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

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?

• 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 $D^2 u$? Apr 20 '20 at 12:54
• It is full Hessian. I figured out using the orthonormal basis anyway... Apr 20 '20 at 12:56
• OK. I edited your question to make this clear, since this notation is not completely standard. Apr 20 '20 at 12:59

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.