# If $u\in H^2(R^n)$, how to prove $\|D^2u\|_{L^2}$ is equal to $\|\Delta u\|_{L^2}$ using Fourier transforms?

Problem:

If $u\in H^2(R^n)$, how to prove $\|D^2u\|_{L^2}$ is equal to $\|\Delta u\|_{L^2}$ using Fourier transforms?

My first question: Is it right to prove this using integration by parts as follows? $$\|\Delta u\|_{L^2}^2=\int\Delta u \Delta u dx=\sum_{i,j=1}^{n}\int \partial_i \partial_i u \partial_j \partial_j u dx=\sum_{i,j=1}^{n}\int \partial_i \partial_j u \partial_i \partial_j u dx=\|D^2u\|_{L^2}$$ My second question: How to prove this using Fourier transform?

I use $$\hat u(\xi) =\int e^{-ix\xi}u(x)$$ as the definition of Fourier transform. $$\partial_ju = i\xi_j\hat{u}$$. Then $$\partial_k\partial_ju = i^2\xi_k\xi_j\hat{u} = \frac{\xi_k}{|\xi|}\frac{\xi_j}{|\xi|}(i^2|\xi|^2\hat{u}) = \frac{\xi_k}{|\xi|}\frac{\xi_j}{|\xi|} \widehat{\Delta u}.$$

To conlude take the $L^2$ norm on both sides and use the obvious inequality $\frac{\xi_j}{|\xi|} \le 1$.

If I take the $L^2$ norm on both sides then I get $\sum_{k,j} \int | \widehat {\partial_k \partial_j u} |^2=\sum_{k,j}\int (\frac{\xi_k \xi_j}{|\xi|^2}\widehat{\Delta u})^2$. Then how can I conclude $\|D^2u\|_{L^2} \le \|\Delta u\|_{L^2}$?

Could anyone kindly help? Thanks very much!

• What is $D$ here? Mar 29 '15 at 21:07
• The approach with Fourier transform looks basically like the approach with integration by parts: when you take the Laplacian you wind up multiplying in Fourier space by $|\xi|^2$, now rearrange the product of the sums into a sum of products so that it looks more like the Fourier result of $D^2u$.
– Ian
Mar 29 '15 at 21:07
• @CameronWilliams $D$ is (possibly weak) differentiation. Sherry is probably working out of Evans PDE.
– Ian
Mar 29 '15 at 21:08
• @user01123581321345589144... Sorry about that. I also got two downvotes. Someone seems is downvoting all questions. Mar 30 '15 at 1:33
• It is bad form to edit your post this heavily. If you have another question, please make another post and ask it there. Do not add onto your question after receiving a legitimate answer. This is likely why you are being downvoted. Mar 30 '15 at 3:10

By definition $$\hat u(\xi)=\int e^{-ix\xi}u(x)dx$$ We also know $\widehat{\partial^ku(\xi)}=i^{|k|}\xi^k\widehat{u(\xi)}$ Similar to the integration by part step posted in my question, we can get $$\int(\Delta u)^2=\sum_{i,j=1}^n\int(\widehat {u_{x_i x_i}} \widehat{u_{x_j x_j}})=\sum_{i,j=1}^n\int (-\xi_i \xi_i \hat u)( -\xi_j \xi_j \hat u)=\sum_{i,j=1}^n\int (-\xi_i \xi_j \hat u)( -\xi_i \xi_j \hat u)=\sum_{i,j=1}^n\int(\widehat {u_{x_i x_j}} \widehat{u_{x_i x_j}})=\int|\widehat {D^2u|^2}$$ Thus $$\|\widehat {\Delta u}\|_{L^2}=\|\widehat {D^2 u}\|_{L^2}$$. By plancherel's theorem, $$\|\Delta u\|_{L^2}=\| {D^2 u}\|_{L^2}$$