The convexity of a function $e^{x+y}$

Question

The question is to text the function $$e^{x+y}$$ and determine if it's convex, concave or not.

As $(e^x)' = e^x$, all four second derivatives in a Hessian matrix will be the same: $$H= \begin{bmatrix} e^{x+y} & e^{x+y} \\ e^{x+y} & e^{x+y} \\ \end{bmatrix}$$

And so the determinant will be naturally equal to zero.

Does it mean that the function is neither concave nor convex? Or it only means I should use another test to determine the convexity or concavity?

Thanks, Paul

Answer 1

Apply a change of coordinates in the following way:

$$X=x+y \quad Y=x-y.$$

So your function become $e^X$, which is constant along straight lines parallel to $X-$axis.

Now the Hessian matrix is $$ \begin{pmatrix} e^X & 0\\ 0 & 0 \end{pmatrix}, $$

which is diagonal and its eigenvalues are $0$ and $e^X$ which are non negative for every point $(X_0,Y_0)$, so the matrix is positive semidefinite, thus the function in convex.

Answer 2

$f: z \mapsto e^z$ is a convex function on $\mathbb R$ and $g: (x, y) \mapsto x + y$ is a linear mapping from $\mathbb R^2$ to $\mathbb R$. What can you say about the composition $f \circ g$ ?