Dear experts I have a fixed point problem of the type:
$ \phi(x) = F(\phi(x)) $, where $x \in \mathbb R^3 $.
$\phi(x)$ is differentiable non-negative function on a given domain. I am trying to find when $\frac{\delta F(\phi(x))}{\delta\phi(y)}$ evaluated at the fixed point $\phi^*$ is greater than zero.
I have arrived at the following result. If the operator is defined as $F(\phi) = \int K(x',x)\phi(x')dx'$, using the definition of a functional derivative
$\frac{\delta F(\phi(x))}{\delta\phi(y)} = \lim_{\epsilon\to0}\frac {1}{\epsilon} F(\phi(x)+\epsilon\delta(x-y))-F(\phi(x)) = $
$=\lim_{\epsilon\to0}\frac {1}{\epsilon} \int K(x',x)\delta(x'-y)dx'= K(y,x) > 0 $. If $K(x,x') > 0$ , $ \forall x,x'$.
It seems that the properties of the functional derivative depend on the properties of the Kernel.Is this type of argumentation correct? However, for the general case I can't apply the argument with the functional derivative.
$ \phi^*(x) = F(\phi^*(x)) $, where $x \in \mathbb R^3 $.
Here $F(\phi^*(x))$ is no longer linear integral operator. $\phi(x)$ is again defined to be positive and differentiable. Is it possible in this general case to determine when $\frac{\delta F(\phi(x))}{\delta\phi(y)}$ evaluated at the fixed point $\phi^*$ is greater than zero.
Thanks,