$e^{-1/f(x)}$ convexity

Question

Given:

$$g(x)=e^{-1/f(x)}, \textbf{dom} g=\{x|f(x)<0\}$$

where $f$ is convex.

How can I prove that $g(x)$ is convex, concave or neither?

I have tried to find a counter example and failed. So based on those examples I've tried, I think $g$ is convex but not sure how to prove it.

Tried using the definition of convexity where $ 0 \le \theta \le 1$ and I know that: $$f(x_0) < 0, f(x_1) < 0$$ $$f(\theta x_0 + (1-\theta) x_1) \le \theta f(x_0) + (1-\theta) f(x_1)$$

and need to show that: $$g(\theta x_0 + (1-\theta) x_1) \le \theta g(x_0) + (1-\theta) g(x_1)$$ $$e^{-1/f(\theta x_0 + (1-\theta) x_1)} \le \theta e^{-1/f(x_0)} + (1-\theta) e^{-1/f(x_1)}$$

But I am stuck here. Taking the log of both sides didn't get me very far.

Answer 1

Let $\phi(t) = e^{-{1 \over t}}$ for $x <0$ and $+\infty$ otherwise, it is straightforward to check that $\phi$ is convex and non decreasing.

The composition of a non decreasing convex function with a convex function is convex.

Here $g= \phi \circ f$.

Answer 2

$$g(x)=e^{-\frac{1}{f(x)}}\implies g''(x)=\frac {e^{-\frac{1}{f(x)}} } {[f(x)]^4}\Big[f(x)^2 f''(x)+(1-2 f(x)) f'(x)^2\Big]$$