How to prove that the following function is convex?

Question

I want to prove convexity of the following function:

$$f(x) = log_x \left(1 + \frac{(x^a-1)(x^b - 1)}{x-1}\right)$$

for any fixed $a, b \in (0, 1)$ and:

1. $x\in(0,1)$
2. $x\in(1, \infty)$

I'm trying to solve it for a very long time, tried to investigate the sign of second derivative, but it leads to very ugly equation which I am unable to handle. Any hints are appreciated. Thanks!

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Edit:

$f''(x)$ equals to: $$-\frac{{\left(\frac{x^{x - 1} {\left(p^{y} - 1\right)} x}{p - 1} + \frac{{\left(p^{x} - 1\right)} x^{y - 1} y}{p - 1} - \frac{{\left(p^{x} - 1\right)} {\left(p^{y} - 1\right)}}{{\left(p - 1\right)}^{2}}\right)}^{2}}{{\left(\frac{{\left(x^{x} - 1\right)} {\left(p^{y} - 1\right)}}{p - 1} + 1\right)}^{2} \log\left(x\right)} +\\ + \frac{\frac{x^{x - 2} {\left(p^{y} - 1\right)} {\left(x - 1\right)} x}{p - 1} + \frac{2 \, p^{x - 1} x^{y - 1} x y}{p - 1} + \frac{{\left(p^{x} - 1\right)} p^{y - 2} {\left(y - 1\right)} y}{p - 1} - \frac{2 \, x^{x - 1} {\left(p^{y} - 1\right)} x}{{\left(p - 1\right)}^{2}} - \frac{2 \, {\left(p^{x} - 1\right)} x^{y - 1} y}{{\left(p - 1\right)}^{2}} + \frac{2 \, {\left(p^{x} - 1\right)} {\left(x^{y} - 1\right)}}{{\left(p - 1\right)}^{3}}}{{\left(\frac{{\left(p^{x} - 1\right)} {\left(x^{y} - 1\right)}}{p - 1} + 1\right)} \log\left(p\right)} -\\ - \frac{2 \, {\left(\frac{x^{x - 1} {\left(p^{y} - 1\right)} x}{p - 1} + \frac{{\left(p^{x} - 1\right)} x^{y - 1} y}{p - 1} - \frac{{\left(p^{x} - 1\right)} {\left(p^{y} - 1\right)}}{{\left(p - 1\right)}^{2}}\right)}}{{\left(\frac{{\left(x^{x} - 1\right)} {\left(p^{y} - 1\right)}}{p - 1} + 1\right)} x \log\left(p\right)^{2}} + \frac{\log\left(\frac{{\left(p^{x} - 1\right)} {\left(p^{y} - 1\right)}}{p - 1} + 1\right)}{p^{2} \log\left(x\right)^{2}} + \frac{2 \, \log\left(\frac{{\left(p^{x} - 1\right)} {\left(p^{y} - 1\right)}}{p - 1} + 1\right)}{p^{2} \log\left(p\right)^{3}} $$

Some elements of that equation are positive, some of them are negative - I do not see any way of how to prove that $f''(x) > 0$.

I have also tried to find counter example by testing the Jensen's inequality on randomly generated data, but everything seems OK - the function seems to be convex, but I am unable to prove it.

Answer 1

You can write $$ f(x) = \log_x \left(1 + \frac{(x^a-1)(x^b - 1)}{x-1}\right) = \frac{\ln \left(1 + \frac{(x^a-1)(x^b - 1)}{x-1}\right)}{\ln x}, $$ and prove that $f''(x) > 0$ where needed.

Answer 2

We have $$f(x) = \log_x \left(1 + \frac{(x^a-1)(x^b - 1)}{x-1}\right) = \frac{\ln \left(1 + \frac{(x^a-1)(x^b - 1)}{x-1}\right)}{\ln x} =\ln \left(\left(1 + \frac{(x^a-1)(x^b - 1)}{x-1}\right)^{1/\ln x}\right)$$

Then we may show Log-Convexity:

For $f(x)=\log g(x)$ (on convex domains $x\in(0,\infty)/1$): $$ f(x) \text{ is convex.}\Leftrightarrow g(x)\text{ is log-convex.}$$

1) $g(x)$ log-convex iff: $$g(\theta x + (1 - \theta) y) \leq g(x)^{\theta} g(y)^{1 - \theta}\qquad\theta\in(0,1),x,y\in{(0,\infty)/1}$$

2) $g(x)$ log-convex iff $h(x)=-g(x)$ is log-concave:

$$h(x)h''(x) \leq h'(x)^2$$

Set $$g(x)=\left(1 + \dfrac{(x^a-1)(x^b - 1)}{x-1}\right)^{1/\ln x} \qquad a, b \in (0, 1).$$