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?



Approximating functions in $H^1_0(U) \cap H^2(U)$ via $H^1$ norm and $L^2$ projection
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.


