If $|x| := \sqrt{|x|_1^2 + |x|_2^2}$ on $E$, then $\|f\|^2 = \min_{g\in E'} \left [ \|f - g \|_1^2 + \|g\|_2^2 \right ]$ on its dual $E'$ In solving this exercise, I come across below question.

Let $(E, |\cdot|_1)$ be a 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 $E'$. We define a new norm $|x| := \sqrt{|x|_1^2 + |x|_2^2}$ on $E$ and let $\| \cdot \|$ be its dual norm on $E'$. Then
$$
\|f\|^2 = \min_{g\in E'} \left [ \|f - g \|_1^2 + \|g\|_2^2 \right ].
$$
If ether $\| \cdot \|_1$ or $\| \cdot \|_2$ is strictly convex, then so is $\| \cdot \|$.

Could you have a check on my attempt?
I post my proof separately as an answer below. This allows me to subsequently remove this question from unanswered list.
 A: First, we need some auxiliary results.

Let $(E, |\cdot|)$ be a normed linear space and $(E', \| \cdot \|)$ its dual. Let $\varphi, \psi:E \to (-\infty, +\infty]$ such that $\varphi, \psi \not\equiv +\infty$. The conjugate $\varphi':E' \to (-\infty, +\infty]$ of $\varphi$ is defined as
$$
\varphi'(f) := \sup_{x\in E} [ \langle f, x \rangle - \varphi(x)], \quad \forall f \in E'.
$$
The conjugate $\psi'$ of $\psi$ is defined similarly. The inf-convolution of $\varphi$ and $\psi$ is defined by
$$
(\varphi \nabla \psi)(x) := \inf_{y\in E} [\varphi(x-y) + \psi(y)], \quad \forall x\in E.
$$
Lemma 1: $\|f\|^2 = \sup_{x\in E} \left [ 2 \langle f, x \rangle - |x|^2 \right ]$ for all $f\in E'$. A proof can be found here.
Lemma 2: Assume $\varphi, \psi$ are convex, and there exists $x_0 \in E$ such that $\varphi(x_0), \psi(x_0) \neq +\infty$ and that $\varphi$ is continuous at $x_0$. Then
$$
(\varphi+\psi)' = ( \varphi' \nabla \psi') \quad \text{on} \,\, E'.
$$
This is Ex 1.23.3 in Brezis's book of Functional Analysis.
Lemma 3: The space $E$ is strictly convex if and only if the map $x \mapsto |x|^2$ is strictly convex. A proof is given here.

We have $|\cdot|_1$ is equivalent to $|\cdot|_2$ if and only if $\|\cdot \|_1$ is equivalent to $\|\cdot\|_2$. There exist $\alpha, \beta >0$ such that $\alpha \|f\|_2 \le \|f\|_1 \le \beta \|f\|_2$. Let $\varphi(x) := |x|_1^2$ and $\psi(x) := |x|_2^2$. By Lemma 1,
$$
\|f/2\|^2 = \sup_{x\in E} \left [ \langle f, x \rangle - |x|^2 \right ] = \sup_{x\in E} \left [ \langle f, x \rangle - \varphi(x) - \psi(x) \right ] = (\varphi + \psi)'(f).
$$
Also by Lemma 1, $\varphi'(f) = \sup_{x\in E} \left [ \langle f, x \rangle - |x|_1^2 \right ] = \|f/2\|_1^2$. Similarly, $\psi'(f) = \|f/2\|_2^2$. By Lemma 2:
$$
(\varphi + \psi)'(f) = ( \varphi' \nabla \psi')(f) = \inf_{g\in E'} \left [ \|f/2 - g/2 \|_1^2 + \|g/2\|_2^2 \right ].
$$
It follows that
$$
\|f\|^2 = \inf_{g\in E'} \left [ \|f - g \|_1^2 + \|g\|_2^2 \right ].
$$
Consider $\Psi_f: g \mapsto  \|f - g \|_1^2 + \|g\|_2^2$. If $\|g\|_2 \ge \|f\|_2$, then
$$
\Psi_f(g) \ge \alpha^2 \|f - g \|_2^2 + \|g\|_2^2 \ge (\alpha^2+1) \|g\|_2^2 - \alpha^2 \|f\|_2^2 \ge \|f\|_2^2 = \Psi_f(f).
$$
Let $G_f:=\{g\in E' ; \|g\|_2 \le \|f\|_2\}$. Then
$$
\|f\|^2 = \inf_{g\in G_f} \Psi_f(g).
$$
Because the norm of dual space is l.s.c. in weak$^\star$ topology, so is $\Psi_f$. By Banach-Alaoglu theorem, the set $G_f$ is compact in weak$^\star$ topology, so
$$
\|f\|^2 = \min_{g\in G_f} \Psi_f(g) = \min_{g\in E'} \left [ \|f - g \|_1^2 + \|g\|_2^2 \right ].
$$
Now we assume that $\| \cdot\|_1$ is strictly convex. By Lemma 3, we need to prove that the map $f \mapsto \|f\|^2$ is strictly convex. Let $t \in (0, 1)$ and $f,h\in E'$ such that $f \neq h$. Let $g,g'$ be the minimizers of $\Psi_f$ and $\Psi_h$ respectively. Then
\begin{aligned}
  & \|tf+(1-t)h\|^2 \\
\le & \left \|[tf+(1-t)h] - [tg+(1-t)g'] \right\|_1^2 + \|tg+(1-t)g'\|_2^2 \\
= & \left \| t(f-g)+(1-t)(h-g') \right\|_1^2 + \|tg+(1-t)g'\|_2^2 \\
< & t \| f-g\|_1^2 + (1-t) \| h-g' \|_1^2 + t\|g\|_2^2 + (1-t)\|g'\|_2^2  \text{ because } \| \cdot\|_1 \text{ is strictly convex}\\
= & t\|f\|^2 + (1-t) \|h\|^2.
\end{aligned}
This completes the proof.
