f(x) is a non increasing function with range (1,0]. Now can I prove analytical that \log(f(x)) is a convex function of x?

Question

I have a function f(x). And it is non-increasing with respect to x, for $x>0$. And also I know that f(x) is convex. Now the range of f(x) is bounded by $(1,0]$. Now, I want to prove analytically $log(f(x))$ is convex. I am not finding out any theorem regarding that type of problem. I tried to solve by definition that $log(f(\lambda*x1+(1-\lambda)*x2))\le \lambda*log(f(x1))+(1-\lambda)*log(f(x2))$. But it was too complex because of the nature of my $f(x)$. I also tried by taking second derivative of $log(f(x))$ but no luck because of the nature of $f(x)$. So, I am looking for any theorem or special case that I can apply on that. Thanks.

Answer 1

Take $$f(x)=x^2$$ with $f(1)=\frac 12$.

$f $ is not increasing at $(0,1]$ and

$$f((0,1])=(0,1]$$

but

$$x\mapsto \ln(f(x))=2\ln(x)$$ is not convex at $(0,1)$.