Fréchet derivative of a function of functions Let $$P:C[0,1]\to C[0,1],P(f)(x)=1+kxf(x)\int_0^1 \frac{f(s)}{x+s}ds$$
I want to find $P'(f)$. Now, I'm somewhat confused about how one uses the definition of the Fréchet derivative in this case to differentiate with respect to $f$. Clearly, we need $$P(f+h)-P(f)-o(h)=Ah,$$
where $h$ is a function. But if this is the case, then the $P(f)$ is Fréchet differentiable means it satisfies
$$\lim\limits_{h(x)\to 0} \frac{\|P(f(x)+h(x))-DP(f(x))h(x)\|}{\|h(x)\|}=0.\mbox{  $\textbf(*)$ }$$
Then this would mean that $h(x)$ is approaching the constant function $h(x)\equiv 0$. Is this correct?
Also, is my understanding correct that $o(h)$ means the error term for the expression in $\mbox{$\textbf{(*)}$}$?
Thanks in advance for your insight.
 A: $$P(f)(x) = 1 + kxf(x)\int_0^1 \frac{f(s)}{x+s}ds$$
Plugging in $f+h$, we have:
$$P(f+h)(x) = 1 + kx(f(x)+h(x)) \int_0^1 \frac{f(s)+h(s)}{x+s}ds$$
$$ = 1 + kxf(x)\int_0^1 \frac{f(s)}{x+s}ds + xh(x) \int_0^1 \frac{f(s)}{x+s} ds + xf(x) \int_0^1 \frac{h(s)}{x+s} ds +  xh(x) \int_0^1 \frac{h(s)}{x+s}ds$$
Now subtract $P(f)(x)$:
$$P(f+h)(x) - P(f)(s) = xh(x) \int_0^1 \frac{f(s)}{x+s} ds + xf(x) \int_0^1 \frac{h(s)}{x+s} ds  + xh(x) \int_0^1 \frac{h(s)}{x+s} ds$$
We are looking for something that is linear in $h(x)$. The last term seems to depend on $h$ quadratically, indeed by the mean value theorem for integrals we have that there is an $\eta \in [0,1]$ such that $$xh(x) \int_0^1 \frac{h(s)}{x+s} ds = x\frac{h(x)h(\eta)}{x+\eta},$$ this is assuming $x > 0$.
With that in mind, let's guess that the linear operator in $h$ is $$Ah(x) = xh(x) \int_0^1 \frac{f(s)}{x+s} ds + xf(x) \int_0^1 \frac{h(s)}{x+s} ds.$$
Indeed, $Ah(x)$ is linear in $h$, so it is a viable candidate. Moreover,
$$P(f+h)(x) - Pf(x) - Ah(x) = xh(x)\int_0^1 \frac{h(s)}{x+s} ds.$$
Hence you are left with demonstrating $$\lim_{h \to 0} \frac{xh(x) \int_0^1 \frac{h(s)}{x+s} ds}{h(x)} = 0.$$
