What is the dual of $(\mathbb{R}^n,∥\cdot∥_{\infty})$

Question

I know that the dual of $(\mathbb{R}^n,∥\cdot∥_{p})$ is $(\mathbb{R}^n,∥\cdot∥_{q})$ with $\frac{1}{p}+\frac{1}{q}=1$ But does this also hold when $p=\infty$ and if so what is the proof?

Answer 1

Every linear functional $f$ on $\mathbb R^n$ is determined by its values on the standard basis $e_1,\dots,e_n$. And conversely, for any $n$-tuples $(y_1,\dots,y_n)$ there is a linear functional $f$ such that $f(e_j)=y_j$. This sets up a bijection between the dual of $\ell_\infty^n$ and the space $\ell_1^n$. It remains to check the bijection is an isometry: $$\sup_{\|x\|_\infty\le 1 } x\cdot y = \|y\|_1$$ The inequality $\le $ follows from the triangle inequality. For $\ge$, use the vector $x$ such that $x_j=\operatorname{sign} y_j$ for all $j$.