I am trying to show: $$\|f_\alpha \| = \| \alpha \|_{\ell^\infty} $$ Given that: $$f_\alpha:\mathbb{R}^2 \rightarrow \mathbb{R}, \quad f_\alpha = \alpha_1 x_1 + \alpha_2 x_2$$
($\mathbb{R^2}$ is equiped with the $\|.\|_\infty$)
I am struggling to get passed the definition of the operator norm:
$$\|f\|= \sup \limits_{\substack{x \in \mathbb{R}^2\\ \|x\|_{\ell^\infty} = 1}} |\alpha_1 x_1 + \alpha_2 x_2 | = \sup \limits_{\substack{x \in \mathbb{R}^2\\ x \neq 0}} \frac{ |\alpha_1 x_1 + \alpha_2 x_2 |}{\|x \|_{\ell^\infty}}$$ Could someone point me in the right direction?
As @Alonso Delfín mentioned, the desired conclusion is wrong. Actually this question is related to the so-called dual norm. By Hölder inequality, for $a,x\in\mathbb{R}^n$, there holds $$ \sum |a_ix_i|\leq\|a\|_{\ell^p}\|x\|_{\ell^q}, $$ where $p^{-1}+q^{-1}=1$ for $1\leq p,q\leq\infty$ and for each $p,q$ and $a$ there exists $x$ establishing the equality. Therefore we have $$ \|\langle a,\cdot\rangle\|_{\ell^p}=\|a\|_{\ell^q}, $$ and that's why it is called the dual norm. As your case, it actually should be $$ \|\langle a,\cdot\rangle\|_{\ell^\infty}=\|a\|_{\ell^1}. $$