I want to prove bu using spectral decomposition that $${\left\| {{\partial _x}{{(1 - \partial _x^2)}^{ - 1}}{\partial _x}h} \right\|_{{L^2}(0,L)}} \le C{\left\| h \right\|_{{L^2}(0,L)}}$$ it is well known that the operator $${T:=(1 - \partial _x^2)^{ - 1}}:{L^2}(0,L) \to H_0^1(0,L) \cap {H^2}(0,L) \subset {L^2}(0,L)$$ is linear continuous, self-adjoint and compact and positive definite, so there exists a Hilbertian basis of eigenvectors of $T$. So for all function belonging to ${L^2}(0,L)$ we can write $$f = \sum\limits_{n \ge 0} {{a_n}\sin ({{n\pi x} \over L})} $$ and define the operator $$A: = {\partial _x}{(1 - \partial _x^2)^{ - 1}}{\partial _x}$$ so $$Af = \sum\limits_{n \ge 0} {{{\lambda _n^2} \over {\lambda _n^2 + 1}}{a_n}\sin ({{n\pi x} \over L})} $$ where ${({\lambda _n})_{n \ge 0}}$ is a sequence of eigenvalues of $T$.
My question is: how can I prove the inequality above by using this decomposition? and is this decomposition is correct?
Use Parseval's identity an the fact that $$ \Bigl|\frac{\lambda_n^2}{\lambda_n^2+1}\,a_n\Bigr|\le|a_n|. $$