Let $\Gamma _g(f)=\int_{\mathbb R}fg$. Prove that $\|\Gamma _g\|=\|g\|_{L^{p'}}$.

Question

Let $p>1$ and $p'$ s.t. $\frac{1}{p}+\frac{1}{p'}=1$. Let $g\in L^{p'}$ and $\Gamma _g\in(L^p)'$ defined by $$\Gamma _g(f)=\int_{\mathbb R}fg.$$ Define $$\|\Gamma _g\|:=\sup_{\|f\|_{L^p}=1}|\Gamma _g(f)|.$$ How can I prove that $\|\Gamma _g\|=\|g\|_{L^{p'}}$ ? If $p=2$, then I have that that $|\Gamma _g(g/\|g\|_{L^{p'}})|=\|g\|_{L^{p'}}$ and thus $$\|\Gamma _g\|=\|g\|_{L^{p'}}.$$

But if $p>1$ and $p\neq 2$, I don't see how I can do because $g\notin L^p$ a priori. Any idea ?

Answer 1

Hint

If $f=g^{\frac{p'}{p}}$, then $f\in L^p$ and $fg=g^{p'}$.