I'm trying to solve an exercise that asks if a function $f: \mathbb{R}^2 \to \mathbb{R}$ is $\alpha$-convex. The function in question is defined by: $f(x,y):=x^2 + 2x + xy + y^2 - 4y + 3.$
I know the for convexity, we need to calculate the Hessian of $f$ and see if it's non-negative definite (and positive-definite for strict convexity), but upon an internet search, I could locate anything useful for when a function $f$ is $\alpha$-convex. So I'd appreciate a formal definition and the solution to the exercise above.
P.S. My guess would be if a function $f$ is $\alpha$-convex, $\alpha > 0,$ then $Hess(f) - \alpha I$ is non-negative definite, but I'm not sure!
Yes. A function $f$ is $\alpha$-convex, if and only if $\nabla^2f(x) - \alpha I \succeq 0 \ \forall x$.
We can easily deduce that this $\alpha$ correspond to the minimum possible eigen value of the Hessian.
Hint: $\nabla^2f(x) - \alpha I \succeq 0 \ \implies y^T (\nabla^2f(x) - \alpha I) y = y^T\nabla^2f(x)y - \alpha \geq0$. Can you find the $y$ for which we get the minimum value?
In your case, $$ \lambda_{min} = 1 $$
Hence, the function $f$ is $1$-convex.