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From Tao's blog: Let ${d=1}$. For any ${r > 0}$, show that the functional ${\lambda_r}$ defined by the formula $$\displaystyle \langle f, \lambda_r \rangle := \int_{|x| < r} \frac{f(x)-f(0)}{|x|}\ dx + \int_{|x| \geq r} \frac{f(x)}{|x|}\ dx$$

is a distribution that does not arise from either a locally integrable function or a Radon measure. Note that any two such functionals ${\lambda_r, \lambda_{r'}}$ differ by a constant multiple of the Dirac delta distribution.

Question. Why any two such functionals ${\lambda_r, \lambda_{r'}}$ differ by a constant multiple of the Dirac delta distribution?

eraldcoil
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1 Answers1

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It might be of interest to look at the notes at the end of THIS ANSWER and THIS ONE where this distribution arises in the Fourier transform of $\log(|x|)$.

Without loss of generality, let $r'<r$. Then, we have

$$\begin{align} \langle f,\lambda_r\rangle-\langle f,\lambda_{r'}\rangle&=\int_{|x|\in (r',r)}\frac{f(x)-f(0)}{|x|}\,dx-\int_{|x|\in (r',r)}\frac{f(x)}{|x|}\,dx\\\\ &=-2f(0)\int_{r'}^r\frac1x\,dx\\\\ &=-2\log(r/r')f(0) \end{align}$$

Hence, in distribution we see that

$$\lambda_r=\lambda_{r'}-2\log(r/r')\delta_0$$

as was to be shown.

Mark Viola
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