Does a strongly convex function in multiple variables have a "contraction" property?

Contraction is probably a bad term but I mean the following property: if $$f(x)$$ is $$\lambda$$-strongly convex and $$x^*$$ is its minimizer, then for all $$x$$ we have $$\|x-x^*\|^2\leq \frac{2}{\lambda}(f(x)-f(x^*))$$. What you get is that if the function value is close to the minimum, then the variables are also close.

I'm wondering what if $$f(x,y)$$ is strongly convex in $$x$$ and $$y$$, that is, for any $$x$$, $$f(x,\cdot)$$ is strongly convex and same hold for any fixed $$y$$ (but not necessarily jointly convex)? Say I find $$x,y$$ s.t $$f(x,y)\leq f(x^*,y^*)+\epsilon$$, where $$(x^*,y^*)$$ minimizes $$f$$. Can we say anything about $$x$$ being close to $$x^*$$ or $$y$$ being close to $$y^*$$? Thanks.

No. The contraction property is a direct consequence of strong convexity. Specifically, from the first order character of strongly convex functions: $$f(\mathbf{y})\geq f(\mathbf{x})+\partial f(\mathbf{x})^T(\mathbf{y}-\mathbf{x})+\frac{\lambda}{2}\|\mathbf{y}-\mathbf{x}\|^2.$$ Without strong convexity, there is no contraction. Furthermore, convexity in individual coordinates might sound like a "nice" property, however it's almost meaningless from the perspective of convex analysis. Think of the function $$f(x,y) = x^22^{y^2+1}$$ which is strongly convex for $$x,y$$ separately, but is a completely messy function otherwise.