I'm trying to bound the quantity $\langle \nabla \Psi(x),\bar{x}-x \rangle$ above, with the bound depending on $\|x-\bar{x}\|$ and perhaps also of $\|x-y\|$ for fixed (but not varying) points $y$. Where here $\Psi(x):X\rightarrow \mathbb{R}$ with $X$ a finite dimensional Banach space (or $\mathbb{R}^n$, whatever) And $\Psi(x)$ is a $\mu$-strongly convex function (with $\mu$>0) that can be written as $\Psi=f+g$ with $f$ convex and differentiable with $\nabla f$ $L$-Lipschitz continuous and $g$ $\mu$-strongly convex.
I know that if $\Psi$ was differentiable and its gradient was $L$-Lipschitz continuous one could fix some point $x^*$ on the optimal set and bound as
$\langle \nabla \Psi(x), \bar{x}-x \rangle \leq \|\nabla \Psi(x)\|\|\bar{x}-x\| = \|\nabla \Psi(x)-\nabla \Psi(x^*)\|\|\bar{x}-x\| \leq L\|x-x^*\|\|\bar{x}-x\|$
And the bound is done. So my question is, is there an analogous of this property on the non-differentiable case? Like, I know that I can pick a point $x^*$ on the optimal set such that $0 \in \partial \Psi(x^*)$, but then can I say that for a $v \in \partial \Psi(x)$ it holds
$\|v\| = \|v-0\| \leq L\|x-x^*\|$ or something on that line?
Any help is appreciated
The answer is no. On the real line consider $\Phi(x)=|x| $ (and add some smooth convex function with minimum in zero if you like). Then the minimum is in zero but the subgradient at any positive point is about 1.