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a) Let $\psi \in \mathcal{D}(\mathbb{R})$ be such that $\psi = 1$ in a neighbourhood of the origin. Show that the linear functional defined by \begin{equation} T_{[\psi]}(\phi) = \int \frac{\phi(x) -\phi(0)\psi(x)}{|x|} dx , \phi \in \mathcal{D}(\mathbb{R}) \end{equation} defines a distribution.

b) Show that there exists $\psi \in \mathcal{D}(\mathbb{R})$ as above such that \begin{equation} \lim_{\epsilon \downarrow 0} \left(\frac{1}{|x|+\epsilon} -2 |\log \epsilon|\delta\right) = T_{[\psi]}. \end{equation}

I'm able to do that first part by showing, \begin{equation} T_{[\psi]}(\phi)=\int_{-\infty}^{0} \log |x|[\phi'(x)-\phi(0)\psi'(x)]dx - \int_{0}^{\infty} \log |x|[\phi'(x)-\phi(0)\psi'(x)]dx \end{equation}

However, I have no idea how to do the second part. Any help will be appreciated.

1 Answers 1

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Let $\psi$ be such that $\psi \equiv1$ on $[-1,1]$, support $\psi \subset[-b,b] \ (b>1)$ and $\int_1^b \frac{\psi(x)}{x}=0$.

By symmetry, $\int_{-b}^{-1}\frac{\psi(x)}x=0$. Let $\epsilon < 1.$

Let $\phi \in \mathcal{D}(\mathbb{R})$ and support $\phi \subset [-a,a].$

$$\left(\frac{1}{|x|+\epsilon}\right)(\phi)=\int_{\mathbb{R}}\frac{\phi(x)}{|x|+\epsilon}\ dx = \int_{-a}^0\frac{\phi(x)}{\epsilon-x}\ dx + \int_{0}^a\frac{\phi(x)}{\epsilon+x}\ dx\\=2|\log \epsilon|\delta(\phi)+\int_{-\infty}^0\phi'(x)\log(\epsilon-x)\ dx-\int_0^{\infty}\phi'(x)\log(\epsilon+x)\ dx$$

Thus, $$\lim_{\epsilon \downarrow 0} \left(\frac{1}{|x|+\epsilon} -2 |\log \epsilon|\delta\right) -T_{[\psi]}=\phi(0)\left[\int_{-\infty}^0\log|x|\psi'(x)\ dx - \int_0^{\infty}\log|x|\psi'(x)\ dx\right]\\=\phi(0)\left[\int_{-b}^{-1}\log|x|\psi'(x)+\int_{-1}^0\log|x|\psi'(x)-\int_0^1\log|x|\psi'(x)-\int_1^b\log|x|\psi'(x)\right]=0 $$