If $v\in H^1(]0,1[)$, then $|v(\lambda)| \leq ||v||_{L^1(]0,1[)} + ||v'||_{L^2(]0,1[)}$

Question

Let $v\in H^1(]0,1[)$.

I want to prove for all $\lambda \in [0,1]$ that,

$$|v(\lambda)| \leq ||v||_{L^1(]0,1[)} + ||v'||_{L^2(]0,1[)}$$

My idea :

I defined $u(\lambda)= \int_0^{\lambda}v'(t)dt$

then, $|u(\lambda)|\leq ||v'||_{L^2(]0,1[)}$ (Using Cauchy Schwarz inequality).

and we have also $u'=v',$ which implies $v=u+ C$ (in distribution sense), with $C=v(0)$.

then $|v(\lambda)|\leq|u(\lambda)|+|v(0)| \leq ||v'||_{L^2(]0,1[)}+|v(0)|$.

and I'm stuck there ( why is $v$ defined on boundary $\{0,1\} ?$

Answer 1

Functions in the space $H^1(0,1)$ are absolutely continuous so they can be extended continuously up to the boundary of $(0,1)$.

You could have defined $u$ a bit differently, but in fact there is no need to use any functions other than $v$ and $v'$. If $x,y \in [0,1]$ then $$v(x) - v(y) = \int_x^y v'(t) \, dt$$ so that $$|v(x)| \le |v(y)| + \|v'\|_{L^2}.$$ Now integrate with respect to $y$: $$|v(x)| \le \|v\|_{L^1} + \|v'\|_{L^2}.$$

Answer 2

Instead of defining $u(\lambda)=\int_0^\lambda v^\prime(t)\mathrm dt$, you could define $u(\lambda)=\int_x^\lambda v^\prime(t)\mathrm dt$ for a fixed $x\in]0,1[$. Then, Cauchy Shwarz gives $|u(\lambda)|\leq \|v^\prime\|_{L^2}$, but this time $v = u + u(x)$. This is true because $v-u$ has a vanishing derivative so that $v-u$ is constant almost everywhere, but is continuous by the classical Sobolev injection. Your last inequality holds, replacing $0$ by $x$. You can finally integrate from $x=0$ to $x=1$.