Eigenfunctions Laplacian - Bounding the Fourier Coefficients

Let $$\Omega \subset \mathbb{R}^{N}$$ be an open set with boundary of class $$C^{\infty}$$ and let $$\{\lambda_{k}\}$$ and $$\{v_{k}\}$$ be the eigenvalues and eigenvectors of -$$\Delta$$ with Dirichlet condition in $$\Omega$$, i.e., $$\begin{equation*} -\Delta v_{k} = \lambda_{k} v_k \end{equation*}$$ in which $$\Delta$$ is the Laplacian Operator. We assume that $$\{v_{k}\}$$ are normalized in $$L^{2}(\Omega)$$, so that $$\{v_{k}\}$$ is a Hilbert Basis for $$C^{\infty}(\Omega)$$, i.e., $$\begin{equation*} u = \sum_{k=1}^{\infty} \langle u,v_{k} \rangle v_{k}(x) \ \ \ \ \text{ (in } L^{2} \text{)} \end{equation*}$$ for all $$u \in C^{\infty}(\Omega)$$ and $$\begin{equation*} \langle v_{k},v_{j} \rangle = \begin{cases} 0, & \text{ if } k \neq j \\ 1, & \text{ if } k = j \end{cases} \end{equation*}$$ in which $$\langle u,v \rangle = \int_{\Omega} u(x) v(x) \ dx$$ is the inner product of $$L^{2}(\Omega)$$ (we suppose that $$u$$ and $$v_{k}$$ are real-valued functions).

Now, let $$u \in C^{1}(\bar{\Omega})$$, $$u(x) = 0$$ for $$x \in \partial \Omega$$. Show that there exists a constant $$C > 0$$ such that $$\begin{equation*} \Bigg\lvert \int_{\Omega} u(x) v_{k}(x) \ dx \Bigg\rvert \leq \frac{C}{\lambda_{k}^{1/2}} \end{equation*}$$

From First Greens's Identity, for example, we get that $$\begin{equation*} \lambda_{k} \Bigg\lvert \int_{\Omega} u(x) v_{k}(x) \ dx \Bigg\rvert \leq \Bigg\lvert \int_{\Omega} \langle\nabla u(x),\nabla v_{k}(x)\rangle \ dx \Bigg\rvert \end{equation*}$$ but I cannot figure out how to proceed from here.

EDIT: The result follows by applying the Cauchy-Schwarz inequality $$\begin{equation*} \lambda_{k} \Bigg\lvert \int_{\Omega} u(x) v_{k}(x) \ dx \Bigg\rvert \leq \int_{\Omega} \lvert\langle\nabla u(x),\nabla v_{k}(x)\rangle \rvert\ dx \leq \int_{\Omega} |\nabla u(x)||\nabla v_{k}(x)| dx \leq \lVert (|\nabla u(x)|) \rVert_{L^{2}} \lVert(|\nabla v_{k}(x)|) \rVert_{L^{2}} \end{equation*}$$ and First Green's identity $$\begin{equation*} \int_{\Omega} v_{k}(x) \Delta v_{k}(x) dx = -\int_{\Omega} |\nabla v_{k}(x)|^{2} dx \implies \lVert(|\nabla v_{k}(x)|) \rVert_{L^{2}} = \sqrt{\lambda_{k}} \end{equation*}$$ so the result follows taking $$C = \lVert (|\nabla u(x)|) \rVert_{L^{2}} < \infty$$. Notice the difference between the norm $$| \cdot |$$ of vectors in $$\mathbb{R}^{N}$$ and the norm $$\lVert \cdot \rVert_{L^{2}}$$ of $$L^{2}$$.

From the inequality you have obtained: $$\begin{equation*} \lambda_{k} \Bigg\lvert \int_{\Omega} u(x) v_{k}(x) \ dx \Bigg\rvert \leq \Bigg\lvert \int_{\Omega} \nabla u(x) \cdot \nabla v_{k}(x) \ dx \Bigg\rvert \end{equation*}$$ you can first use Cauchy-Swharz: $$\lambda_{k} \Bigg\lvert \int_{\Omega} u(x) v_{k}(x) \ dx \Bigg\rvert \leq \|\nabla u\|_{L^2} \|\nabla v_k \|_{L^2}$$ then you can use the crucial property:

$$\|\nabla v_k\|_{L^2}= \sqrt{\lambda_k}$$

as by Green: $$\|\nabla v_k\|_{L^2}^2=\int_\Omega \nabla v_k \cdot \nabla v_kdx = -\int_\Omega v_k \nabla\cdot(\nabla v_k)dx= -\int_\Omega v_k \Delta v_kdx=\lambda_k \int_\Omega v_k v_k dx =\lambda_k \|v_k\|_{L^2}^2=\lambda_k$$

So finally: $$\lambda_{k} \Bigg\lvert \int_{\Omega} u(x) v_{k}(x) \ dx \Bigg\rvert \leq \sqrt{\lambda_k} \|\nabla u\|_{L^2}$$ which is the expected result (as $$u \in C^1$$, the quantity $$\|\nabla u\|_{L^2}$$ is finite).

• Thanks. I had applied the Cauchy-Swharz ineuqality but have not been able to bound the norm of the gradient of $v_{k}$. Oct 27, 2018 at 18:50
• @marcstyle to bound the norm of the gradient of $v_k$ the idea is to notice that by integration by parts (or Green): $$\int \nabla v_k \nabla v_k = -\int \nabla \cdot (\nabla v_k) v_k=-\int \Delta v_k v_k$$ and $-\delta v_k =\lambda_k v_k$. Oct 28, 2018 at 12:47
• Indeed, but it just seem that it is missing a vector norm. In your answer I believe we have to show that $\lVert (|\nabla v_{k}|) \rVert_{L^{2}} = \sqrt{\lambda_{k}}$, i.e., the $L^{2}$ norm of the gradient vector norm equals the square root of the eigenvalue. In fact, $\lVert \nabla v_{k} \rVert_{L^{2}}$ does not make sense as $\nabla v_{k}$ is a vector in $\mathbb{R}^{N}$ and not a function in $L^{2}$. I have incorporated your answer to my question with this correction. Oct 30, 2018 at 19:24
• By the $L^2$ norm of the vector field I mean:$$\left(\int_\Omega \nabla v_k \cdot \nabla v_k \right)^\frac{1}{2}$$ where $\cdot$ is the inner product of $\mathbb{R}^N$. If you need additional explanations I can edit my answer :-). Oct 30, 2018 at 20:32
• Yeah, I figured, thanks. Oct 30, 2018 at 23:00