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.

Answer 2

As I am too just a student actually learning this topic below a more details answer that I hope will help you and other students.
As my answer is long and with a lot of details beware, maybe there is some mistakes if you see one please write it in the comment.

Background

To prove that the dual of $l^p_n$ identifies itself with $l^q_n$, where $1<p,q<∞$ and $ \frac{1}{p} + \frac{1}{q} = 1 $ we utilize the concept of dual spaces and properties of norms in $l_p$ spaces . The dual space of a normed vector space $E$, denoted as $E^∗= L(E; \mathbb{R})$, consists of all bounded linear mapping from $T:E \to\mathbb{R}$.
So we want to show that $(l^p_n)^*$ are isometric isomorph to $l^q_n$ (includes all the requirements of isomorphism but adds the preservation of distances or norms).

Prove

1- $ \forall \underline{y} = ( y_1; y_2; ... ; y_n) \in l^p_n $ we define the following linear mapping to $ \mathbb{R} $ $T_{\underline{x}}(\underline{y})= <\underline{x}|\underline{y}> = \sum_{1 \leq i \leq n}x_i \cdot y_i$ with $\underline{x} \in l^p_n$.
As you can see there is an infinity of such linear mapping each one caraterize by his "parameter" $\underline{x} $

2- The linear mapping $T_{\underline{x}}(.)$ verify the Holder inequality $ |<\underline{x}|\underline{y}>| \leq ||\underline{x}||_q ||\underline{y} ||_p$
This allow us to prove that $T_{\underline{x}}(.)$ is continuous and bounded.
We write $ || . ||_* $ the norm of the dual space $(l^p_n)^*$. Hence we have that every $ ||T_{\underline{x}}(.) ||_* \leq ||x||_q$. It means that the norm of $T_{\underline{x}}(.)$ in the dual space is at most $||x||_q$

3- Now let prove the isomorphism between $(l^p_n)^*$ and $l^q_n$ as follow.
$T_{\underline{x}}(.) \mapsto \underline{x}$ it means we match every $ T_{\underline{x}} \in (l^p_n)^*$ (by definition) to the corresponding $\underline{x} \in l^q_n $. By definition this mapping is injective and surjective (trivial) thus it is bijective and this is a morphism.

4- Now let prove that this isomorphism is an isometry.
By definition the norm on $(l^p_n)^*$ is $||.||_{\infty}$ (as it is the norm in the space of linear mapping). Thus by definition if we want to show that this two space are isometric we need to show that $ \forall T_{\underline{x}}(.) \in (l^p_n)^* \Rightarrow ||T_{\underline{x}}(.) ||_* = || T_{\underline{x}}(.) ||_{\infty} = || \underline{x} ||_q ^n $
By definition $|| T_{\underline{x}}(.) ||_{\infty} = Sup \{ |<\underline{x} | \underline{y}>| : ||\underline{y}||_p \leq 1\}$

5- Now let's prove that for any given $\underline{y} \in \mathbb{R}^n $ different from zero, there exists at least one $ \underline{x} \in \mathbb{R}^n $ different from zero such that $ |x \cdot y | = || x ||_p ||y ||_q $
Constructing $x$
For each component $y_i$ of $y$, define $x_i = |y_i|^{q-1} \epsilon_i$ with $ \epsilon_i = 1 $ if $y_i \geq 0 $ and $\epsilon_i = -1 $ if $y_i < 0 $
Proving the Equality
1.Calculate $||x||_p^p = \sum \frac{|y_i|^{pq}}{|y_i|^p} = \sum \frac{|y_i|^{p+q}}{|y_i|^p} = \sum |y_i|^p = ||y||_q ^q$ from $\frac{1}{p} + \frac{1}{q} = 1 = \frac{p+q}{pq}$.
2.Calculate $|x \cdot y| = \left|\sum_{i=1}^{n} x_i y_i\right| = \left|\sum_{i=1}^{n} |y_i|^{q} \text{sgn}(y_i) y_i \epsilon_i \right| = \sum_{i=1}^{n} |y_i|^q$.
And this finish the proof.

Q.E.D.