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One define the operator $T$ as :$$T: = (I - {{{\partial ^2}} \over {\partial {x^2}}}):H_0^1(0,L) \cap {H^2}(0,L) \to {L^2}(0,L) $$ let $f \in H_0^2(0,L) \cap {H^4}(0,L)$. What can we say about ${T^{ - 1}}{{{\partial ^4}f} \over {\partial {x^4}}}$ ? In which space is it?

My attempt is as following : ${T^{ - 1}}:{L^2}(0,L) \to H_0^1(0,L) \cap {H^2}(0,L)$ so ${{{\partial ^4}f} \over {\partial {x^4}}} \in {L^2}(0,L)$ So $${\left\| {{T^{ - 1}}{{{\partial ^4}f} \over {\partial {x^4}}}} \right\|_{{L^2}}} \le c{\left\| {{{{\partial ^2}f} \over {\partial {x^2}}}} \right\|_{{L^2}}}. $$ Is this write ? Thank you.

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  • $\begingroup$ Yes, $(1-D^2)^{-1}D^2$ is bounded on $L^2$ with operator norm $=1$. This is immediate from the fact that $-D^2\ge 0$ if you impose Dirichlet boundary conditions. $\endgroup$ Commented Mar 1, 2017 at 23:46
  • $\begingroup$ @Christian Remling.Hello sir. have we the same result here?math.stackexchange.com/questions/2297354/… $\endgroup$
    – Gustave
    Commented May 26, 2017 at 15:10

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