# 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.

• Your question is phrased as an isolated problem, without any further information or context. This does not match MSE quality standards, so it may attract downvotes, or be closed. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. Making these improvements will attract more appropriate answers and make the question more valuable for future MSE visitors. – Lord_Farin Apr 27 '13 at 10:39
• i edit my message. can you help me please? – jijiii Apr 27 '13 at 10:45
• Thanks for providing context. What did you try to do so far for part 4? (Please edit the question to include your work.) – Lord_Farin Apr 27 '13 at 10:49
• i edit my message to write my work. Can you help me please. – jijiii Apr 27 '13 at 11:10
• Thank you! This way we can provide you optimal assistance. – Lord_Farin Apr 27 '13 at 11:15

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:
$$\|u\|^2_{H^2} \le (2+2\lambda^{-1})\|u\|_*^2$$
• by 3 we have $$||\Delta u||^2 + ||\nabla u||^2 + ||u||^2 \leq 2(||\Delta u||^2 + ||\Delta u||^2)$$ how we deduce the last relation with $\lambda$? – jijiii Apr 27 '13 at 16:12
• I don't found $$\|u\|^2_{H^2} \le (2+2\lambda^{-1})\|u\|_*^2$$ i found $$\|u\|^2_{H^2} \le (2+\lambda)\|u\|_*^2$$ what's the good result? – jijiii Apr 27 '13 at 18:40