Continuity of improper integral with a continuous integrand.

Question

I am a newbie in analysis and am trying to wrap my head around some continuity/compactness/finiteness concepts.

Let $f(x,y):\mathbb{R}^2\mapsto\mathbb{R}$ be a continuous function in both $x$ and $y$ and define $$ F(x) = \int_0^{\infty}f(x,y)dy. $$ Assume that $F(x)<\infty$ for all $x>0$. Can I show that $F(x)$ is continuous for $x>0$?

(The original question is different (see details below), but this is the most important part of the proof (thanks to the hint given by Brian))

Given a compact set $\mathcal{X}$ and suppose $0$ is not a member of $\mathcal{X}$. Can I show $\sup_{x\in\mathcal{X}} F(x) <\infty $?

Here are what I tried so far:

1. Since $\mathcal{X}$ is compact so $\sup_{x\in\mathcal{X}} F(x) = \max_{x\in\mathcal{X}} F(x)$. Moreover, since $F(x)<\infty$ for any $x>0$ and $0\notin \mathcal{X}$, then $\max_{x\in\mathcal{X}} F(x)<\infty$. (This proof seems okay to me. But the continuity of $f(x,y)$ is not used anywhere, which makes me a little puzzled).

2. Since $f(x,y)$ is continuous function and $F(x)<\infty$ for all $x>0$, then $F(x)$ is a continuous function for any $x>0$ (is this true?). Since $\mathcal{X}$ is compact so $\sup_{x\in\mathcal{X}} F(x) = \max_{x\in\mathcal{X}} F(x)$. The maximization is finite because it is the maximum of a continuous function in a compact set. (This proof does use all given conditions, but the first assessment is not clear to me. Any reference to existing theorem is appreciated.)

Any help is highly appreciated!

Answer 1

For part 1, only continuous functions are guaranteed to attain their maximum on compact sets.

Answer 2

I think you need the extra assumption that the integral is uniformly convergent. Altough I do not have a counterexample, the following argument seems to show where this is needed:

For if we write $\vert F(x)-F(z)\vert \leq \vert \int_0^{M}(f(x,y)-f(z,y))dy\vert +\vert \int_M^{\infty }(f(x,y)-f(z,y))dy\vert$ and let $\epsilon >0$ and then choose $\delta_1 >0$ so that $\vert f(x,y)-f(z,y)\vert <\epsilon $ whenever $d((x,y),(z,y))<\delta_1 .\ $

Then in this case,

$\tag 1\vert \int_0^{M}(f(x,y)-f(z,y))dy\vert\leq M\epsilon$.

Now, $\vert \int_M^{\infty }(f(x,y)-f(z,y))dy\vert \leq \vert \int_M^{\infty }f(x,y)\vert +\vert \int _M^{\infty }f(z,y)dy\vert$.

If the convergence of the integral is uniform we may choose

choose $M$ such that $\tag2\vert \int_M^{\infty }f(x,y)dy\vert<\epsilon $ and $\tag3\vert \int _M^{\infty }f(z,y)dy\vert<\epsilon$.

Take $\delta =\delta_1/2$ and combine $1)$, $2)$, and $3)$ to finish the proof.

If the convergence is not uniform then I do not see how to get $\vert \int_M^{\infty }(f(x,y)-f(z,y))dy\vert< \epsilon $

Answer 3

This fails in general. For example, define

$$f(x,y) =\begin{cases} 0, & x\le 1\\ = \frac{1}{1+(y-1/(x-1))^2}, & x>1 \end{cases}$$

Then $f$ is continuous on $\mathbb R^2,$ but $\lim_{x\to 1^+}F(x) = \pi,$ while $F(1) = 0.$