Is exp(-x) convex?

Question

Is $f(x)=e^{-x}$ a convex function?

I know that $e^x$ is convex. If I take the second order derivative of $f(x)$:

$$f''(x)=e^{-x}$$

Then we can see for all the $x$, $f''(x)>0$. I'm not sure about the case $f''(\infty)=0$. But it looks satisfy the convex definition.

Answer 1

Convexity is a purely local property. We call a function convex if it is convex at every point in its domain, and $\pm\infty$ is not usually considered part of the domain.

Answer 2

We have $f''(x) = e^{-x} > 0$, so yes, $f(x) = e^{-x}$ is convex. We don't look at $f(\pm \infty)$, because $+\infty$ and $-\infty$ are not real numbers.