Let $f$ be a smooth real function on $R^2$ such that $$\int_{R^2} f dxdy$$ exists. Let $a\in R$. Must $\int_R f(a,y) dy$ exist?
I believe this is true, but I don't know how to prove it.
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JSCB
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Which integral is it? What do you mean by existence? Ever heard of Fubini? – user251257 Nov 25 '16 at 12:54
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By existence, I mean convergence. Does fubini applies here? – JSCB Nov 25 '16 at 12:57
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Lebesgue or Riemann integral? – user251257 Nov 25 '16 at 12:58
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Riemann integral – JSCB Nov 25 '16 at 13:28
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For illustration a simple but non-smooth example first:
Let $h:\mathbb R\to \mathbb R$ be the hat function, that's is $h(x) = \max(0, 1 - |x|)$. Now, let $$ f(x, y) = h(x \cdot \exp(|y|)). $$ Then, we have $$ \int f(x,y) dx dy = \int \exp(-|y|) dy = 2. $$ But, $f(0, y) = 1$ is obviously not integrable on $\mathbb R$.
For a smooth example:
Let $$g(x) = \frac1{\sqrt{2\pi}} \exp(-x^2)$$ be the density of the standard normal distribution. Now, consider $$ f(x,y) = g(x /g(y)). $$
For a symmetric example: $$ \tilde f (x, y) = f(x,y) + f(y,x). $$
user251257
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Thank you for your answer. What if I add the condition that $f$ is symmetric? (i.e. $f(x,y)=f(y,x)$) – JSCB Nov 25 '16 at 14:36
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Suppose $f(x):\mathbb{R}\to \mathbb{R}$ is an odd, nonzero and smooth function such that the integral exists (In a Riemann sense). Then $F(x,y):\mathbb{R}^2\to \mathbb{R}$ given by $F(x,y)=f(x)$ is such a counter example, just take $a$ to be some value such that $f(a)\neq 0$.
Edit: so the above conditions on $f$ aren't exactly what I was picturing in my head. What I mean is a smooth function $f$ such that it is positive for $x>0$ and for $x<0$ it is given by $f(x):= -f(-x)$. This way the function cancels itself out when taking the integral and so it is zero.
Leon Sot
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True, I meant this in a Cauchy principle way with square domain given that the above poster said in a convergent sense. – Leon Sot Nov 25 '16 at 14:24
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the principle value equals the improper Riemann integral if the later exists, not the other way around. – user251257 Nov 25 '16 at 14:27