Proving an alternative norm on Sobolev space is equivalent to usual norm 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.
 A: 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:
$$\|u\|^2_{H^2} \le (2+2\lambda^{-1})\|u\|_*^2$$
