$ \lVert {\bf f} \rVert_{p} \leqslant (m(E))^{1-1/p} \lVert {\bf f} \rVert_{\infty} $ for any $ f \in L_{\infty}(E) $

Question

If $m(E)$ is finite and ${\bf f}\in L_\infty(E)$ then for any $p\geqslant 1$, $$ \lVert {\bf f} \rVert_{p} \leqslant (m(E))^{1-1/p} \lVert {\bf f} \rVert_{\infty} .$$

I tried to apply Hölder's inequality here but I am not sure whether that is the right path or not. Any hints are appreciated.

Answer 1

$$ \int_E \lvert f \rvert^p \cdot 1 \leqslant \left(\sup_E \lvert f \rvert^p \right) \int_E 1 = \lVert f \rVert_{\infty}^p m(E), $$ since $\lVert \lvert f \rvert^p \rVert_{\infty} = \lVert f \rVert_{\infty}^p$. Therefore, taking $p$th roots, $$ \lVert f \rVert_{p} \leqslant m(E)^{1/p}\lVert f \rVert_{\infty} $$ It has to be this because the equality case will occur if $f=1$; your inequality would give $m(E)^{1/p} \leqslant m(E)^{1-1/p}$, which is not true for every $p$ or possible value of $m(E)$.