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Prove that exist a function $f:\mathbb{R}\rightarrow \mathbb{R}$ infinitely differentiable, such that

$$\int_{\mathbb{R}}f(t)\,dt=1\;\mbox{ and }\; \int_{\mathbb{R}}t^nf(t)dt=0,\, \forall\, n\geq 1\; \mbox{integer}.$$

Some ideas? Thank you.

ElliptCg
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    $f$ cannot be compactly supported, otherwise using Stone-Weirstrass' theorem along with $\int_{\mathbb{R}}t^nf(t)\mathrm{d}t=0,n\geqslant 1$, it would follow that $f$ is identically null which is excluded by $\int_{\mathbb{R}}f(t)\mathrm{d}t=1$. – C. Falcon Dec 13 '15 at 23:12
  • The first condition I can use mollifier functions, but I not find the equilibrio with the second condition. – ElliptCg Dec 13 '15 at 23:13
  • Yes C. Falcon is correct. – ElliptCg Dec 13 '15 at 23:15

1 Answers1

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Choose $\phi\in C_c^\infty (\Bbb{R})$ with $\phi \equiv 1$ in a neighborhood of $0$.

Let $f=\mathcal{F}^{-1}(\phi)$ be the inverse Fourier transform of $\phi$.

Then $\int f \,dx =\mathcal{F}(0)=\phi(0)=1$.

Furthermore, $$ \int t^n f(t)\,dt = c_n \frac{\partial^n}{\partial x^n}\mathcal{F}(f)(0)= c_n \frac{\partial^n}{\partial x^n}\phi(0) =0 $$ for some constant $c_n \in \Bbb{C}$.

EDIT: Since you want $f$ to be real-valued, you have to do one of two things:

1) Take $\phi$ to be real valued and symmetric (i.e. $\phi(-x) = \phi(x)$ for all $x$), or

2) Just consider the real part ${\rm Re}(f)$ instead of $f$ itself.

PhoemueX
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