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.
$$ 0= \int_{-\infty}^{\infty}e^{-x^2-y^2+2xy}f(x)dx \\= e^{-y^2}\int_{-\infty}^{\infty}e^{+2xy}e^{-x^2}f(x)dx,\;\; y\in\mathbb{R}. $$ 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.