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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?

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I see that you know some measure theory, because you talk about "measurable" functions. Assuming that, this is much easier. For example, first argue that $$ g(z)=\int_{0}^{1}\frac{f(x)}{(x-z)^{2}}dx $$ is continuous on $\mathbb{R}\setminus[0,1]$. This is where the dominated convergence theorem comes in handy. Suppose first that $a$ and $z$ are in $(0,\infty)$. Then Fubini's theorem allows you to interchange orders of integration to obtain: $$ \begin{align} \int_{a}^{z}g(z')dz' & =\int_{a}^{z}\left(\int_{0}^{1}\frac{f(x)}{(x-z')^{2}}dx\right)\,dz'\\ & =\int_{0}^{1}f(x)\left(\int_{a}^{z}\frac{1}{(x-z')^{2}}\,dz'\right)\,dx \\ & =\int_{0}^{1}\frac{f(x)}{x-z}dx-\int_{0}^{1}\frac{f(x)}{x-a}\,dx. \end{align} $$ The left side is a continuously differentible function of $z\in(0,\infty)$ because $g$ was argued to be continuous. So the right side must be continuously differentiable in $z$ (and you don't have to check). Furthermore, the derivative of the left side with respect to $z$ is $g$ because $g$ is continuous. Therefore, $$ g(z) = \frac{d}{dz}\int_{0}^{1}\frac{f(x)}{x-z}\,dx, $$ which is what you wanted to show.

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  • $\begingroup$ The point is exactly this: how can you say that you can apply Fubini? The function is not positive, hence she has to be in $L^1$ fi you want to apply Fubini, but I'm not shure if the proof that I made allows me to say that $f \in L^1$. If my proof is wrong then how can I find a dominating function of $f(x,z)$, respectively $\frac {\delta}{\delta z} \frac {f(x)}{(x-z)^2}$? $\endgroup$ – Bman72 Jan 14 '14 at 22:11
  • $\begingroup$ @Ale: Integrable normally means $L^{1}$ (i.e., absolutely integrable.) You can apply Fubini's theorem because you're integrating over a region where the denominator is non-zero. That is, if $f$ is measurable, then $f(x)/(x-z')^{2}$ is jointly measurable and absolutely integrable on $(x,z') \in [0,1]\times [a,x]$ (or $[0,1]\times [x,a]$ depending on $a < x$ or $x \le a$.) $\endgroup$ – DisintegratingByParts Jan 14 '14 at 22:56
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    $\begingroup$ @Ale: Your proof of direct differentiability takes some work because of having to find uniform bounds on the fractions leading to the partial derivatives. You can avoid that work by starting with what you know to be the derivative then integrating instead. You only have to demonstrate continuity for one function, which is an easier problem when it comes to bounding expressions. $\endgroup$ – DisintegratingByParts Jan 14 '14 at 23:02

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