Frechet Derivatives of a nonlinear integral operator The nonlinear integral operator $P:C[0,1]\to C[0,1]$ is defined as follow:
$$P(f)(x)=1+kxf(x)\int_0^1\frac{f(s)}{x+s}ds$$
In order to obtain the Frechet derivative of the operator, I start with:
$$P(f+h)(x)-P(f)(x)=kxf(x)\int_0^1\frac{h(s)}{x+s}ds+kxh(x)\int_0^1\frac{f(s)}{x+s}ds+kxh(x)\int_0^1\frac{h(s)}{s+x}ds$$
It seems the last term is of order $\lVert h(x)\lVert _\infty ^2$ but how to prove it? Note that the function inside the integral is no longer continuous.
 A: First, in order that $P(f)$ be defined for $x\in[0,1]$, I prefer to define $P$ as follows:
$$P(f)(x)=\cases{1&if $\quad x=0$\cr
\displaystyle1+kxf(x)\int_0^1\dfrac{f(s)}{x+s}ds&if $\quad x\in(0,1]$}
$$
In this case $P(f)\in C([0,1])$ for every $f\in  C([0,1])$. Indeed, we only need to prove the continuity of $x\mapsto P(f)(x)$ at $x=0$, and this follows from the inequality
$$\forall x\in(0,1],\quad |P(f)(x)-1|\le |k| g(x)\Vert f\Vert_\infty^2
$$
where $g(x)=x\int_0^1\frac{ds}{x+s}=x\log(\frac{x+1}{x})$. So, 
$\lim\limits_{x\to 0}P(f)(x)=1=P(f)(1)$ because 
$\lim\limits_{x\to 0}g(x)=0$.
Now, if for a given $f\in C([0,1])$ we consider $L_f:C([0,1])\to C([0,1])$, defined by
$$L_f(h)(x)=\cases{0&if $\quad x=0$\cr
\displaystyle kxf(x)\int_0^1\dfrac{h(s)}{x+s}ds+
kx h(x)\int_0^1\dfrac{f(s)}{x+s}ds&if $\quad x\in(0,1]$}$$
Then $L_f$ is a bounded linear operator, because clearly we have
$$\forall x\in[0,1],\qquad |L_f(h)(x)|\le 2g(x)|k|\,\Vert f\Vert_\infty \Vert h\Vert_\infty \le
2|k|(\log2) \Vert f\Vert_\infty \Vert h\Vert_\infty $$
where we used the easy to prove inequality $\Vert g\Vert_\infty=g(1)=\log2$.
Finally
$$\Vert P(f+h)-P(f)-L_f(h)\Vert_\infty\le \vert k\vert \Vert g\Vert_\infty
\Vert h\Vert_\infty^2=\vert k\vert(\log 2)\Vert h\Vert_\infty^2$$
So $L_f$ is the differential of $P$ at $f$.
