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

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 .

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