How can I ensure the existence of this right-handed limit of the form $\lim_{\phi\to0^+}\{\frac1{\phi}\frac{\partial{F(\phi,\psi)}}{\partial\phi}\}$?

Question

I'm working on a class of models defined by a certain function $F(\phi,\psi)$ (there is a lot of freedom in this choice). Here, $\phi$ and $\psi$ are in fact real functions, which I know are non-negative, have at least two derivatives, and are bounded. For simplicity, let's say both have a maximum at one, so that $\phi, \psi\in[0,1]$. However, I can't constrain the behavior of $\phi$ and $\psi$, so I'd rather think of F as as function $F:[0,1]\times[0,1]\to\mathbb{R}^{+}$ for the purposes of this question. I have assumed that $F$ is bounded and twice differentiable (although I think I must use semi-derivatives at the boundary to make this rigorous, since the interval is closed). At one point in my calculations, the fraction $\frac{1}{\phi}\frac{\partial{F(\phi,\psi)}}{\partial\phi}$ appears. The problem is I that know $\phi$ and $\psi$ have a zero, so I'd like to ensure, under as few assumptions as possible, that this won't diverge when $\phi$ approaches zero from the right. In other words want to know what assumptions I must make about $F$ in order to ensure that the right-handed limit $$\lim_{\phi\to 0+}\left\{\frac{1}{\phi}\frac{\partial{F(\phi,\psi)}}{\partial\phi}\right\} $$ exists for every $\psi$ (it doesn't have to be the same number for every $\psi$, I just don't want it to blow up). Now, I'm not very familiar with multi-variable analysis, semi-derivatives etc, so I'm not sure on how to proceed. I thought about assuming the second derivatives $\frac{\partial{^2F(\phi,\psi)}}{\partial\phi^2}$ and $\frac{\partial{^2F(\phi,\psi)}}{\partial\phi\partial\chi}$ are bounded (which would make physical sense in this case) to get a Lipschitz condition on the and partial derivatives. Then I could impose $\lim_{\phi\to 0^+}\frac{\partial{F(\phi,\psi)}}{\partial\phi}=0$ or something similar to make the derivative bounded. But I don't know if there is a version of this argument that makes sense when $\phi=0$, since the usual derivative is not defined there. So, could someone help me find some simple (as in not too restrictive) conditions that would make this limit well defined?

Answer 1

Since you are taking the derivative with respect to $\phi$ it seems like the fact that $\phi$ and $\psi$ are actually functions is irrelevant. In general if you have some differentiable function $G(\phi)$ then the only way that $\phi^{-1}G(\phi)$ could blow up would be if $G(0)\neq 0$. One way to see this is that $G'(0)=\lim_{\phi\to 0} \phi^{-1}G(\phi)-\phi^{-1}G(0)<\infty$. So if $\phi^{-1}G(\phi)\to\infty$ then the only way that the difference could be finite is if $\phi^{-1}G(0)\to\infty$, which happens exactly when $G(0)\neq 0$.

So in this case you want ${\frac {\partial F(\phi,\psi)}{\partial\phi}}|_{\phi=0}=0$. But this is not exactly right because as you have mentioned there is the issue of the derivative on the boundary. So a more careful way to say this condition would be that there exists some $\epsilon>0$ and some twice differentiable function $\tilde{F}: [-\epsilon,1]\times [-\epsilon,1]\to\mathbb{R}$ such that ${\frac {\partial \tilde{F}(\phi,\psi)}{\partial\phi}}|_{\phi=0}=0$ and $\tilde{F}(\phi,\psi)=F(\phi,\psi)$ whenever $\phi,\psi\in [0,1]\times [0,1]$.