Explain the convexity by looking the hessian matrix of a function

Question

The hessian matrix of a function is given by, $$ H = \begin{bmatrix} a & b & c \\[0.3em] b & b & 0\\[0.3em] c & 0 & c \end{bmatrix} $$

where, $a,b,c>0$.

How do we show that the function given my this matrix is convex or concave?

Answer 1

Twice differentiable function is convex on a convex set if and only if its Hessian matrix is positive semidefinite on the interior of the convex set. You can use Sylvester’s criterion to check the positive semidefiniteness: $$ \begin{cases} a \ge 0, \\ ab - b^2 \ge 0, \\ abc - b^2c-c^2b \ge 0. \end{cases} $$ Using your conditions $a, b, c > 0$ we get that if $a \ge b + c$ then you function is convex (but not if and only if, Sylvester’s criterion is only sufficient condition but not necessary).

Similarly you can do for concavity.