Given $\Omega=[a,b] \subset \mathbb R$ and the Sobolev space $H_0^1(\Omega)$
Let $v \in H_0^1(\Omega) $ prove the following inequality
$$\Vert v \Vert_{L^2(\Omega)}\le\frac{\Vert \dot v \Vert_{L^2(\Omega)}}{\sqrt2}(b-a)$$
My thoughts/ideas: I looked at the case that $\ v(x)=\int_a^x \dot v(t)dt$
By Schwarz inequality I get the following: $v(x)^2\le (x-a)\Vert \dot v \Vert_{L^2(\Omega)}^2$
If I integrate both sides and take the square root I get exactly what I wanted to show. However, $\ v(x)=\int_a^b \dot v(t)dt$ isn't necessarily true
I would appreciate any help because I am stuck. I am not too familiar with Sobolev spaces because I couldn't really understand the definition.
$\int_a^b \dot{v} (t) \, dt $ is not a function of $x$, so your identity is in fact never true (well, almost never). Did you intend to write $\int_a^x$?
Try to prove the inequality for smooth functions. If you can do that show it's stable under approximation in the space $H^1_0$ to prove it in the general case (this assumes you know that smooth functions are dense in that space).