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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.

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    It is just an instance of Wirtinger's inequality (second version): https://en.wikipedia.org/wiki/Wirtinger%27s_inequality_for_functions – 2017-01-17

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$\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).

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    Since smooth functions are dense in $H$ , does it follow that there exit a sequence of $v_n$ such that $v_n\ \rightarrow v$ ? – 2017-01-17
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    @XPenguen Yes, with smooth $v_n$, and "$\rightarrow$" refers to the $H^1$ norm. – 2017-01-17
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    Can you give me a hint on how to prove this for smoth functions ? – 2017-01-17
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    @XPenguen have a look at your own question, and replace $\int_a^b$ with $\int_a^x$ where appropriate. – 2017-01-17
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    But I would need $v(a)=0$ . Otherwise I couldnt write $v(x)=\int_a^x \dot v(t)dt$ ? Or is that true if $v \in H^1$ ? – 2017-01-17
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    @XPenguen you are working in $H^1_0$, aren't you? Note the lower index $0$. How did you define this? – 2017-01-17