I'd like to know whether $f(x,y)=(x+1)(y+1)$ is strictly convex over $\mathbb{R^2_+}$.
I try to work with the Hessian, I have: $Hf(x,y)=\begin{bmatrix}0&1 \\ 1&0\end{bmatrix}$
This gives me the eigenvalues $\lambda=-1$ and $\lambda=1$. Since they are not both greater than $0$, I would think it's not strictly convex. But is it the right conclusion considering that the function is only defined on $\mathbb{R^2_+}$? How could I take the domain into account?
It's not even convex. $f(1,9) = f(9,1) = 20$ but $f(5,5) = 36 > 20$.
If there is anywhere that the Hessian is not positive semidefinite, that means it can't be convex or strictly convex, as there are points near there violate the convexity condition.