I'm trying to solve this exercise
Let $f:[0,1]\to \mathbb{R} \space$ an integrable function, show:
$$g(z):=\int_{0}^1\frac{f(x)}{x-z}dx$$
is a continuous differentiable function on $\mathbb R \backslash[0,1]$.
This is what I've done: we have seen two theorems in class. One allows me to see whether an integral is differentiable and calculate its derivative, while the other one allows me to see whether the integral is continuous (If you need them I can post them). Therefore I've decided to check first whether the integral was differentiable and after that whether the derivative of the integral was continuous. To check whether the integral is differentiable I've to check 1) the function inside the integral is in $L^1$, 2) that she is differentiable with derivative $\frac {\delta f(x,z)}{\delta z}$, where $f(x,z):=\frac{f(x)}{x-z}$. 3) I have to check whether the derivative has a dominating function. Here are my calculation:
1) I was unsure about this point. I've considered the fact that $z \in (-\infty,\epsilon) \cup (1+\epsilon, +\infty)$ where $\epsilon \gt0$ and that $x \in [0,1]$ to point out that the denominator must be bounded by some constant $C\gt0$ (is this true? Can I do it?). If I do like this I get that the denominator is never 0 and this implies that the function $\frac{f(x)}{x-z}$ is integrable.
2)I've calculated the derivative (since the derivative exist by the previous point at every point z)
$$\frac {\delta f(x,z)}{\delta z}= \frac {\delta}{\delta z} \frac{f(x)}{x-z}=\frac {f(x)}{(x-z)^2}$$
3) by doing the same reasoning as in 1) I've thought that the denominator can't vanish and hence there exist a dominating function which is in $L^1$ The derivative of g(z) will be
$$g(z)'= \int_{0}^1 \frac {f(x)}{(x-z)^2}dx$$
Now what I've done is to check that this function is continuous on $\mathbb {R}\backslash [0,1]$ (I have to check 1') the function inside the integrale is measurable, 2') she is continuous for x fixed and 3') there exist a dominating function of the function inside the integrale)
1) the function $\frac {f(x)}{(x-z)^2}$is measurable since $f(x)$ measurable and $(x-z)^2$ measurable (since is never 0 and never $\infty$, hence is a number, hence is continuous)
2) the function is continuous since composition of continuous functions (same argument as above)
3) same argument as above with dominating function h defined as:
$$h(x) =\begin{cases} f(x) & \text{if } (x-z)\le 1 \\ (\frac {1}{C}+\epsilon) & \text{if } (x-z)\lt 1 \end{cases}$$
where $C = (x-z)$ and $\epsilon \gt 0$
Did I do everything wrong? Is there a simple way to see if an integral depending on a parameter is continuous / differentiable / continuously differentiable?