Details on equivalent strong convexity Show $f(x)$ ($C^1$ smooth) is strongly convex iff $g(x)=f(x)-\frac{\alpha}{2}\|x\|^2$ is convex.
I'll use the definition of convexity, $f(y)\ge f(x)+\langle\nabla f(x), y-x\rangle$.
$f(y)-\frac{\alpha}{2}\|y\|^2\ge f(x)-\frac{\alpha}{2}\|x\|^2+\langle\nabla(f(x)-\frac{\alpha}{2}\|x\|^2),y-x\rangle$
I can distribute the $\nabla$ and break apart the inner product.
$f(y)-\frac{\alpha}{2}\|y\|^2\ge f(x)-\frac{\alpha}{2}\|x\|^2+\langle\nabla f(x),y-x\rangle-\langle\nabla\frac{\alpha}{2}\|x\|^2,y-x\rangle$
I can move the $\frac{\alpha}{2}\|y\|^2$ to the other side.
$f(y)\ge f(x)+\langle\nabla f(x),y-x\rangle+\frac{\alpha}{2}(\|y\|^2-\|x\|^2)-\langle\nabla\frac{\alpha}{2}\|x\|^2,y-x\rangle$
I'm not sure what to do with the last two terms to move toward the definition of strong convexity, namely $f(y)\ge f(x)+\langle\nabla f(x), y-x\rangle+\frac{\alpha}{2}\|y-x\|^2$.
 A: I think the "trick" you're missing is the identity
$$
\|x-y\|_2^2=\|x\|_2^2+\|y\|_2^2-2\langle{y,x\rangle}.
$$
($\Rightarrow$) Suppose $f$ is a strongly convex function with constant $\alpha>0$. Then
\begin{align*}
g(x)-g(y)&=f(x)-f(y)-\tfrac{1}{2}\alpha\big(\|x\|^2-\|y\|^2\big)\\
&\geq\langle\nabla{f}(y),x-y\rangle+\tfrac{1}{2}\alpha\|x-y\|^2-\tfrac{1}{2}\alpha\big(\|x\|^2-\|y\|^2\big) &&\text{b/c $f$ strongly convex}\\
&=\langle\nabla{f}(y),x-y\rangle+\tfrac{1}{2}\alpha\|x\|^2+\tfrac{1}{2}\alpha\|y\|^2-\alpha\langle{x,y}\rangle-\tfrac{1}{2}\alpha\big(\|x\|^2-\|y\|^2\big)  &&\text{ suggested identity}\\
&=\langle\nabla{f}(y),x-y\rangle-\alpha\langle{y,x}\rangle+\alpha\|y\|^2\\
&=\langle\nabla{f}(y)-\alpha{y},x-y\rangle\\
&=\langle\nabla{g}(y),x-y\rangle
\end{align*}
and thus $g$ is convex.
($\Leftarrow$) Now suppose there exists $\alpha>$ such that $g$ is convex. Then
\begin{align*}
f(x)-f(y)&=g(x)-g(y)+\tfrac{1}{2}\alpha\big(\|x\|^2-\|y\|^2\big)\\
&\geq\langle\nabla{g}(y),x-y\rangle+\tfrac{1}{2}\alpha\big(\|x\|^2-\|y\|^2\big)&&\text{b/c/ $g$ convex}
\end{align*}
Using the suggested trick, you can show that this last line is equal to 
$$
\langle\nabla{f}(y),x-y\rangle+\tfrac{1}{2}\alpha\|x-y\|^2,
$$
and thus $f$ is strongly convex. I'll leave the details to you.
