There are other ways a function can be a continuous extension, but probably the most basic way (and likely about the only way you'll see in elementary calculus) is that you have a function …

May 10, 2019 · This fact is useful to resolve this natural question: Let $\{X_i\}_{i=1}^{\infty}$ be i.i.d. random variables uniform over $[-1,1]$.

Oct 15, 2016 · A piecewise continuous function doesn't have to be continuous at finitely many points in a finite interval, so long as you can split the function into subintervals such that each …

Jan 27, 2014 · I understand the geometric differences between continuity and uniform continuity, but I don't quite see how the differences between those two are apparent from their definitions. …

$\begingroup$ @user135626: What I wrote is correct. You are misreading it. I'm not saying the derivative is zero, I'm saying that if the derivative exists, the numerator of the difference …

Aug 4, 2022 · Isn't this violating the definition of continuous stochastic process or is it that I have to keep $\omega$ constant throught out the process ? Also, is $\omega$ in the definition of …

Prove that if $ f $ is continuous on $ \left[ 0, \infty\right)$ and uniformly continuous on $[b, \infty ...

Oct 8, 2021 · The regularity of Brownian paths is expressed quite precisely by it being $\alpha$-Holder continuous for all $\alpha< \frac 12$ but not $\alpha$-Holder continuous for $\alpha …

Added @Dimitris's answer prompted me to mention, beyond the fact that the implication on normed spaces indeed is an equivalence, that it's the converse which holds in the wider …

May 21, 2012 · We have thus (intuitively) shown that the only way to make a continuous function on a closed set produce a non-closed image is to make the domain unbounded. In other …