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Advanced Math/differentiation with respect to integral boundaries.

Question
QUESTION: Hi there:

I have the following math question.

Let $f$, $g$  and $h$ be continuous real valued function with domain $X \times Y$ where $X$ and $Y$ are compact sets.
Let me define the set
$S(x) = \{ y : h(y,x) \geq 0 \}$
and
$\bar{S}(x) = \{ y : h(y,x) \leq 0 \}$

Also for any $(x,y) \in X \times Y$ such that $h(y,x)=0$, it holds $f(y,x)=g(y,x)$

Let the function $m$ be defined as:
\begin{equation*}
m(x) = \int_{y \in S(x)} f(y,x) dy + \int_{y \in \bar{S}(x)} g(y,x) dy
\end{equation*}

I can differentiate the term inside the integral if  $f$ and $g$ and their partial derivatives are continuous in $x$ and $y$

What I do not know is the assumptions required on $h$  to have $m$ differentiable.
My intuition is that since at every point where  $h(y,x)=0, f(y,x)=g(y,x)$ differentiation with respect to the limits of the integral is not a problem but I want to make the argument formal.

Thanks,

Nick

ANSWER: Go back to basics. Write down m(x+h)-m(x) . Rearrange the limits/sets of integration and the two integrals so that you are integrating f+g over S(x+h)\S(x) and again over S(x)\S(x+h) (and subtracting the two).

Rewriting the difference quotient this way may help. Then construct a function q[n](x) as the difference quotient when h=1/n. Use one of your standard theorems to prove q[n] converges to some q and then, of course, q = m'. You are on compact sets -- this should allow you to do what you want.

This is the kind of question where I think you want some guidelines, pointers, hints, or at least a place/idea/method to start with. I won't solve out the whole thing or do the work even on my own here -- but if you need more pointers or if there is a more specific issue, please ask a follow up. If this a homework question, though, it will help me to know the context (on the other hand, I will not give much more of an answer/hint for a homework problem).

---------- FOLLOW-UP ----------

QUESTION: Hi:

Thank you again for your suggestion. It works under the assumption that the set
$S(x) = { y : h(y,x) geq 0 }$ is a closed and bounded interval. For example [a1(x),a2(x)]. But what would happen if the set was more complicated.  What assumption do we need on $h$ to be able to write $S(x)$ as the union of closed intervals?

Thanks,

Nick

ANSWER: But your function is h is continuous (for a fixed x, it's a continuous function in only the variable y). I think that should be plenty. You pull back a closed interval, you should get a set of closed intervals. Either I've missed something, or this last detail is easy to tidy up.

---------- FOLLOW-UP ----------

QUESTION: All my apologies. I miswrote what I meant.

I know that the set S(x) is closed and bounded. But it does not necessarily write as a connected interval [a1(x),a2(x)]. It is less clear whether the argument goes through if the set is more complicated (but still bounded and closed!).
For example imagine the set S(x) writes [a1(x),a2(x)] for some value of x but by changing x slightly the set S(x) becomes the union of [a1(x),a2(x)]  and [a3(x),a4(x)].
And maybe there are even stranger examples to think about.

Thanks,

Nick

But h is continuous, and the set S is the pullback of [0,&inf;). It must be closed because h is continuous, and for any set on which h is one-to-one, the pullback is an interval (because there, h is a homeomorphism from that set into [0,&inf;). Thus the set is a union of closed intervals (and you may assume they are disjoint).

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