# Convexity of an exponential-based function

Given $$a, b, c, t > 0$$, let function $$f : \Bbb R_{>0} \times \Bbb R_{>0} \to \Bbb R_{>0}$$ be defined by

$$f(x,y) := b \exp{\left(\frac{a x} {t x- c }\right)} \left[\exp{\left( \frac{ay}{ty - c} \right)} - 1 \right]$$

where $$t x- c > 0$$ and $$ty - c >0$$. I want to determine whether function $$f$$ is convex.

I tried the get the second derivative but the derivatives are too complicated to show the convexity. The exponential function is convex and if the function inside the $$\exp$$ if convex then $$\exp$$ (the function) is convex. In my problem, the function inside the exp is a linear fractional which is quasiconvex. So, I am not sure which operation preserves convexity I can use. Suggestions would be welcome.

• This function is convex if and only if $g:x\to \exp\left( \frac{ax}{tx-c} \right)$ is convex. Commented Aug 28, 2021 at 6:44

If $$g:x\to \exp(\frac{ax}{tx-c})$$ is convex then $$f$$ is convex.
From wolfram alpha we get that the second derivative of $$g$$ is $$g''(x)=\frac{ac g(x) (ac+2t(tx-c))}{(c-tx)^4}$$ which is positive. Therefore $$f$$ is convex.