For a project at my University, I have to prove that in the space of real numbers, if the Fourier transform of a measure is zero, then the measure is zero. I found a post with a solution to this problem
If the Fourier transform of a signed measure is identically zero, is the same true of the measure?
but I don't understand how to do step two, if anyone could help me with the demonstration I would be very grateful, I have the following sketch of proof
Let $\mu$ be a signed measure. Assuming $\hat \mu (u) = \int exp(iux) d \mu (x) = 0$
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If $f$ is continuous and periodic with some period $M$, then $\int f d\mu = 0$. (Stone-Weierstrass theorem.)
If $f$ is continuous and periodic, we have that by the Stone-Weierstrass theorem, there is a sequence of trigonometric polynomials $P_n$ that approximate $f$ for $n$ large enough, then by the dominated convergence theorem $$\int P_n d \mu \to \int f d \mu$$ and as $P_n$ can be expressed as the integral of exponential functions, the integral $\int P_n d \mu$ becomes zero, as it converges to $\int f d \mu$, the last integral is also zero
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If $f$ is continuous and compactly supported, then $\int f d\mu = 0$. (Approximate $f$ by periodic functions with very long period.)
I have trouble proving this step, can it be said that if $f$ is continuous and compactly supported, then it can be approximated by periodic functions? I don't know a result that justifies that
- By the Riesz theorem, we have that $F$ linear functional with locally compact support can be represented as $F(f) = \int f d \mu$, with $\mu$ unique for all $f$, taking $F=0$ we have that $\mu = 0$ and again by the Riesz theorem $|F|=|\mu|$ implies that $\mu = 0$
Thank you in advance.