If for all $f\in L_p$, $g\in L_q$ with $q=p/(p-1)$ satisfies $G(f)=\int fg d\mu$, then $\|G\|=\|g\|_q$. Where we define $\|G\|=\sup \left\{|G(f)|:f\in L_p,\;\|f\|_p\leq1 \right\}$.

My attempt:

$|G(f)|=|\int fg d\mu|\leq\int |fg| d\mu\leq \|g\|_q\|f\|_p\leq\|g\|_q$

Taking the sup over $f$, we get $\|G\|\leq\|g\|_q$

Now I need to prove the converse inequality and this is where I get stuck. However, if I set $q=p=2$, I can do it easily by evaluating $G(\cdot)$ at $g/\|g\|_2$:

$|G(g/\|g\|_2)|=|\int \frac{g^2}{\|g\|_2} d\mu|=\int \frac{g^2}{\|g\|_2} d\mu=\|g\|_2$

Since $\|(g/\|g\|_2)\|=1$ this implies that $\|G\|\geq\|g\|_2$

Hence, for $p=q=2$ we get that $\|G\|=\|g\|_2$

Any hint on how to deal with the general case will be greatly appreciated.

After much thought and seeing this post I realize that I can use:



Define $G(f)\equiv \int fg\,d\mu$. In order to show that $g\in L_q$ we begin by showing that $G$ is a bounded linear functional. Indeed,\begin{align*} G(af_1+bf_2)&=\int(af_1+bf_2)g\,d\mu\\ &=a\int f_1g\,d\mu+b\int f_2g\,d\mu\\ &=aG(f_1)+bG(f_2) \end{align*} thus $G$ is linear. Now, by Hölder's inequality,$$ |G(f)|\leq\int |fg|\,d\mu\leq\|f\|_p\|g\|_q,\text{ with }\frac{1}{p}+\frac{1}{q}=1. $$ Hence, $|G(f)|\leq M\|f\|_p$ with $M=\|g\|_q$ and $G$ is a bounded linear functional. By Riesz Representation Theorem, $g\in L_q$. Now, set$$ \|G\|=\sup \{|G(f)|:f\in L_p,\|f\|_p\leq 1\}. $$ Then$$ \|G\|\leq |G(f)|\leq \|f\|_p\|g\|_q\leq \|g\|_q. $$ On the other hand, set$$ f=\text{sign}(g)|g|^{q/p}\|g\|^{-q/p}. $$ Then, by a simple computation, we get that$$ |G(f)|\geq \|g\|_q. $$ Hence, $\|g\|_q\leq \|G\|$ and thus $\|G\|=\|g\|_q$.


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