Sobolev space and norms Hello I'd like to know the solution to this question in my Numerics of PDE's class. Here goes: 
Given $V := \lbrace v \in H^1(0,1): v(0) = 0 \rbrace \subset H^1(0,1)$
Show that the $H^1$-seminorm $|\cdot|_1$ is equivalent to the $H^1$-norm $||\cdot||_1$ on V. 
We have naturally that $|v|_1\leq||v||_1$, but the other inequality, I can't seem to find a good way.
Thanks!
 A: Let me call attention to a technical (but important) detail.  Since functions in Sobolev classes are technically only defined "almost everywhere", how can we impose a condition $v(0)=0$ in a meaningful way?  You have tried to get the required estimate (using the fact that $(0,1)$ has length one), assuming "naively" that we can compute $v(x) = v(0) + \int_0^x v'(t) dt$.  There is something to be said about why this is justified.
It might not be expected on an exam to regurgitate all the "machinery" of Sobolev norm definitions, but a key idea to keep in mind is that $H^1(0,1)$ is defined in such a way that $C^\infty[0,1]$ functions will be dense in the topology given by the $H^1$ norm.
So it suffices to prove the estimate $||v||_1 \le C|v|_1$ for some constant $C \gt 0$ (and all $v\in V$, $C$ not depending on $v$) by proving the estimate only for "nice" smooth functions, about which the meaning of $v(0)=0$ has no ambiguity.
Then the gist of your appeal to Cauchy-Schwartz to prove the estimate is correct.
A: Pick $f\in V$, then
\begin{align*}
\|f\|_{H^1}^2 
&= \int_{0}^1 f(x)^2\, \mathrm{d} x + \int_{0}^1 f'(x)^2\, \mathrm{d} x\\
&= \int_{0}^1 f'(x)\left(\int_{x}^1 f(\xi)\, \mathrm{d} \xi \right)\, \mathrm{d} x + \int_0^1 f'(x)^2\, \mathrm{d} x\\
&\le \left(\int_0^1 f'(\xi)^2\, \mathrm{d}x \right)^{1/2} \left(\int_0^1 \left( \int_x^1 f(\xi)\, \mathrm{d} \xi \right)^2\, \mathrm{d} x\right)^{1/2} + \int_0^1 f'(x)^2\, \mathrm{d} x\\
& \le \left(\int_0^1 f'(\xi)^2\, \mathrm{d}x \right)^{1/2}  \left(\int_0^1 \left( \int_x^1 f(\xi)^2\, \mathrm{d} \xi\right)\left(\int_{x}^1 1 \, \mathrm{d} \xi  \right)\, \mathrm{d} x \right)^{1/2} + \int_0^1 f'(x)^2\, \mathrm{d} x\\
& \le \left(\int_0^1 f'(\xi)^2\, \mathrm{d}x \right)^{1/2}  \left(\int_0^1 \left( \int_x^1 f(\xi)^2\, \mathrm{d} \xi\right)(1-x)\, \mathrm{d} x \right)^{1/2} + \int_0^1 f'(x)^2\, \mathrm{d} x\\
& \le \left(\int_0^1 f'(\xi)^2\, \mathrm{d}x \right)^{1/2}  \left(\int_0^1 \int_x^1 f(\xi)^2\, \mathrm{d} \xi\, \mathrm{d} x \right)^{1/2} + \int_0^1 f'(x)^2\, \mathrm{d} x\\
&= \left(\int_0^1 f'(\xi)^2\, \mathrm{d}x \right)^{1/2}  \left(\int_0^1 \int_0^\xi f(\xi)^2\, \mathrm{d} x\, \mathrm{d} \xi \right)^{1/2} + \int_0^1 f'(x)^2\, \mathrm{d} x\\
&= \left(\int_0^1 f'(\xi)^2\, \mathrm{d}x \right)^{1/2}  \left(\int_0^1  f(\xi)^2 \int_0^\xi\, \mathrm{d} x\, \mathrm{d} \xi \right)^{1/2} + \int_0^1 f'(x)^2\, \mathrm{d} x\\
&= \left(\int_0^1 f'(\xi)^2\, \mathrm{d}x \right)^{1/2}  \left(\int_0^1  f(\xi)^2 \xi \right)^{1/2} + \int_0^1 f'(x)^2\, \mathrm{d} x\\
&\le \left(\int_0^1 f'(\xi)^2\, \mathrm{d}x \right)^{1/2}  \left(\int_0^1  f(\xi)^2 \right)^{1/2} + \int_0^1 f'(x)^2\, \mathrm{d} x\\
&= |f|_{H^1} \|f\|_{L^2}+ |f|_{H^1}^2\\
&= |f|_{H^1} (\|f\|_{L^2} + |f|_{H^1})\\
&\le |f|_{H^1} \sqrt{2} ( \|f\|_{L^2}^2 + |f|_{H^1}^2)^{1/2}\\
&=\sqrt{2}  |f|_{H^1} \|f\|_{H^1} \\
\end{align*}\
So comparing the first line with the last and canceling a factor of $\|f\|_{H^1}$ we get $\|f\|_{H^2} \le \sqrt{2}|f|_{H^1}$  Since I wrote every single step, I leave it to the reader to supply the justifications.
