I have a function $\;g(x)-x\;$ on the domain $\;0<x<\frac{1}{3}\;$, where
$g(x)=\frac{\left(9 x^2+1\right) \cosh ^{-1}\left(\frac{36 x^2+\left(1-9 x^2\right)^2 \cosh (2 \pi x)}{\left(9 x^2+1\right)^2}\right) \sqrt{81 x^4+54 x^2+\left(1-9 x^2\right)^2 \cosh (2 \pi x)+1}}{2 \sqrt{2} \pi \cosh (\pi x)}+\frac{18 \sqrt{2} x^2 \left(9 \pi x^3+4 \tanh (\pi x)\right)}{2 \sqrt{2} \pi }.$
I want to prove that $\;g(x)-x\neq 0\;$, or if possible, prove that it is always greater than zero $\;g(x)-x>0\;$ (I know this by numerical data).
Proof:
Assuming the proposition to be false, I have
$g(x)\leq x \quad Eq. (1)$.
On the other hand, values of $\;g(x)\;$ at the endpoints of the domain are as follows:
$\lim_{x\to 0} \, g(x)=0 \quad and \quad \lim_{x\to \frac{1}{3}} \, g(x)=1.33$
Now, considering the equal sign in Eq. (1) for $\;x=\frac{1}{3}\;$, we get $\;1.33=\frac{1}{3}\;$ which is a contradiction. If the inequality sign holds in (1), it gives the range of $\;g(x)\;$ as $\;(0,\frac{1}{3})\;$, while from the limit above, we see that $\;g(\frac{1}{3})=1.33\;$ which is greater than $\;\frac{1}{3}\;$.
Question: Are these arguments sufficient to conclude that the proposition $\;g(x)\leq x\;$ is false and $\;g(x)-x\;$ is always greater than zero? Or, the argument only brings a counterexample and it is not sufficient as proof.
Any comment is appreciated.
Your argument does not work.
Your proposition is that for all $x$ with $0<x<1/3$ we have $g(x)-x>0$. Assuming it to be false would be to say there exists a $x$ with $0<x<1/3$ and $g(x)-x\leq 0$.
I.e under the assumption that the proposition is false we can have $g(x)-x>0$ for all $x$ with $0<x<1/3$ except one