A inequality involving $L^p$ norm and $L^1$ norm.

Question

Prove \begin{equation} \lVert f\rVert_p\leq \sup_{\lVert g\rVert_q =1}\lVert fg\rVert_1 , \end{equation}where \begin{equation} \dfrac {1} {p} +\dfrac {1} {q}=1.\end{equation}

Answer 1

When $1\leq p< \infty$, Hölder gives the reverse inequality. Which is therefore an equality, where the sup turns out to be a a max. That's the isometric embedding of $L^p$ into $(L^q)^*$. It is always surjective when $1<p<\infty$. For $p=\infty$, if the measure is $\sigma$-finite, this still yields an isometry between $L^\infty$ and $(L^1)^*$.

So to get that inequality (assuming of course $\|f\|_p>0$), you need to take the right $g$. Try something based on $f$.

For $p=1$, just take $g=1$. For $1< p<\infty$, take $$g=\frac{f^{p-1}}{\|f\|_p^{p-1}}.$$ For $p=\infty$, take a measurable set $A$ of positive measure on which $|f(x)|\geq (1-\epsilon) \|f\|_\infty$. Then set $$g=\frac{1_A}{\mu(A)}$$ to get that the rhs is $\geq (1-\epsilon) \|f\|_\infty$ for every $\epsilon>0$.