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Let $m\in C[a,b]$ and consider $$ T:(C[a,b],\Vert\cdot\Vert_1) \to (C[a,b],\Vert\cdot\Vert_1): f\mapsto m\cdot f $$ Provided $\Vert T\Vert = \Vert m\Vert_\infty$, I want to show $\|T\|$ is attained iff there exi st $f\in C[a,b]\setminus\{0\}$ such that $\Vert m\Vert_\infty|f| = |m||f|$.

The 'if' direction is trivial, how to deal with the 'only if' part?

Thank you in advance.

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1 Answer 1

up vote 2 down vote accepted

If $\Vert T\Vert$ is attained then for some $f\in C([a,b])\setminus\{0\}$ holds $$ \Vert T(f)\Vert_1=\Vert m\Vert_\infty\Vert f\Vert_1 $$ which is equivalent to $$ \int\limits_a^b\Delta(x)dx=0 \tag{1} $$ where $\Delta(x)=\Vert m\Vert_\infty|f(x)|-|m(x)||f(x)|$. Obviously $\Delta\in C([a,b])$ and $\Delta(x)\geq 0$ for all $x\in[a,b]$. So $(1)$ holds only if $\Delta(x)= 0$ for all $x\in[a,b]$. This is equivalent to $|m||f|=\Vert m\Vert_\infty|f|$.

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