I was playing around with the strong convexity definition and got stuck at some point. I was wondering if someone could kindly help me out.
We say that function $f$ is strongly convex if
$1) f(x) \geq f(y) + \xi^T(x- y) + \frac{\mu}{2}||x-y||^2$ for $\mu >0$ and $\forall x, y \in dom f$
Now, suppose $f$ is convex and let
$h(x) = f(x) -\frac{\mu}{2}||x||^2$
which is also convex. If $\xi \in \partial f(y)$, then $\xi - \mu y \in \partial h(y)$.
Then, I can use the definition of convexity for function $h$.
$h(x) \geq h(y) + (\xi - \mu y)^T (x-y)$. Now, let us replace $h$ by $ f -\frac{\mu}{2}||x||^2$.
$f(x) -\frac{\mu}{2}||x||^2 \geq f(y) -\frac{\mu}{2}||y||^2 + (\xi - \mu y)^T (x-y)$
$f(x) \geq f(y) + \frac{\mu}{2}||x||^2 -\frac{\mu}{2}||y||^2 + \xi^T(x-y) - \mu y^Tx + \mu ||y||^2$
This is where I got stuck. I was expecting to obtain the inequality (1), however, I am making a mistake somewhere.
Just observe that $$ \frac{\mu}{2}\|x\|^2-\frac{\mu}{2}\|y\|^2+{\mu}\|y\|^2=\frac{\mu}{2}\|x\|^2+\frac{\mu}{2}\|y\|^2 $$ and $$ \frac{\mu}{2}\|x\|^2+\frac{\mu}{2}\|y\|^2-\mu y^Tx=\frac{\mu}{2}\|x-y\|^2. $$