Is $l_2$ on $\mathbb{R}^n$ the only norm for which it is equal to its dual norm?

Question

Given any norm $\|.\|$ on $\mathbb{R}^n$, its dual norm $\|.\|^D$ is defined as the following: $\|v\|^D = \sup_{\|x\|\leq 1} |(v,x)|$, where $(,)$ is the standard Euclidean Inner product. Under that definition, it turns out that the dual norm of the $l_p$ norm is $l_q$, when $1/p+1/q=1$. That implies that for $l_2$, it is equal to the dual norm. Now I wanted to think about the converse: If $\|.\|$ is a norm such that $\|.\| = \|.\|^D$, then is $\|.\|=l_2$? After being unable to find anything online, I wrote the following justification:

Consider any vector in the boundary of the ball of radius $1$ in $\|.\|$ : \begin{align*} 1 = \|x\| = \|x\|^D = \sup\left\{\left|\sum^n_i x_iv_i\right|: \|v\| \leq 1\right\} \end{align*} Using C-S Inequality (with $(,)$ denoting the usual inner product) we have: \begin{align*} |(x,v)| \leq \|x\|_2 \|v\|_2 \end{align*} where equality holds if and only if $x,v$ are linearly dependent. Considering that, let $v = x$. Then $\|v\| = 1$ and we have: \begin{align*} |{(x,v)}| = \|x\|_2^2 \implies \|x\|^D = \|x\| = 1 \geq \|x\|^2_2 \implies 1 \geq \|x\|_2 \end{align*} Suppose $\exists v_0$ such that $\|v_0\|<1$, and $\|x\|^D \leq |{(x,v_0)}|$ . Then consider $v' = \frac{v}{\|v_0\|}$. We will have $\|{(x,v)}\| \leq \|{x}\|^D \leq |(x,v_0)| \implies \|v_0\|\geq 1$. Hence we have a contradiction.

Consider the functional defined by $f: \overline{B_1}(0)\rightarrow \mathbb{R}$ as $f(v) = |(x,v)|$, where $B$ denotes the ball in $\|.\|$. Then by the compactness of the unit ball in any norm, and continuity of $f$, it attains maxima; let that point be denoted by $v_0$. Then by the earlier paragraph, $\|v_0\|=1$. However, we showed that whenever $\|v_0\| =1$, we should get $\|v_0\|_2 \leq 1$. Hence we get: \begin{align*} \left|{\sum^n_i x_iv_i}\right| \leq |{(x,v_0)}| \leq \|x\|_2\|v_0\|_2 \leq \|x\|_2 \implies 1 = \|x\|^D \leq \|x\|_2 \end{align*} Hence we get the following for any unit vector: \begin{align*} \|x\| = \|x\|_2 \end{align*} Now given a general non zero vector, consider: \begin{align*} 1 = \|{\frac{x}{\|x\|}}\| = \|{\frac{x}{\|x\|}}\|_2 \implies \|x\|= \|x\|_2\ \square \end{align*}

1. Is the proposed solution correct? Is there a better way to see the result?
2. Can this notion be meaniningfully extended in any sense to infinite dimensions?

Answer 1

Let $V$ denote a subspace of square summable real valued sequences with a norm $\|\cdot \|.$ For example $V= \mathcal{F}$, the subspace of all the sequences with finitely many nonzero terms or $V=\ell^2.$ Assume the norm is self-dual, i.e. $$\|x\|=\sup\left \{ \left |\sum_{j=1}^\infty x_jy_j\right | \,:\, \|y\|\le 1\right \}\quad (*)$$ In particular $(*)$ means that the RHS is finite. Let $\|x\|=1.$ Then $$1=\sup\left \{ \left |\sum_{j=1}^\infty x_jy_j\right | \,:\, \|y\|\le 1\right \}$$ Hence plugging in $y=x$ gives $$1\ge \sum_{j=1}^\infty x_j^2=\|x\|_2^2$$ Thus $\|x\|_2\le 1=\|x\|.$ By homogeneity this implies $$\|x\|_2\le \|x\|$$ Fix $x\neq 0.$ There exists $y^{(n)}$ such that $\|y^{(n)}\|\le 1$ and $$ \left |\sum_{j=1}^\infty x_jy^{(n)}_j \right | \ge \|x\|-{1\over n} $$ By the Cauchy-Schwarz inequality we get $$\|x\|-{1\over n} \le \|x\|_2\|y^{(n)}\|_2\le \|x\|_2\,\|y^{(n)}\|\le \|x\|$$ Hence $$\|x\|_2\ge \lim_n\|x\|_2\|y^{(n)}\|=\|x\|$$ Thus we have obtained $\|x\|=\|x\|_2.$