# Deriving $|u(x)-u(y)|\le|x-y|^{1-\frac 1p}\left(\int_0^1 |u'|^p \, dt \right)^{1/p}$

Assume $n=1$ and $u \in W^{1,p}(0,1)$ for some $1 \le p < \infty$.

(a) Show that $u$ is equal a.e. to an absolutely continuous function and $u'$ (which exists a.e.) belongs to $L^p(0,1)$.

(b) Prove that if $1 < p < \infty$, then $$|u(x)-u(y)|\le|x-y|^{1-\frac 1p}\left(\int_0^1 |u'|^p \, dt \right)^{1/p}$$ for a.e. $x,y \in [0,1]$.

PDE Evans, 2nd edition: Chapter 5, Exercise 4

Note: I figured out part (a) already, thanks to this question. I am looking for help on part (b), though.

Part (a) established already that $u' \in L^p(0,1)$. So Hölder's inequality can be used, which means we can say $\int_0^1 |u'| \, dt \le \left(\int_0^1 |u'|^p \, dt \right)^{1/p}$. But I got nowhere after doing all this: $$u(1)-u(0)=\int_0^1 u'(t) \, dt \le \int_0^1 |u'| \, dt \le \left(\int_0^1 |u'|^p \, dt \right)^{1/p}.$$

How can I approach this? By the way, I do notice that the desired inequality of part (b) bears resemblance to Morrey's inequality for $n=1$.

## 1 Answer

This is indeed a consequence of Hölder's inequality: $$|u(x)-u(y)| \leq \int_y^x |u'(t)|dt \leq \left( \int_y^x dt\right)^{(p-1)/p} \left( \int_y^x |u'|^p dt \right)^{1/p}$$ $$\leq|x-y|^{1-1/p} \left( \int_0^1 |u'|^p dt \right)^{1/p}.$$