How can I show if the eigenvalues of the Hessian is greater than $\alpha$, then the function is strongly convex?

Question

If $f$ is strongly convex then there exists an $\alpha >0$ and a convex function $g$ such that $$f(x) = g(x) + \frac{\alpha}{2} \|x\|^{2}$$

How can I show that if the eigenvalues of $H_f(x) \geq \alpha$, then $f$ is strongly convex?

Answer 1

Your wording is a little ambiguous, here is the precise claim I believe you are asking about:

Claim. Let $h\colon \mathbb R^n\to \mathbb R$ be a twice differentiable function on $\mathbb R^n$ whose Hessian's eigenvalues are all at least $\alpha$. Then there exists a convex function $k(x)$ such that $h(x)=k(x)+\tfrac{\alpha}{2}\|x\|^2$.

Proof. Observe that $$ \frac{\partial }{\partial x^i}\frac{\partial }{\partial x^j}\bigl(h(x)-\frac{\alpha}{2}\|x\|^2\bigr)=\frac{\partial }{\partial x^i}\frac{\partial }{\partial x^j}h(x) - \alpha I_{ij}, $$ where $I$ is the $n\times n$ identity matrix. Since adding or subtracting a multiple of the identity matrix has the effect of shifting the eigenvalues (since that is the effect it has on the characteristic polynomial), it follows that the Hessian of $h(x)-\tfrac{\alpha}{2}\|x\|^2$ has all eigenvalues non-negative, meaning that the function $h(x)-\tfrac{\alpha}{2}\|x\|^2$ is convex. Finally, we set $k(x)$ to equal this function.