Let $(E, |\cdot|_1)$ be a normed vector space (n.v.s) and $(E', \| \cdot \|_1)$ its dual. Let $|\cdot|_2$ be an equivalent norm of $|\cdot|_1$ on $E$, and $\| \cdot \|_2$ its dual norm on $E'$. In a previous thread, I proved that
If we define a new norm $|x| := \sqrt{|x|_1^2 + |x|_2^2}$ on $E$, and let $\| \cdot \|$ be its dual norm on $E'$, i.e., $$ \|f\| := \sup_{0\neq x\in E} \frac{\langle f, x\rangle}{|x|}, \quad \forall f\in E'. $$ Then $$ \|f\|^2 = \min_{g\in E'} \left [ \|f - g \|_1^2 + \|g\|_2^2 \right ]. $$ Moreover, if ether $\| \cdot \|_1$ or $\| \cdot \|_2$ is strictly convex, then so is $\| \cdot \|$.
As such, I try to prove below duality result
If we define a new norm $\|f\| := \sqrt{\|f\|_1^2 + \|f\|_2^2}$ on $E'$, and let $| \cdot |$ be its dual norm on $E$, i.e., $$ |x| := \max_{0\neq f\in E'} \frac{\langle f, x\rangle}{\|f\|}, \quad \forall x\in E. $$ Then $$ |x|^2 = \inf_{y\in E} \left [ |x-y |_1^2 + |y|_2^2 \right ]. $$ Moreover, if ether $| \cdot |_1$ or $| \cdot |_2$ is strictly convex, then so is $| \cdot |$.
However, I'm stuck at the ending (and essential) step. I would like to ask if this duality result is true and if it's possible to fix my below failed attempt.
My attempt: First, we need some auxiliary results.
Let $(E, |\cdot|_1)$ be a n.v.s and $(E', \| \cdot \|_1)$ its dual. Let $J:E \to E'', x \mapsto \hat x$ be the canonical isometric linear embedding. Let $\varphi, \psi:E \to (-\infty, +\infty]$ such that $\varphi, \psi \not\equiv +\infty$. We define $$ \begin{align} \varphi':E' \to (-\infty, +\infty], f \mapsto \sup_{x\in E} [ \langle f, x \rangle - \varphi(x)] \\ \varphi'':E'' \to (-\infty, +\infty], \lambda \mapsto \sup_{f\in E'} [ \langle \lambda, f \rangle - \varphi'(f)]. \end{align} $$
The conjugates $\psi',\psi''$ are defined similarly. The inf-convolution is defined by $$ \varphi \square \psi:E \to [-\infty, +\infty], x \mapsto \inf_{y\in E} [\varphi(x-y) + \psi(y)]. $$
Lemma 1: Assume there exists $f_0 \in E'$ such that $\varphi' (f_0), \psi' (f_0) \neq +\infty$. Then $$ (\varphi \square \psi)' = \varphi'+\psi'. $$ This is Ex 1.23.1 in Brezis's book of Functional Analysis.
Lemma 2: If $\varphi$ is convex and l.s.c., then $\varphi (x) = \varphi'' (\hat x)$. This is Fenchel–Moreau theorem.
Lemma 3: $\|f\|^2 = \sup_{x\in E} \left [ 2 \langle f, x \rangle - |x|^2 \right ]$ for all $f\in E'$. A proof can be found here.
First, let $\varphi(x) := |x|_1^2$ and $\psi(x) := |x|_2^2$. By Lemma 3, $\|f\|_1^2 = \sup_{x\in E} \left [2 \langle f, x \rangle - |x|_1^2 \right ] = \varphi'(2f)$. Similarly, $\|f\|_2^2 = \psi'(2f)$. By a variant of Lemma 3, $$ \begin{align} |x|^2 &= \sup_{f\in E'} \left [ 2\langle f, x \rangle - \|f\|^2 \right ] \\ &= \sup_{f\in E'} \left [ 2\langle f, x \rangle - \|f\|_1^2 - \|f\|_2^2 \right ] \\ &= \sup_{f\in E'} \left [2 \langle f, x \rangle - \varphi'(2f) - \psi'(2f) \right ] \\ &= \sup_{f\in E'} \left [ \langle f, x \rangle - \varphi'(f) - \psi'(f) \right ] \\ &= \sup_{f\in E'} \left [ \langle \hat x, f \rangle - \varphi'(f) - \psi'(f) \right ] \\ & = (\varphi' + \psi')'(\hat x)\\ & = \big((\varphi \square \psi)'\big)'(\hat x) \quad \text{by Lemma } 1 \\ & = (\varphi \square \psi)'' (\hat x) . \end{align} $$
Our proof will be complete if we can show that $(\varphi \square \psi)'' (\hat x) = (\varphi \square \psi)(x)$. This would be true by Lemma 2 if $\varphi \square \psi$ is l.s.c. Clearly, $\varphi \square \psi$ is u.s.c. In this direction, we need to prove $\varphi \square \psi$ is continuous, which may be not true.