The tightest upper bound on $-\left\langle B(y-x),\nabla f(A x)\right\rangle$

Question

Let $f$ be a $C^2$ convex function with Hessian $H$ such that $lI\mathcal\le H(x)\le LI,\ \forall x\in\mathbb{R}^n$. Then using Taylor's theorem it follows that $$f(x)-f(y)+\frac{l}{2}\|y-x\|_2^2\le-\langle y-x,\nabla f(x)\rangle\le f(x)-f(y)+\frac{L}{2}\|y-x\|_2^2$$ where $\|\cdot\|_2$ is the $l_2$ norm. I need to find an upper bound on the following expression $$-\left\langle B(y-x),\nabla f(A x)\right\rangle$$ where $A,B$ are $n\times n$ matrices. Note that here the notation $\nabla f(Ax)$ means evaluating $\nabla f$ at $Ax$.

Using the preceding inequalities I could derive the following bound:$$-\left\langle B(y-x),\nabla f(A x)\right\rangle\le f(Bx)-f(By)+\frac{L}{2}\|By-Ax\|_2^2-\frac{l}{2}\|Bx-Ax\|_2^2$$ However, though the LHS of the inequality beocmes $0$ when $x=y$, the RHS doesn not, in general, which makes me conclude that this must be a loose inequality.

Can anybody help refer me to any existing technique that might help me finding the tightest inequality? Basically, I need the convex function-type inequality as in the first inequalities, and Cauchy-Scwartz type inequality is not useful to me.

Please do not provide a full answer, and give only suggestions, or references to useful techniques. Thanks in advance.

I asked this question here in math.stackexchange too, but nobody responded.

Answer 1

If $A$ is invertible and $BA^{-1}$ is, by any chance, symmetric positive definite, then you could write $$ \langle B(y-x),\nabla f(Ax)\rangle = \langle BA^{-1}A(y-x),\nabla f(Ax)\rangle $$ and use that $\langle BA^{-1}y,x\rangle =: \langle y,x\rangle_{BA^-1}$ defines an inner product. Use your first inequality in this new inner product and norm to get $$ -\langle B(y-x),\nabla f(Ax)\rangle \leq f(Ax)-f(Ay) + \frac{L}2\|A(y-x)\|_{BA^{-1}}^2. $$