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Let $U$ be a bounded domain in the Euclidean space with sufficiently smooth boundary. Let $\{f_i\}$ be a orthonormal basis of $H^1_0(U)$ satisfying $-\Delta f_i = \lambda_i f_i$ where $\lambda_i \leq \lambda_{i+1}$.

For fixed $N\in\mathbb N$ let $V_N$ be the subspace of $H^1_0(U)$ spanned by $\{f_1, \dots , f_N\}$. Let $u$ be an element of $H^1_0(U) \cap H^2(U)$ and $\pi_N(u)$ be its $L^2$ orthogonal projection onto $V_N$.

I want to show that $$ \int_U \mid \nabla u - \nabla \pi_N(u) \mid ^2 \leq C \int_U \mid \nabla \nabla u \mid ^2 $$ where $C$ is specifically chosen to be $\frac{1}{\lambda_N}$ and $\nabla\nabla u$ is the Hessian of $u$.

How is this possible? I can show that there exists such a $C$ via argument by contradiction. But, I cannot find a way to 'construct' a specific $C$. Could anyone help me?

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  • $\begingroup$ Is the double $\nabla\nabla u$ a typo in your statement and did you mean $(\dots)\leq C\int_U |\nabla u|^2$, or do you write $\nabla\nabla u$ the full second-order hessian $D^2 u$? $\endgroup$ Apr 20 '20 at 12:54
  • $\begingroup$ It is full Hessian. I figured out using the orthonormal basis anyway... $\endgroup$
    – Isaac
    Apr 20 '20 at 12:56
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    $\begingroup$ OK. I edited your question to make this clear, since this notation is not completely standard. $\endgroup$ Apr 20 '20 at 12:59
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Yes, this is possible and actually true, up to a slight shift in the index and a missing constant: $$ \|\nabla u-\nabla \pi_N(u)\|^2_{L^2}\leq \frac{d}{\lambda_{N+1}}\| D^2 u\|^2_{L^2}. $$ Here $d$ is the dimension of $U\subset \mathbb R^d$.

Proof Let me first remind a classical fact: Since the $f_i$'s are orthogonal, so are the gradients $\nabla f_i$ and the Laplacians $\Delta f_i$ (in the $L^2$ sense). By this I mean $$ (\nabla f_i,\nabla f_j)_{L^2}=0 \quad \mbox{and} \quad (\Delta f_i,\Delta f_j)_{L^2}=0 \qquad \mbox{for }i\neq j. $$ Writing $u=\sum_{i\geq 1} u_i f_i$, we have $\pi_N(u)=\sum_{i\leq N}u_i f_i$ thus $$ \|\nabla u-\nabla\pi_N(u)\|^2_{L^2} = \left\|\sum_{i\geq N+1}u_i\nabla f_i\right\|^2_{L^2}=\sum_{i\geq N+1} u_i^2\|\nabla f_i\|^2_{L^2}. $$ Now using $-\Delta f_i=\lambda_i f_i$ it is eas to see that $$ \|\nabla f_i\|^2_{L^2}=\lambda_i \|f_i\|^2=\lambda_i \|\frac{1}{\lambda_i}\Delta f_i\|^2_{L^2}=\frac{1}{\lambda_i}\|\Delta f_i\|^2_{L^2}, $$ hence \begin{multline*} \|\nabla u-\nabla\pi_N(u)\|^2_{L^2}=\sum_{i\geq N+1}u_i^2\frac{1}{\lambda_i}\|\Delta f_i\|^2_{L^2} \\ \leq \frac{1}{\lambda_{N+1}}\sum\limits_{i\geq N+1}u_i^2\|\Delta f_i\|^2_{L^2} \leq \frac{1}{\lambda_{N+1}}\sum_{i\geq 1}u_i^2\|\Delta f_i\|^2_{L^2} \\ = \frac{1}{\lambda_{N+1}}\left\|\sum_{i\geq 1}u_i\Delta f_i\right\|^2_{L^2} =\frac{1}{\lambda_{N+1}} \|\Delta u\|^2_{L^2}. \end{multline*} Using the convexity inequality $|\sum_{k=1}^d a_k|^2\leq d \sum_{k=1}^d a_k^2$ gives $\|\Delta u\|^2_{L^2}\leq d \|D^2u\|^2_{L^2}$ and the result follows.

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  • $\begingroup$ Thank you for your detailed answer. Could you help me with the link below as well? I would enormously appreciate it.. $\endgroup$
    – Isaac
    Apr 20 '20 at 13:52
  • $\begingroup$ math.stackexchange.com/questions/3632210/… $\endgroup$
    – Isaac
    Apr 20 '20 at 13:52

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