# 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!

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.

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.