Let $f \in \mathcal{C}^2(\mathbb{R}^n).$ Recall that we defined $f$ to be strongly convex if there exists $\beta > 0$ such that $\langle D^{2}f|_{x}y,y\rangle\ge\beta$ for every $x, y \in \mathbb{R}^{n}$ or equivalently if the function $g(x)=f(x)-\frac{\beta}{2}\|x\|^{2}$ is convex. Show that if there exists $\gamma>0$ such that $$f(tx+(1-t)y) \leq tf(x)+(1-t)f(y)-\gamma t(1-t)\|x-y\|^{2}\tag{*}$$ for all $x,y \in \mathbb{R}^{n}, t \in [0,1]$ then the function $g(x)=f(x)-\gamma\|x\|^2$ is convex.
I tried showing that the function $g(x)=f(x)-\gamma \|x\|^2$ is convex using the above assumption but did not strike any luck. Any hints on how to proceed for this problem are appreciated. Than you for you help.
"Equivalently" is not really right here as the definition of strong convexity does not assume $f \in \mathcal{C}^2$. However, $f$ satisfies (*) iff $g$ is convex.
The proof: (*) is equivalent to $$ g(tx+(1-t)y)\le tg(x)+(1-t)g(y)+\gamma\text{Res} $$ where $$ \text{Res}=t\|x\|^2+(1-t)\|y\|^2-\|tx+(1-t)y\|^2-t(1-t)\|x-y\|^2. $$ It is straightforward calculation to show that $\text{Res}=0$ for $\ell^2$-norm. For example, the $\|x\|^2$ term is $$ t\|x\|^2-t^2\|x\|^2-t(1-t)\|x\|^2=(t-t^2-t+t^2)\|x\|^2=0. $$ I leave it for you to verify the rest.