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Consider a continuous function $f: X \times Y \rightarrow \mathbb{R}_{} \geq 0$, where $X \subset \mathbb{R}^n$ is compact, and $Y \subseteq \mathbb{R}^m$ is closed.

Define $\hat{f}:X \rightarrow \mathbb{R}_{\geq 0}$ as the parametric integral

$$ F(x) \ := \ \int_Y f(x,y) dy $$

Assume that $X$ is such that $\sup_{x \in X} F(x) < \infty $.

QUESTION: is $F(\cdot)$ continuous? If not (counterexample please), under which additional conditions we can have continuity?

Examples. $f(x,y)=y^{-x}$ with $Y = [1,\infty)$: $F(x)=1/(x-1)$, $X=[1+\epsilon,M]$, so $\lim_{\delta \rightarrow 0} ( F(x)-F(x+\delta))= \lim_{\delta \rightarrow 0} \frac{\delta}{(x-1)(x+\delta-1)} = 0$. With $f(x,y)=e^{-xy}$ and $Y = [0,\infty)$ we have $F(x) = 1/x$, over $X=[\epsilon,M]$, so $\lim_{\delta \rightarrow 0} ( F(x)-F(x+\delta))=0$ as well. Of course all the cases of the kind $f(x,y)=g(y)h(x)$ with $g(\cdot)$ continuous.

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1 Answer 1

up vote 3 down vote accepted

Try $f(x,y) = x^2 y e^{-xy}$ with $X = [0,1]$, $Y = [0,\infty)$, which has $F(x) = 1$ if $x > 0$, $F(0) = 0$.

Conditions that imply continuity will typically rely on the Dominated Convergence Theorem. Thus $F$ is continuous if $\int_Y sup_{x \in X} |f(x,y)|\ dy < \infty$.

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Thanks. Can you please formally prove the claim ($f(\cdot)\geq 0$): "$\int_Y \sup_{x \in X}f(x,y) dy < \infty \ \Rightarrow \ F \text{ continuous}$"? Also, you may be interested in math.stackexchange.com/questions/142824/… –  Adam May 9 '12 at 17:17
Suppose $x_n \to x$ and use the Lebesgue Dominated Convergence Theorem to show $F(x_n) \to F(x)$. –  Robert Israel May 9 '12 at 18:42
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