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I am having trouble with the following question.

Let $\mu$ be finite measure on $\mathbb{R}$ and let $\hat{\mu}(\xi) = \int_{-\infty}^\infty e^{-ix \xi} d\mu(x)$ be its Fourier transform. Prove that

$$|\mu(\{x\})| \le \limsup_{|\xi| \rightarrow \infty} |\hat{\mu}(\xi)|$$

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1 Answer 1

We use the following result, which is a consequence of Fubini's theorem: $$\mu(\{x\})=\lim_{T\to +\infty}\frac 1{2T}\int_{-T}^Te^{-itx}\widehat\mu(t)dt.$$ We pick $T_n\uparrow+\infty$ such that $$\left|\mu(\{x\})-\frac 1{2T_n}\int_{-T_n}^{T_n}e^{-itx}\widehat\mu(t)dt\right|\leq \frac 1n.$$ We can assume $T_n\geq n^2$. Removing $[-n,n]$, and using intermediate value theorem ($\widehat \mu$ is continuous), we can write $$\mu(\{x\})=\frac{\widehat\mu(s_n)+\widehat\mu(t_n)}2,$$ where $s_n\to +\infty$ and $t_n\to -\infty$. This gives the result.

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