Showing $H^1(I) \subset C^{1/2}(\overline{I}) $

Question

Hoi, let $H_1(I)$ the sobolev space on the interval $I = (\alpha,\beta)\subset \mathbb{R}$

http://en.wikipedia.org/wiki/Sobolev_space See here for more on sobolev spaces.

$H_m(I)$ contains all $u\in L_2(I)$ such that $\partial u,\partial^2 u,\cdots, \partial^m u$ exist weakly in $L_2$

I want to show that $H_1(I) \subset C^{1/2}(\overline{I})$ where the latter space is the space of Holder-continuous functions with exponent $\alpha= 1/2$.

Holder continuity is $|u(x)-u(y)| \leq C|x-y|^{\alpha}$ for a $C$ depending on $u$. (for all $x,y$ )

Can anyone provide any hint, as how to do this? Thanks for any help.

Not sure if it helps: But $H_1(I)$ is endowed with norm $$|u|=(\int|u|^2+\int|\partial u|^2 )^{1/2} $$and the Holder space with:

$$ |u| = \sup_{x\neq y}\frac{|u(x)-u(y)|}{|x-y|^{1/2}}+ \sup_x|u(x)|$$

Answer 1

Can you use the fact that functions $u \in H_1(I)$ are absolutely continuous on $I$? If so you can use some calculus of AC functions. For example, if $x < y$ then $$ |u(x) - u(y)| \le \int_x^y u'(s) \, ds \le \sqrt{y-x} \|u'\|_{L^2(I)} $$ A similar argument will work to bound $\sup_x |u(x)|$ by $\|u\|_{H_1(I)}$.