you get what you are working with?

$\endgroup$

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.