prove $\|v \|_{L^2} + \|v' \|_{L^2} \leq 2\|v \|_{H^1}$

Question

let $\Omega = (0,1)$ and $H^1 (\Omega)$ be the Sobolev space defined as follows :

$$ H^1 (\Omega) = \{v \in L^2(\Omega )\; ;\; v' \in L^2(\Omega )\}$$

where $v'$ is the weak derivative of $v$. meaning :

$v' \in L^2(\Omega ) $ and $\int_0^1 v(x) \phi'(x)dx = - \int_0^1v'(x)\phi(x)dx$

for every continuously differentiable function $\phi$ of compact support.

the associated norm is defined as follows : $\|v \|^2_{H^1} = \|v \|^2_{L^2} + \|v' \|^2_{L^2}$

I was reading a mark scheme of a certain exam and they used the following inequality : $\|v \|_{L^2} + \|v' \|_{L^2} \leq 2\|v \|_{H^1}$ $(\star)$

while I was trying to show that I accidentally showed that

$\|v \|_{H^1} \leq \|v \|_{L^2} + \|v' \|_{L^2} $

so how do you prove $(\star)$ ?

Answer 1

we have, $$ 0\leq\parallel v\parallel^2_{L^2}\leq \parallel v\parallel^2_{L^2}+\parallel v'\parallel^2_{L^2}$$ $$ \parallel v\parallel_{L^2}\leq\parallel v\parallel^2_{H^1}$$ using the same argument we obtain,$$ \parallel v'\parallel_{L^2}\leq \parallel v\parallel^2_{H^1}$$ Hence, $$ \parallel v\parallel_{L^2}+\parallel v'\parallel_{L^2}\leq2\parallel v\parallel_{H^1}$$