# Spectral decomposition of differential operator

Let $\Omega$ a bounded open set of $\mathbb R^n$ and let $\{ {e_n}\} _{n = 1}^\infty$ an orthonormal basis of ${L^2}(\Omega )$, let $\{ {\lambda _n}\} _{n = 1}^\infty$ the eigenvalues of the Laplacian Dirichlet operator.

It is well know that if $u \in H_0^1(\Omega )$ then $u = \sum\limits_{n \ge 1} {{\lambda _n}(} {e_n},u){e_n}$ where $(\cdot,\cdot)$ is the inner product in ${L^2}(\Omega )$, and for $f \in {H^{ - 1}}(\Omega )$ we have $f = \sum\limits_{n \ge 1} {{1 \over {{\lambda _n}}}(} {e_n},f){e_n}$.

Now I want to estimate the duality bracket $\langle f,u \rangle$ by using spectral decomposition. We have:$$\langle f,u \rangle = \sum\limits_{n \ge 1} ( {e_n},f)({e_n},u)$$ How can I prove the continuity of $f$ on $H_0^1(\Omega )$? My second question: Let $${\partial _x}{(I - \partial _x^2)^{ - 1}}{\partial _x}:{L^2}(0,1) \to {L^2}(0,1)$$ How can I use spectral decomposition to prove the continuity of this operator ? Thank you.

• I think "$u = \sum_{n \ge 1} {\lambda _n}( {e_n},u){e_n}$" should be $u = \sum_{n \ge 1} ( {e_n},u){e_n}$ or other changes should be made to make things consistent. For the first one, using $(\cdot,\cdot )$ for $L^2$-inner product and $\{e_n\}$ being orthonormal in $L^2$, don't you have \begin{align} |\langle f,u \rangle| &= \Big|\sum_{n \ge 1} \frac1{\lambda_n^{\frac12}}( {e_n},f)\lambda_n^{\frac12}({e_n},u)\Big|\\ &\le \Big(\sum_{n \ge 1} \frac1{\lambda_n}(e_n,f)^2\Big)^{\frac12} \Big(\sum_{n \ge 1} \lambda_n({e_n},u)^2\Big)^{\frac12} =\|f\|_{H^{-1}}\|u\|_{H^1_0} \end{align}? – shall.i.am May 31 '17 at 9:50
• Yes sir. I was wrong, so generally if we have $f \in {H^{ - m}}(\Omega )$ and $u \in H_0^m(\Omega )$ then $$(f,u) = \sum\limits_{n \ge 1} {(f,{e_n})(u,{e_n}) \le } \sum\limits_{n \ge 1} {{1 \over {\lambda _n^{{m \over 2}}}}(f,{e_n})\lambda _n^{{m \over 2}}(u,{e_n}) \le {{\left\| f \right\|}_{{H^{ - m}}(\Omega )}}} {\left\| u \right\|_{H_0^m(\Omega )}}$$ ?For the second statement, how can I start ? thanks for your answers sir. – Gustave May 31 '17 at 12:49
• One thing to note is that we have $(I-d^2/dx^2)^{-1}u=\sum_{n=1}^{\infty}\frac1{1+\lambda_{n}}(u_n,e_n)e_n$ for $u\in L^2(0,1)$. – shall.i.am Jun 1 '17 at 1:49
• Thank you shall.i.am ..my problem is how to deal with the derevative and its expansion.... – Gustave Jun 1 '17 at 2:16
• What is $e_n$ here, in the case of $L^2((0,1))$? For example if you have the zero Dirichlet boundary condition, it's just $\sqrt{2}\sin(n\pi\cdot)$, isn't it? – shall.i.am Jun 1 '17 at 2:26

For the first one, we have \begin{align} |\langle f,u \rangle| &= \Big|\sum\limits_{n \ge 1} \frac1{\lambda_n^{\frac12}}( {e_n},f)\lambda_n^{\frac12}({e_n},u)\Big|\\ &\le \Big(\sum\limits_{n \ge 1} \frac1{\lambda_n}( {e_n},f)^2\Big)^{\frac12} \Big(\sum\limits_{n \ge 1} \lambda_n({e_n},u)^2\Big)^{\frac12} =\|f\|_{H^{-1}}\|u\|_{H^1_0} \end{align} For the second one, we note we have $e_n=\sqrt{2}\sin(n\pi\cdot)$ under the zero Dirichlet condition.