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)$?
 A: Two inequalities:


*

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

*$|\varphi (x)- \varphi (0)|\le 2\|\varphi|_\infty.$
Thus
$$\int_0^\infty\left |\frac{\varphi (x)- \varphi (0)}{x^{3/2}}\right |\,dx$$ $$ \le \left (\int_0^1x^{-1/2}\,dx \right)\|\varphi'\|_\infty + \left (\int_1^\infty x^{-3/2}\,dx\right)\cdot 2\|\varphi\|_\infty.$$ 
A: As $\varphi \in C_0^\infty(\mathbb{R})$, you can write
$$\varphi(x) - \varphi(0) = x \varphi^\prime(0) + x \epsilon(x)$$ with
$\lim\limits_{x \to 0} \epsilon(x) = 0$. As $\int_0^1 \frac{dx}{x^{1/2}}$ converges, $\int\limits_0^1 \frac{\varphi(x)-\varphi(0)}{x^{3/2}} dx$ converges.
Then for $x$ large enough, say $ x >M >0$ , $\varphi(x)$ vanishes. And as $\int_M^{\infty} \frac{\varphi(0)}{x^{3/2}} \ dx$ converges, $\langle T, \varphi \rangle$ is well defined.
A: My first impulse is to do what you said and then apply Fubini:
$$\begin{align}\int_0^\infty\frac{\phi(x)-\phi(0)}{x^{3/2}}\,dx&=\int_0^\infty x^{-3/2}\int_0^x\phi'(t)\,dt\,dx
\\&=\int_0^\infty\phi'(t)\int_t^\infty x^{-3/2}\,dx\,dt
\\&=2\int_0^\infty t^{-1/2}\phi'(t)\,dt.\end{align}$$So $$|\langle T,\phi\rangle|\le c\rho(\phi),$$where $$\rho(\phi)=\sup(1+|t|)|\phi'(t)|$$ and $$c=2\int_0^\infty\frac{t^{-1/2}}{1+t}\,dt.$$Since $\rho$ is one of the seminorms in the definition of the Schwarz space this shows that not only is $T$ a distribution, it's actually a tempered distribution.
(If you really want the specific inequality you mentioned, it's clear that if $\phi$ is supported on $K$ then $\rho(\phi)\le C_K\sup_K|\phi'|$.)
