How to prove the following function as convex?

Question

Let $k > 0$ be given. Let $$f(x)=\left(2^\frac{k}{1-x}-1\right)\left(\frac{1-x}{x}\right)$$ where $0 < x < 1$. Simulation results shows that the function is having only one minima. I want to prove $f(x)$ is a convex function mathematically. Here $h(x)=\left(2^\frac{k}{1-x}-1\right)$ is a increasing function and $g(x)=\left(\frac{1-x}{x}\right)$ is a monotonically decreasing function. As a result $g'(x)<0$. These are observed by matlab simulations. But when I am considering these functions together I am failing to prove that as a convex function.

Answer 1

Note that $$f(x) = \sum_{j=0}^{\infty}\frac{a^j}{j!}\frac{1}{x(1-x)^{j-1}},$$ where $a=k\ln2$. Note that for all $j\ge 0$, the function $h(x)=x(1-x)^{j}$ satisfies the following:$$h'(x)=(1-x)^{j-1}(1-(j+1)x),\\ h''(x)=-(1-x)^{j-2}(2j-j(j+1)x),\\ h''(x)h(x)-2(h'(x))^2=-(1-x)^{2j-2}((1-x)^2+jx^2).$$ Assuming that my calculations above are correct, we find that the function $1/h(x)$ is convex in $[0,1]$ since $$\frac{d^2(1/h(x))}{dx^2}=-\frac{h''(x)h(x)-(h'(x))^2}{h^3(x)}\ge 0.$$ Therefore, when $k\ge 0$, $f$ is a sum of convex functions and is itself a convex function.

Answer 2

Problem: Let $k > 0$. Let $$f(x) = \Big(2^{k/(1-x)} - 1\Big)\frac{1-x}{x}.$$ Prove that $f(x)$ is convex on $(0, 1)$.

Proof: Let $h(x) = \ln (\mathrm{e}^{x} - 1) - \ln x$ and $g(x) = \frac{k\ln 2}{1-x}$. We have $$f(x) = \mathrm{e}^{h(g(x)) - \ln x + \ln (k\ln 2)}.$$

We have the following results (their proof is given later).

Fact 1: $h(x)$ is convex and nondecreasing on $(0, +\infty)$.

Fact 2: $g(x)$ is convex on $(0, 1)$.

From Facts 1 and 2, we know that $h(g(x))$ is convex on $(0, 1)$. See page 85 in [1]. Indeed, let $F(x) = h(g(x))$ and we have $F''(x) = h''(g(x))g'(x)^2 + h'(g(x)) g''(x)\ge 0$.

Thus, $h(g(x)) - \ln x + \ln (k\ln 2)$ is convex on $(0, 1)$.

Thus, $f(x)$ is convex on $(0, 1)$. Here we have used the fact that if $p(x)$ is convex, then $\mathrm{e}^{p(x)}$ is also convex. Indeed, $(\mathrm{e}^p)'' = \mathrm{e}^p (p')^2 + \mathrm{e}^p p'' \ge 0$.

We are done.

[1] Boyd and Vandenberghe, "Convex optimization". http://web.stanford.edu/~boyd/cvxbook/bv_cvxbook.pdf

Proof of Fact 1: We have $$h'(x) = \frac{\mathrm{e}^x x - \mathrm{e}^x + 1}{(\mathrm{e}^x-1)x}$$ and $$h''(x) = \frac{\mathrm{e}^{2x} - \mathrm{e}^x x^2 - 2\mathrm{e}^x + 1}{(\mathrm{e}^x - 1)^2x^2}.$$

Let $h_1(x) = \mathrm{e}^x x - \mathrm{e}^x + 1 $. We have $h_1'(x) = \mathrm{e}^x x \ge 0$ on $[0, +\infty)$ and $h_1(0) = 0$. Thus, we have $h_1(x) \ge 0$ on $[0, +\infty)$. Thus, $h'(x) \ge 0$ on $(0, +\infty)$.

Let $h_2(x) = \mathrm{e}^{2x} - \mathrm{e}^x x^2 - 2\mathrm{e}^x + 1$. We have $h_2'(x) = \mathrm{e}^x (2\mathrm{e}^x - x^2 - 2x - 2)$. Let $h_3(x) = 2\mathrm{e}^x - x^2 - 2x - 2$. We have $h_3'(x) = 2(\mathrm{e}^x - 1 - x) \ge 0$ on $[0, +\infty)$ and $h_3(0) = 0$. Thus, $h_3(x) \ge 0$ on $[0, +\infty)$. Thus, $h_2'(x) \ge 0$ on $[0, +\infty)$. Note also that $h_2(0) = 0$. Thus, $h_2(x) \ge 0$ on $[0, +\infty)$. Thus, $h''(x) \ge 0$ on $(0, +\infty)$.

Thus, $h(x)$ is convex and nondecreasing. We are done.

Proof of Fact 2: Since $g''(x) = \frac{2k\ln 2}{(1-x)^3} > 0$ on $(0, 1)$, $g(x)$ is convex on $(0, 1)$. We are done.