Inspired by the recent post "Limit of integral", I propose the following problem (hoping it will not turn out to be too easy). Suppose that $g:[0,1] \times [0,1] \to {\bf R}$ is continuous in both variables separately. Is it true that, for all $x_0 \in [0,1]$, $$ \lim \limits_{x \to x_0 } \int_0^1 {g(x,y)\,{\rm d}y} = \int_0^1 {g(x_0 ,y)\,{\rm d}y} . $$
Limit of integral - part 2
3 Answers
The answer is no. In fact, it is related to Leibniz's rule. The counterexample below is from 'Counterexamples in Analysis' and was recently posted here by Jonas Meyer. Consider the function
$$ g(x,y) = \left\\{ \begin{array}{lr} \left (\frac{3x^2}{y^2} - \frac{2x^4}{y^3} \right)e^{-x^2/y} & y>0, \\\ 0 & y=0. \end{array} \right.$$
It's easy to see that $g$ is continuous in both variables separately. We have that $\int_{0}^1 g(0,y) dy = 0 $ and $$ \int_{0}^1 g(x,y) dy = e^{-x^2}(1-2x^2)$$ if $ x \ne 0$. So $$\lim_{x \rightarrow 0} \int_{0}^1 g(x,y) dy = 1 \ne \int_{0}^1 g(0,y) dy.$$
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0Yes, $g(x,y)$ is the partial derivative of the function $f(x,y)$ given there, and this inequality is precisely why that other example works. – 2010-11-08
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0@Shai Covo: I agree! Joint continuity does not imply continuity, I removed my erroneous answer. – 2010-11-08
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0@Jonas Meyer: I ended up copying some of your LaTeX. Hope you don't mind. – 2010-11-08
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0@Nuno: Of course I don't. I do that all the time. – 2010-11-08
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0@AD.: I think you meant, "Separate continuity does not imply continuity." "Joint continuity" is synonymous with continuity. – 2010-11-08
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0@Jonas Meyer: Yes, that was what I meant. Thanks. – 2010-11-09
I think I have another counterexample. Define $f(x)=\int_{0}^x e^{-1/t}dt$ for $x\gt0$. This is chosen because it goes to zero as $x$ goes to zero from the right and because experimentation led me to the differential equation $\frac{f''(x)}{f'(x)}=\frac{1}{x^2}$ as a sufficient condition for the following to work. Define
$$g(x,y) = \left\{ \begin{array}{lr} \frac{xy}{f(x)+y^2} & \text{if } x\gt0, \\ 0 & \text{if } x=0, \end{array} \right.$$
for $(x,y)$ in $[0,1]\times[0,1]$, and let $x_0=0$. The right hand side of your tentative equation is obviously $0$. The left hand side is $$\lim_{x\to0+}\frac{x}{2}\log\left(1+\frac{1}{f(x)}\right),$$ which comes out to $\frac{1}{2}$ after $2$ applications of l'Hôpital's rule, if I did it correctly.
(I started by playing with the standard example of a discontinuous but separately continuous function on $\mathbb{R}^2$, $f(x,y)=\frac{xy}{x^2+y^2}$ when $x$ or $y$ is nonzero, $f(x,y)=0$ when $x=y=0$. Then I tried to see how the $x^2$ in the denominator could be modified to give a counterexample here, by replacing it with an unknown $f(x)$ that goes to $0$ at $0$ and seeing what further properties of $f(x)$ would make it work. As I mentioned, this led in particular to the sufficient condition $\frac{f''(x)}{f'(x)}=\frac{1}{x^2}$, and hence to this example. Unfortunately, I can't offer any real intuition.)
Added
I decided to look at this a little more, and came up with an example simpler than the other one I gave. With $$g(x,y)=\frac{-y}{\log(x/2)(x+y^2)}$$ for $x\gt0$ and $0$ otherwise, the same result as above holds.
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0Very nice example! – 2010-11-09
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0Nice one! The limit is indeed 1/2. – 2010-11-09
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0@Jonas Meyer: Your last example is very elegant! Apparently, you meant $\log(x)$ in the denominator. – 2010-11-12
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0@Shai Covo: Thank you. I did mean $\log(x/2)$, but it makes no difference for getting the left hand side to come out to $1/2$. I just changed it to $\log(x/2)$ so that we don't have to restrict to $[0,1-\epsilon]\times[0,1]$, a minor technically to accommodate the problem as stated. – 2010-11-12
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0@Shai Covo: But if you prefer, I could change it back to $\log(x)$. – 2010-11-14
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0@Jonas Meyer: As you remarked, the $\log(x/2)$ is indeed better. – 2010-11-14
This is true if there is a (measurable) function $f(y)$ such that $$|g(x,y)|\le f(y), \mbox{ for }0\le x\le 1 $$ and $$\int_0^1 f(y)\,dy<\infty.$$ This follows from the Lebesgue dominated convergence theorem.
Note: Every continuous function is measurable.
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0That is true. Can you find such $f$? :) – 2010-11-08
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0Unless $g(x,y)$ is given, not one can find such $f$. – 2010-11-08
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0In Nuno's example, we have $g(\sqrt{y},y)=e^{-1}/y$, which is not integrable over $(0,1)$. – 2010-11-09
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0@AD. If $g(x,y)$ is continuous, one can take $f$ to be the upper bound of $|g(x,y)|$, so the question is true. In general, one can always use $f(y)=\max_{0\le x\le 1}|g(x,y)|$, but that may be hard to calculate. – 2010-11-09