Proving an alternative norm on Sobolev space is equivalent to usual norm

Question

I have this exercice and my problel is only in item 4, and i will desespere.

Let $f \in L^2(\mathbb{R}^n).$

1- Why the equation $\Delta u - u = \dfrac{\partial f}{\partial x_i}$ admits a unique solution $u \in H^1(\mathbb{R}^n)$?

2- Prove that there exist a constant $C \geq 0$ that $||u||_{H^1} \leq C ||f||_{L^2}$.

3- Prove that there exist a constant $M \geq 0$ that for all $u \in H^2(\mathbb{R}^n)$ we have $||u||_{H^2} \leq M (||u||_{L^2})$.

4- We assume that $$\sum_{i,j=1}^n \displaystyle\int_{\mathbb{R}^n} \dfrac{\partial^2 u}{\partial x_i^2} \overline{\dfrac{\partial^2 v}{\partial x_j^2}} \,\mathrm dx + \lambda \displaystyle\int_{\mathbb{R}^n} u \overline{v} \,\mathrm dx$$ represente an scalar product to $H^2(\mathbb{R}^n)$ for all $\lambda > 0.$

• Prove that this scalar product is equivalent to the classical scalar product to $H^2(\mathbb{R}^n)$

  We denote the norm defined by this scalar product $\|\cdot\|_*$. I wan't to prove the existance of two constantes positives $c_1$ and $c_2$ such that $$c_1 \|u\|_{H^2} \leq \|u\|_* \leq c_2 \|u\|_{H^2}.$$ But i can't prove this two inequality.

Okay, so my work for item 4 is:

to prove the second inequality: we have from item 3) that: $||u||_{H^2} \leq M (||u||_{L^2} + ||\Delta u||_{L^2})$ and we know that $\Delta u = \sum_{i=1}^n \dfrac{\partial^2 u}{\partial x_i^2}$ so $$\sum_{i,j=1}^n \displaystyle\int \dfrac{\partial^2 u}{\partial x_i} \overline{\dfrac{\partial^2 v}{\partial x_j}} dx = \displaystyle\int |\Delta u|^2 dx = ||\Delta u||^2_{L^2}$$ but my problem is to use item 3 to deduce the second inequality.

To prove the first inequality, we have $$||u||^2_{H^2} =||\Delta u||^2_{L^2} + ||\nabla u||^2_{L^2} + ||u||^2_{L^2}$$ and we have by Holder and Young inequalities, \begin{align*} \sum_{i,j=1}^n \displaystyle\int \dfrac{\partial^2 u}{\partial x_i^2} \overline{\dfrac{\partial^2 u}{\partial x_j^2}} dx & \leq \sum_{i,j=1}^n ||\dfrac{\partial^2 u}{\partial x_i^2}||^2_{L^2} . ||\dfrac{\partial^2 \overline{u}}{\partial x_j^2}||^2_{L^2}\\ & \leq \dfrac{1}{2} \sum_{i,j=1}^n (||\dfrac{\partial^2 u}{\partial x_i^2}||^2_{L^2} + ||\dfrac{\partial^2 \overline{u}}{\partial x_j^2}||^2)\\ & \leq \sum_{i,j=1}^n (||\dfrac{\partial^2 u}{\partial x_i^2}||^2_{L^2} + ||\dfrac{\partial^2 \overline{u}}{\partial x_j^2}||^2) \end{align*}

and and I'm stuck for the rest

i have difficulties just for the last step, help me please to finish this exercice Thank's for help.

Answer 1

The statement in item 3 looks wrong. Should it be $\|u\|_{H^2} \leq M (\|u\|_{L^2}+\|\Delta u\|_{L^2})$? This is the inequality you quote in the sentence "we have from item 3)".

The norm defined by the inner product with $\lambda$ is: $$\|u\|_*^2 = \|\Delta u\|_{L^2}^2+ \lambda \|u\|^2_{L^2} \tag1$$ According to your post, the "classical norm" is $$\|u\|^2_{H^2} =\|\Delta u\|^2_{L^2} + \|\nabla u\|^2_{L^2} + \|u\|^2_{L^2}\tag2$$ We have the inequality $\|u\|_*^2 \le (1+\lambda) \|u\|^2_{H^2} $ just as a matter of algebra. In the converse direction, the issue is to estimate $\|\nabla u\|^2_{L^2}$ from above using $\|\Delta u\|_{L^2}^2$ and $\|u\|^2_{L^2}$. This is done by integration by parts followed by the famous $xy\le x^2+y^2$ inequality: $$\int_{\mathbb R^n} \nabla u\cdot \nabla u = - \int_{\mathbb R^n} u\,\Delta u \le \int_{\mathbb R^n}(|u|^2+|\Delta u|^2) \tag3$$ (Strictly speaking, one proves (3) for smooth compactly supported functions first, and then uses the fact that they are dense in $H^2(\mathbb R^n)$.) Once we have (3), the rest again reduces to algebra:
$$\|u\|^2_{H^2} \le (2+2\lambda^{-1})\|u\|_*^2$$