I wish to prove an identity in control theory which uses the assertion that, integral of a monotonic odd mapping is always positive definite.
For instance consider an example below:
\begin{equation}
C(u)=\int_0^u\phi^{-1}(v)dv
\end{equation}
In expression above, $\phi^{-1}=\tanh^{-1}$ is odd function and the limit $u$ can be both positive or a negative number.
Certain points to be noted are:
(i) $\phi(.)$ is bounded and
(ii) Its first derivative is also bounded.
(iii) $\phi$ belongs to $L_2(\Omega)$ i.e its Lebesgue integral should be finite.
Please help me prove this assertion! Thanks a lot for your time and consideration.
If $\phi^{-1}$ is odd and monotonically increasing, it holds true that $\phi^{-1}(\theta u)\theta u\geq 0$ for all $\theta,u\in\mathbb{R}$ which in turn yields that $\phi^{-1}(\theta u) u\geq 0$ for all $u\in\mathbb{R}$ and $\theta\in[0,1]$.
Thus if we consider a change of variables $v=\theta u$ in the calculation of the integral $C(u)$ then we have that $$C(u)=\int_0^u{\phi^{-1}(v)dv}=\int_0^{1}{\phi^{-1}(\theta u) u\: d\theta}\geq 0$$ Also, if $C(u)=0$ then $u=0$. A contradiction argument can be used to prove this. Assume therefore that $u\neq 0$. From $C(u)=0$ we have that $\phi^{-1}(\theta u) u=0$ for a.e. $\theta\in[0,1]$ or equivalently $\phi^{-1}(\theta u)=0$ for a.e. $\theta\in[0,1]$ since $u\neq 0$. However, this implies $u=0$ due to monotonicity of $\phi^{-1}$ .
Thus $C(u)$ is indeed a positive definite function.