one dimensional Sobolev inequality

Question

Given $\Omega=[a,b] \subset \mathbb R$ and the Sobolev space $H_0^1(\Omega)$
Let $v \in H_0^1(\Omega)$ be bounded
Prove that the inequality : $\Vert v \Vert_{L^{\infty}}\le \sqrt{b-a}\Vert \dot v \Vert_{L^2}$ holds.

My ideas: $\;\vert u(x)-u(y)\vert \le \int_a^b \dot v(t)dt \le \sqrt{b-a} \;\Vert u \Vert_{L^2}$ is true by Schwartz inequality. Since $v \in H_0^1$ there exists a sequence of smooth functions $v_n \rightarrow v$....

I dont know how to continue from here.

Would appreciate any help !

Answer 1

As you observed, the inequality holds for smooth functions by virtue of the Fundamental Theorem of Calculus and the Cauchy-Schwarz inequality. To extend it to the entire space $H^1_0$, use mollification: the sequence $v*\phi_n$ converges to $v$ in the norm of the space, and also converges to $v$ almost everywhere (the latter is true for any $v\in L^1$). Hence, at almost every point we have the bound $$ |v(x)|\le \limsup_{n\to\infty } \|v*\phi_n\|_{L^\infty} \le \sqrt{b-a}\|\dot v\|_{L^2} $$