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Given that the function $f:\mathbb{R}\rightarrow\mathbb{R}$ is continuous and bounded.
$\int_{-\infty}^{+\infty}e^{-(x-y)^2}f(y)dy=0, \forall x \in\mathbb{R} $
Prove that $f(x)=0, \forall x \in\mathbb{R}$

P/s: I tried to use complex analysis such as Laurent series, Fourier series but I could not prove it.
Please do not choose functions such as $\sin(x-y)$ as an counter-example, because you misunderstood the problem.

WgTTHT
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$$ 0= \int_{-\infty}^{\infty}e^{-x^2-y^2+2xy}f(x)dx \\= e^{-y^2}\int_{-\infty}^{\infty}e^{+2xy}e^{-x^2}f(x)dx,\;\; y\in\mathbb{R}. $$ Now you can use Complex Analysis on the following entire function $F$ that vanishes on the real line: $$ F(z)=\int_{-\infty}^{\infty}e^{2xz}e^{-x^2}f(x)dx = 0,\;\;\; z\in\mathbb{R}. $$ The conclusion is that $F \equiv 0$ everywhere in $\mathbb{C}$, and in particular on the imaginary line, which is a Fourier transform of $e^{-x^2}f(x)$. The conclusion is that $f$ must vanish almost everywhere.

Disintegrating By Parts
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  • "F(z)=0 for all z belongs to R, implies that F(z)=0 for all z belongs to C". That is not a good point, sir. Can you explain more – WgTTHT Jun 11 '21 at 01:39
  • @WgTTHT A holomorphic function on an open connected region of $\mathbb{C}$ which has a zero set that is infinite and has a cluster point in the region must be identically $0$. – Disintegrating By Parts Jun 11 '21 at 02:40
  • I wonder if the condition "f is bounded" and "f is continuous" are necessary conditions? Can you make the problem stronger with weaker condition ? – WgTTHT Jun 11 '21 at 03:37
  • @WgTTHT : The argument should work if you assume that $e^{x^2}f(x)$ is absolutely integrable, or something just a little stronger than that. – Disintegrating By Parts Jun 11 '21 at 23:11
  • How can you know F is holomorphic? – WgTTHT Jun 12 '21 at 10:57
  • @WgTTHT : $F$ is continuous, and if you integrate $F$ around any finite closed path in the plane, you can interchange orders of integration and get $0$. So it is holomorphic by Morera's Theorem. And you can know what the derivative is by a similar procedure using Cauchy's integral. Complex Analysis allows you to turn the more difficult problems of differentiation into integration. – Disintegrating By Parts Jun 12 '21 at 23:51
  • please guide me to clear your point apply Morera's Theorem and interchange the orders of integration. Please noted the condition of interchange the orders of integration – WgTTHT Jun 13 '21 at 04:06
  • please noted that condition of Cauchy's Integral is the function must be holomorphic. We cannot use Cauchy's Integral while we do not know the function if holomorphic or not? – WgTTHT Jun 13 '21 at 04:16
  • if using your explanation of Morera's theorem, can I prove "every Fourier Transform can be differentiable?" – WgTTHT Jun 13 '21 at 04:20
  • https://en.wikipedia.org/wiki/Morera%27s_theorem – Disintegrating By Parts Jun 13 '21 at 04:23
  • https://math.stackexchange.com/questions/1025789/show-that-the-fourier-transform-is-differentiable – WgTTHT Jun 13 '21 at 04:27
  • I have read Morera's Theorem, but I do not how to ensure that we can use Fubini's Theorem to change the order of integration in this problem. Please help me to prove it – WgTTHT Jun 14 '21 at 03:43
  • How can you know $F(z)$ is continuous? @disintegrating by parts – WgTTHT Jun 15 '21 at 10:01
  • @WgTTHT : Dominated Convergence Theorem is one way. – Disintegrating By Parts Jun 15 '21 at 18:29