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Imagine you are given the following condition: $$ a < 0 < b $$ And you have the following operator: $$ L^2([a, b]) \to L^2([a, b]) \\ \; \\ f(x) \to \cos({x})f(x) $$ If you want to bound the norm of the operator: $$ \Vert Tf(x)\Vert = \int_a^b |\cos(x)f(x)|^2dx \leq \max\{ |\cos(x)|^2\}\int_a^b |f(x)|^2dx \leq\int_a^b |f(x)|^2 dx= \Vert f(x)\Vert $$

$$ \Vert Tf(x)\Vert \leq \Vert f(x)\Vert \; \; \Rightarrow \; \; \Vert T\Vert \leq 1 $$ But how would you find the exact norm of $T$. I'm stuck. Any Hint?

J. W. Tanner
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Álvaro Rodrigo
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  • The operator norm can be given by $\sup{\Vert T f\Vert\ ~:~\Vert f\Vert \leq 1}$. But this doesn't help much. – K.defaoite Jan 29 '24 at 20:34
  • You can't have an $x$ outside the integral Replace the max with $1$ before moving it outside. –  Jan 29 '24 at 20:34
  • More generally: for the norm for a "multiplication map" $f \mapsto \phi f$, the norm is the "essential supremum" of $|\phi|$. – GEdgar Jan 29 '24 at 20:48

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