How does $\| \nabla u \|_{L^2}^2 \leq C \|\nabla u \|_{L^2}$ imply $\| u\|_{W^{1,2}}^2 \leq \| u\|_{W^{1,2}}$? So I'm working on some notes and I found this inequality that I really can't make sense of it. This is what is going on:
Let $u \in W_0^{1,2}$ and let $f \in L^2$ both on some $\Omega \subseteq \mathbb{R}^n$ open and bounded. If 
$$ \int |\nabla u|^2 \leq \int |fu|$$
then
$$ \|\nabla u \|_{L^2}^2=\int |\nabla u|^2 \leq (Holder) \leq \|f\|_{L^2} \| u \|_{L^2} \leq (Poincaré) \leq C \|f\|_{L^2} \| \nabla u \|_{L^2}$$
for some constant C. For some reason from this we get
$$ \| u \|_{W_0^{1,2}}^2 \leq C \|f \| \| u \|_{W_0^{1,2}}$$
but I don't get why there is this implication.
Attempt:
We know by definition that $\| u \|_{W^{1,2}} = \| u \|_{L^2} + \| \nabla u \|_{L^2}$. Also from
$$ \|\nabla u \|_{L^2}^2 \leq C \|f\|_{L^2} \| \nabla u \|_{L^2}$$
we can add $\| u \|_{L^2}$ on both sides getting
$$ \|u\|_{L^2} + \|\nabla u \|_{L^2}^2 \leq C \|f\|_{L^2} \| \nabla u \|_{L^2} + \| u \|_{L^2} \leq C \| u \|_{W_0^{1,2}} $$
but how do I get the $W_0^{1,2}$ norm squared?
Thank you!
 A: You have to use the Poincaré inequality again. This gives
$$\lVert u\rVert_{L^2}^2\le C_1\lVert \nabla u\rVert^2_{L^2}\le C_2\lVert f\rVert_{L^2}\lVert \nabla u\rVert_{L^2}, $$ 
so $\lVert u\rVert_{L^2}^2 + \lVert \nabla u\rVert^2_{L^2}\le C_3\lVert f\rVert_{L^2}\lVert \nabla u\rVert_{L^2}$,
for some big enough constant $C_3>0$. This is the inequality you wanted.

Note that here I used the usual definition of $\|u\|_{W^{1,2}}$, which is $$\|u\|_{W^{1,2}}^2:=\|u\|_{L^2}^2 + \|\nabla u\|_{L^2}^2.$$ This produces a norm that is equivalent to the one you gave. This is readily seen using the elementary inequality 
$$
\left( x^2+y^2\right)^{1/2}\le \lvert x \rvert + \lvert y\rvert \le \sqrt{2}\left( x^2+y^2\right)^{1/2}.$$
This norm has also the great benefit of being Hilbertian.
A: By Poincaré, 
$$\int{|u|^2} \leq C\int{|\nabla u|^2} \leq C\int{|fu|}.$$
By Cauchy-Schwarz, 
$$\|u\|^2_{L^2} \leq C\|f\|_{L^2}\|u\|_{L^2}.$$
Thus $\|u\|_{L^2} \leq C\|f\|_{L^2}$. 
Now, with the same estimation,
$$\|\nabla u\|_{L^2}^2 \leq \int{|fu|} \leq \|f\|_{L^2}\|u\|_{L^2} \leq C\|f\|^2_{L^2}.$$
Therefore, 
$$\|u\|^2_{W^{1,2}_0} = \|u\|^2_{L^2}+\|\nabla u\|^2_{L^2} \leq (C^2+C)\|f\|_{L^2},$$
thus $\|u\|_{W^{1,2}_0} \leq (C+0.5)\|f\|_{L^2},$ hence $$\|u\|_{W^{1,2}_0}^2 \leq (C+0.5)\|f\|_{L^2}\|u\|_{W^{1,2}_0},$$ where $C$ is the Poincaré constant.
