How to show that the dual of $(\mathbb{R}^n,\|{\cdot}\|_p)$ is $(\mathbb{R}^n,\|{\cdot}\|_q)$?

Question

I am trying to brush up on my functional analysis and I learn some $L_p$ spaces since I was never formally intrduced to them through courses. I wanted to know if anyone could offer me a proof or give me a resouce that that would have the proof of the fact that the dual space of $(\mathbb{R}^n,\|{\cdot}\|_p)$ is isometrically isomorphic to $(\mathbb{R}^n,\|{\cdot}\|_q)$ whenever $\frac{1}{p}+\frac{1}{q}=1$.

Answer 1

Let $(y_1,\dots,y_n)$ be an element of the dual space. By definition, the dual space norm is $$ \|y\|_* = \sup_{\|x\|_p\le 1} \sum_{i=1}^n x_i y_i \tag1 $$ By Hölder's inequality, $$ \sum_{i=1}^n x_i y_i \le \|x\|_p \|y\|_q $$ Thus, $\|y\|_* \le \|y\|_q$.

To prove the converse inequality, we should estimate the supremum in (1) from below. To this end, consider the vector $\tilde x$ defined by $\tilde x_i=|y_i|^{q-2} y_i/\|y\|_q^{q-1}$. A computation shows that $$\|x\|_p=\frac{1}{\|y\|_q^{q-1}} \left(\sum_{i=1}^n |y_i|^{p (q-1)}\right)^{1/p} =\frac{1}{\|y\|_q^{q-1}} \|y\|_{q}^{q/p} = 1 $$ Thus, $$ \|y\|_* \ge \sum_{i=1}^n \tilde x_i y_i =\frac{1}{\|y\|_q^{q-1}} \sum_{i=1}^n |y_i|^{q} =\frac{\|y\|_q^{q}}{\|y\|_q^{q-1}} = \|y\|_q $$ Done.