Does subdifferential change if the inner product of the spaces changes?

Question

Consider the finite dimension space $\Bbb R^n$ with two different inner products $\langle\cdot,\cdot\rangle_1$ and $\langle\cdot,\cdot\rangle_2$ respectively. Let $f:\Bbb R^n \to \bar{\Bbb R}$ be a convex function which is finite at $x_0$.

Define the convex subdifferential of $f$ at $x_0$ w.r.t to the inner product $\langle\cdot,\cdot\rangle_i$ by

$$\partial_i f(x_0) = \{ v \in R^n \mid \langle v,x - x_0\rangle_i \; \leq \; f(x) - f(x_0) ~~ \forall x \in \Bbb R^n \} $$

Question: What is the link between $\partial_1 f(x_0)$ and $\partial_2 f(x_0)$ ?

My thought: they are probably same up to multiplication a positive definite matrix, since inner product in finite dimension are link in such a way. I mean there exists a positive definite matrix $A$ such that $$\partial_2 f(x_0) = A \;\partial_2 f(x_0)$$

Am I right? Any precise proof would be appreciated .

Answer 1

There exists symmetric positive definite matrices $A_1$ and $A_2$ such that $\forall x,y, \langle x,y\rangle_1=x^TA_1y$ and $\langle x,y\rangle_2=x^TA_2y$.

Let $v\in \partial_1 f(x_0)$. Then for all $x\in \mathbb R^n$, $$\langle A_2^{-1}A_1v, x-x_0 \rangle _2 = (A_2^{-1}A_1v)^TA_2(x-x_0)=\langle v,x-x_0\rangle_1\leq f(x)-f(x_0)$$ thus $A_2^{-1}A_1v\in \partial_2 f(x_0)$, hence $A_2^{-1}A_1 \partial_1 f(x_0) \subset \partial_2 f(x_0)$ and by symmetry $A_1^{-1}A_2 \partial_2 f(x_0) \subset \partial_1 f(x_0)$.

For the reverse inclusion, take $v\in \partial_2 f(x_0)$ and write $v= (A_2^{-1}A_1) (\underbrace{A_1^{-1}A_2v}_{\in \partial_1 f(x_0)})\in A_2^{-1}A_1 \partial_1 f(x_0)$.

Therefore $$A_2^{-1}A_1 \partial_1 f(x_0) = \partial_2 f(x_0)$$