Can we interchange the order of integration of following double integral ?

$$I = \int_{0}^{1} \int_{0}^{\infty} F(x,y) \overline{R(x,y)} - R(x,y) \overline{F(x,y)} \; dx \; dy$$

Where $F(x,y)= \sum_{n=1}^{\infty} f(nx) e^{2 i \pi n y}$

and $R(x,y)= \sum_{n=1}^{\infty} x^2 n^2 f(nx) e^{2 i \pi n y}$

With $f(x)$ a function from $\mathbb{R^+}$ to $\mathbb{C}$, exponentially decreasing at infinity, such that near zero the asymptotic is $f(x)= a.x +O(x^2)$ ($a$ is a complex constant) and $\int_{0}^{\infty} f(x) dx=\int_{0}^{\infty} x^2 f(x) dx=0$

If we could interchange integration order we would have immediatly $I=0$:

$$ I= \int_{0}^{\infty} \sum_{n=1}^{\infty} f(nx) \overline{x^2 n^2 f(nx)} - \sum_{k=1}^{\infty} x^2 k^2 f(kx) \overline{ f(kx)} \; dx =0 $$

The integral $I$ can also be written (for $n$ and $k$ strictly positive integers) :

$$I= \int_{0}^{1} \int_{0}^{\infty} \sum_{n \ne k} f(nx)\overline{k^2 x^2 f(kx)} e^{2 i \pi (n-k) y} - n^2 x^2 f(nx)\overline{ f(kx)} e^{2 i \pi (n-k) y} dx\; dy$$

$$I= \int_{0}^{1} \int_{0}^{\infty} \sum_{n \ne k} (n^2-k^2) x^2 f(nx)\overline{f(kx)} e^{2 i \pi (n-k) y} dx\; dy$$

So showing we can interchange integration is equivalent to show that following integral is well defined:

$$ \int_{0}^{1} \int_{0}^{\infty} | \sum_{n \ne k} (n^2-k^2) x^2 f(nx)\overline{f(kx)} e^{2 i \pi (n-k) y} | dx\; dy$$

But how can this be proved?

I post a similar question here: Is this double integral of Fourier series always real?

Note: $I$ is well-defined as $\sum_{n=1}^{\infty} f(nx) e^{2 i \pi n y} $ and $\sum_{n=1}^{\infty} n^2 x^2 f(nx) e^{2 i \pi n y}$ do not have any singularity for $x$ near zero (this can be shown using Poisson summation formula, even for $y=0$ there is no singularity as we have $\int_{0}^{\infty} f(x) dx=\int_{0}^{\infty} x^2 f(x) dx=0$).

  • $\begingroup$ What else do we suppose on $f$? Integrals $\int f(x),\int x^2f(x)$ converge absolutely or not necessary? $\endgroup$ – Fedor Petrov Oct 8 '17 at 9:36
  • $\begingroup$ $f(x)$ decreases exponentially at infinity so we have absolute convergence of integrals and no problem to define the sums. (I have added this to my question) $\endgroup$ – Bertrand Oct 8 '17 at 16:33
  • $\begingroup$ What does $f(x) \sim_0 x$ mean? Do you assume that $f(x)$ and $x^2f(x)$ are $L^1$? $\endgroup$ – Pietro Majer Oct 8 '17 at 17:25
  • $\begingroup$ It means near zero $f(x)=a.x+o(x^2)$ ($k$ a constant), so with exponential decrease at infinity $f(x)$ and $x^2 f(x)$ are $L^1$. $\endgroup$ – Bertrand Oct 9 '17 at 7:31
  • $\begingroup$ No sorry! It is a $O(x^2)$... I corrected it... but I think this does not change anything to the problem. $\endgroup$ – Bertrand Oct 9 '17 at 7:51

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