Weaker conditions for differentiating under the integral sign Standard theorems of real analysis give conditions under which it holds
$$\int_0^1 \partial_x f(x,y)dy = \frac{d}{dx}\int_0^1 f(x,y)\,.$$
In most of the formulations that I have found, it is required that, for almost every $y$, $f$ is everywhere differentiable. I'm wondering if this condition can be weakened, at least in some particular setting.
Consider an integral operator $F$ on $L^2(0,1)$ which maps an element $\phi$ to
$$ F\phi(x) = \int_0^1 k(x,y)\phi(y)dy\,.$$
$k(x,y)$ is supposed to be some bounded continuous function on $(0,1)^2$.
If $k$ is of class $C^1$, then all functions in the image of $F$ are of class $C^1$. But can we give some weaker condition in order to have an image at least differentiable?
For example if $k(x,y)=|x-y|$, then it can be proven explicitly (just by writing down the definition of derivative and by bounding the remainder) that it holds
$$\frac{d}{dx}F\phi(x) = \int_0^1sign(x-y)\phi(y)dy\,.$$
which is $C^0$ and so $F\phi(x)$ is even $C^1$. Is this a particular case of some general and well known result?
 A: I am not an expert of this topic, and would love to see some nice references to well-known results of this kind. Meanwhile, let me try to first perform a heuristic computation and see what kind of conditions we can pick up to justify each step.
Let $k $ be measurable. Then
\begin{align*}
F\phi(x_1) - F\phi(x_0)
&= \int_{0}^{1} (k(x_1,y) - k(x_0,y))\phi(y) \, \mathrm{d}y \\
&= \int_{0}^{1} \left( \int_{x_0}^{x_1} \partial_x k(x, y) \, \mathrm{d}x \right) \phi(y) \, \mathrm{d}y \tag{1} \\
&= \int_{x_0}^{x_1} \int_{0}^{1} \partial_x k(x, y) \phi(y) \, \mathrm{d}y\mathrm{d}x. \tag{2}
\end{align*}

*

*$\text{(1)}$ is justified if $x \mapsto k(x, y)$ is absolutely continuous on any compact intervals for any $y$.


*$\text{(2)}$ is justified by the Fubini-Tonelli's Theorem if $\int_{x_0}^{x_1} \int_{0}^{1} \left| \partial_x k(x, y) \phi(y) \right| \, \mathrm{d}y\mathrm{d}x < \infty$ for any interval $[x_0, x_1]$. In particular, this occurs if $y \mapsto \partial_x k(x, y)$, regarded as a family of maps indexed by $x$, is dominated by an $L^2$ function.
Under the above conditions, it follows that $F\phi$ is absolutely continuous and
$$ \frac{\mathrm{d}}{\mathrm{d}x}F\phi(x) = \int_{0}^{1} \partial_x k(x, y) \phi(y) \, \mathrm{d}y \tag{*} $$
almost everywhere.

Example 1. Suppose that $k(x, y)$ is uniformly Lipschitz in the variable $x$, i.e., there exists $L \geq 0$ such that $\left| k(x_1,y) - k(x_0,y) \right| \leq L\left|x_1 - x_0\right|$ for any $x_0$, $x_1$, and $y$. Then $x \mapsto k(x, y)$ is absolutely continuous and $\left| \partial_x k(x, y) \right| \leq L$, and so, both conditions are satisfied and the above conclusion holds.


Example 2. Suppose that the kernel is of the form $k(x-y)$. If $k$ is locally absolutely continuous and its derivative is locally $L^2$, then the conditions are satisfied and we have
$$ \frac{\mathrm{d}}{\mathrm{d}x}F\phi(x) = \int_{0}^{1} k'(x - y) \phi(y) \, \mathrm{d}y. $$
Moreover, by the $L^p$-continuity of translation operator, it follows that $\frac{\mathrm{d}}{\mathrm{d}x}F\phi$ is continuous.

