Study convergence in distribution of $T_{n}(x)=\frac{|x|^{\frac{1}{n}-1}}{2n}$

Question

I have a problems in distribution theory but I don't know if my step correct! Study the convergent of $T_{n}$ in $D'(\mathbb{R})$ Where :

$$T_{n}(x)=\frac{|x|^{\frac{1}{n}-1}}{2n}$$ $$k\in \mathbb{N}~,~ x\in \mathbb{R^{*}}$$ My try : Its clearly that : $T_{n}\in L^{1}_{\text{loc}}(\mathbb{R^{*}})$ so : $\langle T_{n} ,\varphi \rangle =\int_{\mathbb{R}}T_{n}(x)\varphi(x)dx$ And we have $\varphi\in D'(\mathbb{R})\implies \exists a>0 ~\text{supp}\varphi \subset [a,a]$ so : $\langle T_{n} ,\varphi \rangle =\int_{|x|≤a}T_{n}(x)\varphi(x)dx≤\frac{a^{\frac{1}{n}-1}}{2n}\int_{|x|≤a}\varphi (x)dx$ and this go to $0$ because $\int_{|x|≤a}\varphi (x)dx<\infty $

Also here we say convergence in $D'(\mathbb{R^{*}}) ~\text{or} ~D'(\mathbb{R})$ ? Also $\varphi \in D(\mathbb{R}) ~\text{or}~D(\mathbb{R^{*}})$

In finally thank you to every teachers see my problems

I'm greatful

---

Thanks!

Answer 1

I get that $T_n \to \delta$: $$ \left< T_{n}, \varphi \right> = \int_{-\infty}^{\infty} \frac{|x|^{\frac{1}{n}-1}}{2n} \, \varphi(x) \, dx = \int_{0}^{\infty} \frac{x^{\frac{1}{n}-1}}{2n} \, (\varphi(x) + \varphi(-x)) \, dx = \{ x = y^n \} \\ = \int_{0}^{\infty} \frac{\varphi(y^n) + \varphi(-y^n)}{2} \, dy = \int_{0}^{1} \frac{\varphi(y^n) + \varphi(-y^n)}{2} \, dy + \int_{1}^{\infty} \frac{\varphi(y^n) + \varphi(-y^n)}{2} \, dy \\ \to \int_{0}^{1} \varphi(0) \, dy + \int_{1}^{\infty} \underbrace{\varphi(\infty)}_{=0} \, dy = \varphi(0) = \left< \delta, \varphi \right> . $$ This is a bit sloppy. Some more work might be needed to justify the limits better.