Finite target bounds do not supply an infinite target bound

Use Lebesgue measure on (0,1) and let T be the identity on complex finite simple classes. It is complex-linear and has $\mathrm{\parallel }Tf{\mathrm{\parallel }}_1=\mathrm{\parallel }f{\mathrm{\parallel }}_1$ and $\mathrm{\parallel }Tf{\mathrm{\parallel }}_2=\mathrm{\parallel }f{\mathrm{\parallel }}_2$. Countable choice supplies the stated earlier Lebesgue-measure result, which gives measure one to (0,1) and measure $1\mathrm{/}n$ to $(0,1\mathrm{/}n)$ for each integer $n\ge 2$.

Define $f_n=n{\mathbf{1}}_{(0,1\mathrm{/}n)}$. It is a finite simple function of finite-measure support. Direct integration gives $\mathrm{\parallel }{f}_n{\mathrm{\parallel }}_1=n(1\mathrm{/}n)=1$ and $\mathrm{\parallel }{f}_n{\mathrm{\parallel }}_2^2=n^2(1\mathrm{/}n)=n$. Its infinity norm is n: n is a pointwise bound, and every smaller nonnegative bound fails on a set of measure $1\mathrm{/}n>0$. Thus an $L^1\to L^{\mathrm{\infty }}$ bound C would require $n=\mathrm{\parallel }T{f}_n{\mathrm{\parallel }}_{\mathrm{\infty }}\le C\mathrm{\parallel }{f}_n{\mathrm{\parallel }}_1=C$ for every $n\ge 2$, impossible for finite C.