I have a task as in the quote block. By checking the Hessian matrices of $f(x,y)$'s. I found Hessian's of a) b) are not semidefinite, and c) is. Explicitly $H=$
a) \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}
b) \begin{pmatrix} -\frac{1}{x^2\ln10} & 0 \\ 0 & -\frac{1}{y^2\ln10} \end{pmatrix}
c) \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}
so I concluded that only c) is a convex function. My question is why the domain $\left\{(x, y) \in R^{2} |x^{2}+y^{2} \leq 4\right\}$ given in the question makes no impact. Did I miss something, or my answer is wrong?
Which of the following functions is convex? Prove your answers directly by verifying the definition, by applying theorems discussed in class or by discussing a counter example. Make sure you check all the parts of the definition.
a) $f:\left\{(x, y) \in R^{2} |x^{2}+y^{2} \leq 4\right\} \rightarrow R, f(x, y)=x y$
b) $f:\left\{(x, y) \in R^{2} | x^{2}+y^{2} \leq 4\right\} \rightarrow R, f(x, y)=\log _{10}(x y)$
c) $f:\left\{(x, y) \in R^{2} | x^{2}+y^{2} \leq 4\right\} \rightarrow R, f(x, y)=\log _{10}\left(e^{x} e^{y}\right)$
Actually in (b), the domain is incorrect: $\log_{10}(xy)$ is not a real number if $xy \le 0$.
But the domain (when it is correct) has no impact here because the Hessian is or is not positive semidefinite on the whole domain. If you had a function such as $x^3 + y^2$, whose Hessian matrix is $\pmatrix{6x & 0\cr 0 & 2\cr}$, that matrix is positive semidefinite if and only if $x \ge 0$. Then the function is convex on a domain in which $x \ge 0$ always, but not if the domain contains points with $x < 0$.