Is $T$ (defined below) a distribution? 
How to show that 
  $$ 
\langle T, \varphi\rangle = \int\limits_{0}^{+\infty} \frac{\varphi(x)-\varphi(0)}{x^{3/2}} dx,\quad \varphi \in C_0^\infty(\mathbb{R}),
$$ 
  is a distribution?

I would first rewrite the definition of $T$, using  $ \varphi(x)-\varphi(0) = \int_0^x \varphi'(t) dt $. But how to get around the fact that $ \int_{0}^{+\infty} x^{-3/2} dx $ is divergent, how to "tame" this integral in order to show that $ \lvert \langle T, \varphi\rangle \rvert \leq C \: \text{sup}_{K} \lvert \varphi'(x) \rvert $ for all $ \varphi \in C_0^\infty(K)$?

Two inequalities:


*

*$|\varphi (x)- \varphi (0)|\le \|\varphi'\|_\infty |x|.$

*$|\varphi (x)- \varphi (0)|\le 2\|\varphi|_\infty.$
Thus
$$\int_0^\infty\left |\frac{\varphi (x)- \varphi (0)}{x^{3/2}}\right |\,dx$$ $$ \le \left (\int_0^1x^{-1/2}\,dx \right)\|\varphi'\|_\infty + \left (\int_1^\infty x^{-3/2}\,dx\right)\cdot 2\|\varphi\|_\infty.$$ 
