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I want to prove that that given $f:R^2 \rightarrow R$ which is continuous with compact support s.t the integral of $f$ for every straight line $l$ is zero ($\int f(l(t))\mathrm{d}t=0$) then $f$ is almost everywhere $0.$

Well I know how to proof it in case that $l = 1_{B(x,r)}$ is measurable and bounded with compact support (from other thread) , but that's not the case here. Any idea? thanks!

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  • $\begingroup$ Did you try a proof by contradiction? Assume $f$ is not almost everywhere $0$ and is continuous ... $\endgroup$ – quarague Jul 12 at 12:07
  • $\begingroup$ @quarague Well I did tried but didnt achived any success doing so... $\endgroup$ – user2323232 Jul 12 at 12:11
  • $\begingroup$ Sounds like Baire theorem might do the trick $\endgroup$ – tommy1996q Jul 12 at 13:55
  • $\begingroup$ @tommy1996q could you show how? $\endgroup$ – user2323232 Jul 13 at 8:55
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    $\begingroup$ @user2323232: What is the "other thread"? And what do you mean by $l = 1_{B(x,r)}$ given that $l$ is a line while $1_{B(x,r)}$ is a function? $\endgroup$ – Alex M. Jul 15 at 15:44
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The problem you pose is the weak version of a classical result in harmonic analysis, which does not require the continuity nor the compactness of the support of the datum $f$: in order to prove this stronger result, first note that $$ \begin{align} \int\limits_{\Bbb R} f\big(l(t)\big)\mathrm{d}t=0\:\text{ for every}&\text{ straight line } l\subsetneq\Bbb R^2\label{1}\tag{1}\\ \Updownarrow&\\ \int\limits_{\sigma(l_0)} f(l)\,\mathrm{d}l=0\:\text{ for every rigid }&\text{motion }\sigma\in\mathsf{Aut}(\Bbb R^2)\label{2}\tag{2} \end{align} $$ where by rigid motion we mean a bijection of $\Bbb R^2$ in itself which is the composition of a rotation and a translation and $l_0\subsetneq\Bbb R^2$ is a given straight line: this is obvious, since substituting a dilatation of a given straight line $l$ in \eqref{1} does not change the value of the integral, thus these linear transformations do not contribute to it and it is not necessary to consider them. Now we can state and prove the following:

Theorem (Cramér & Wald, Newman, Besicovitch). If $f\in L^1(\Bbb R^2)$ is such that, for an arbitrarily given straight line $l_0\subsetneq\Bbb R^2$, equation \eqref{2} holds, then $f\equiv 0$ a.e..
Proof. Let $\xi=(\xi_1,\xi_2)\neq 0$ so that $\Vert \xi\Vert\neq0$ and define $(x,y) \mapsto \sigma_{\Vert \xi\Vert}(x,y)=(u,v)$ as $$ \begin{pmatrix} u\\ v \end{pmatrix}= \begin{pmatrix} \frac{\xi_1}{\Vert \xi\Vert}& \frac{\xi_2}{\Vert \xi\Vert}\\ -\frac{\xi_2}{\Vert \xi\Vert}& \frac{\xi_1}{\Vert \xi\Vert} \end{pmatrix} \begin{pmatrix} x\\ y \end{pmatrix}\quad\forall (x,y)\in\Bbb R^2 $$ The Jacobian of $\sigma_{\Vert \xi\Vert}$ is equal to one therefore, by applying the Fourier transform to $f$, we get $$ \begin{split} \hat{f}(\xi)&=\iint\limits_{\Bbb R^2} e^{-i\langle\xi,(x,y)\rangle}f(x,y)\,\mathrm{d}x\mathrm{d}y\\ &=\int\limits_{\Bbb R} e^{-i\Vert\xi\Vert u} \left[\,\int\limits_{\Bbb R}f\big(\sigma_{\Vert \xi\Vert}^{-1}(u,v)\big)\,\mathrm{d}v\right] \mathrm{d}u=0 \end{split} $$ since the integral inside the square brackets in the right side of the above equality is equal to \eqref{2} for some $\sigma=\sigma_{\Vert \xi\Vert}(u,\cdot)$. By the arbitrariness of $\xi\in\Bbb R^2\setminus\{0\}$, $f\equiv 0$ a.e.. $\blacksquare$

Final notes

  • When I saw the question and the comment by Christian Blatter, I though this was a classical problem for the Radon transform. However, I saw the comment of mathworker21, I remembered a counterexample and the fact that this is in reality a result related to the Pompeiu problem: I found the above simple and beautiful theorem in reference [1] (Theorem 1, p. 26) while looking for that example.
  • Finally, Chakalov's example ([1], p. 27) shows that disks have not the Pompeiu property, i.e. there exists (even very smooth) functions $f$ such that, for many (even infinitely many) $r>0$, $$ \int\limits_{B(x_0,r)} f(y)\,\mathrm{d}y=0\quad \forall x_0\in\Bbb R^2,\; \text{ does not imply }f\equiv 0\, \text{a.e.} $$ where $x_0=(x_{0,1},x_{0,2})$ and $y=(y_1,y_2)$ are points in $\Bbb R^2$.
    Edit. Following the comment of mathworker21, I checked Chakalov's example as reported in [1], and I found that integral formula (2) in the paper is flawed by typos. The correct one is given by Garofalo and Segala in [2], p. 137, and for the completeness I report it here. If we choose $f(x,y)=\sin(ax)$, $a>0$ then we have $$ \int\limits_{B(x_0,r)} f(x,y)\,\mathrm{d}x\mathrm{d}y=\frac{2\pi r}{a}\sin(ax_{0,1})J_1(ar) $$ where $J_1$ is the first kind Bessel function of order $1$. Choosing $a$ such that $ar$ is a zero of $J_1$ makes the above integral zero for any $x_0\in\Bbb R^2$. Note that the example works also in dimension $n>2$, for the higher dimensional analogue of the Pompeiu problem.

References

[1] Nicola Garofalo (1989), "A new result on the Pompeiu problem", Rendiconti del Seminario Matematico, Torino, Fascicolo Speciale "PDE and Geometry", vol. 46, 25-38 (1989), MR1086204, Zbl 0737.35145

[2] Nicola Garofalo and Fausto Segala (1994), "Univalent functions and the Pompeiu problem", Transactions of the American Mathematical Society, 346, 137-146, MR1250819, Zbl 0823.30027.

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    $\begingroup$ 1). $||\xi|| \not = 0$. 2). mathworker21. 3). I'm confused by the Chakalov example. What is wrong with the following argument. For a.e. $x_0 \in \mathbb{R}^2$, $0 = \lim_{r \downarrow 0} \frac{1}{r}\int_{B(x_0,r)} f(y)dy = f(x_0)$. $\endgroup$ – mathworker21 Jul 16 at 21:45
  • $\begingroup$ The keystone is that even if $\int_{B(x_0,r)}f(y)dy$ for infinitely many $r>0$, the limit for $r\to0$ of the quotient of the integral respect to the area of $B(x_0,r)$ is not necessarily zero. As you see from the limiting form of $J_0(x)$ for $x=0$, $$\begin{split}\lim_{r\to0}\frac{1}{\pi r^2}\int_{B(x_0,r)}f(y)dy &= \lim_{r\to0}\frac{1}{\pi r^2}\frac{2\pi r}{a}\sin(ax_{0,1})J_1(ar) \\ &= \lim_{r\to0}\frac{1}{\pi r^2}\frac{2\pi r}{a}\sin(ax_{0,1})\frac{1}{2}ar=\sin(ax_{0,1}).\end{split}$$ Note that the formula in ref. [1] is flawed by typos. $\endgroup$ – Daniele Tampieri Jul 17 at 7:50
  • $\begingroup$ Ohhh, is the "for all" just being applied to $x_0$ and not $r$?? I.e., you have some function $f$ and some $r > 0$ such that for all $x_0$, $\int_{B(x_0,r)} f(y)dy = 0$? $\endgroup$ – mathworker21 Jul 17 at 11:29
  • $\begingroup$ @mathworker21 Exactly: you have infinitely many $r>0$ for which the integral is $0$, corresponding to the zeros of $J_1$. However, $\int_{B(x_0,r)} f(y)dy$ is not zero for all $r\>0$, and this is the essence of Chakalov's example (I apologize for the later in my reply, but I just had lunch :)). $\endgroup$ – Daniele Tampieri Jul 17 at 12:03
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    $\begingroup$ Actually, I would prefer they weren't taken. Feel free to write your own thoughts on the matter and I'd be happy to read them :) $\endgroup$ – mathworker21 Jul 17 at 14:39
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Suppose there exists $p\in\mathbb{R}^2$ with $f(p)\neq 0$, then WLOG (by replacing $f$ by $-f$ if necessary) we might assume $f(p)>0$. But then by continuity of $f$ there exist $\varepsilon>0$ such that $f(x)>f(p)/2$ for all $x$ with $\|x-p\|\leq\varepsilon$, which then gives that the integral of $f$ over every line contained in $B(p,\varepsilon)$ is strictly positive, contradiction.

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    $\begingroup$ I think OP wanted infinite lines. $\endgroup$ – mathworker21 Jul 15 at 11:27

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