Function which is strongly convex and L-smooth

Question

Let $f:\mathbb{R}^{n}\to \mathbb{R}$. We say that $f$ is $\mu$ strongly convex if $$ f(y)\ge f(x)+\nabla f(x)^T(y-x)+\frac{\mu}{2}\lVert x-y \rVert^2$$ and we say that $f$ is $\mu$ smooth if $$ f(y)\le f(x)+\nabla f(x)^T(y-x)+\frac{\mu}{2}\lVert x-y \rVert^2$$ If $f$ is both of them, how can I prove that $$f(x)=\frac{\mu}{2}\lVert x-b\rVert^2+c$$ It is obvius that $$f(y)=f(x)+\nabla f(x)^T(y-x)+\frac{\mu}{2}\lVert x-y \rVert^2$$ but I can not do any more. Any help will be appreciated.

Answer 1

You also don't have to do more, $x$ is fix. In the case $n=1$ it is easiest to see. $$f(x)+ f'(x)(y-x) + \frac{\mu}{2} (x-y)^2 = \frac{\mu}{2}y^2 + (f'(x)+\mu x)y +(f(x)+\frac{\mu}{2}x^2)$$ now complete the square to get something of the form $$\frac{\mu}{2}(y-b)^2 + c.$$ In higher dimensions this is the same (coordinatewise).