So I just started studying distribution theory and I am asked to show that $\left\langle f_a,\varphi\right\rangle = \int_{-\infty}^{-a} + \int_a^{\infty} \! \frac{\varphi(x)}{|x|} \, \mathrm{d}x + \int_{-a}^{a} \! \frac{\varphi(x)-\varphi(0)}{|x|} \, \mathrm{d}x$ is a distribution for any $a>0$.
My book says that any function gives rise to a distribution by setting $\left\langle f,\varphi\right\rangle= \int_\Omega \! f(x)\varphi(x) \, \mathrm{d}x$ and if this integral is defined. So I guess I should check the convergence of the integral above, but how to do this? I guess I still do not understand the concept of a distribution.
You need to estimate your $|\left\langle f_a,\varphi\right\rangle|$ in terms of finite number of norms $\|\varphi\|_\infty$, $\|\varphi' \|_\infty$,..., $\|\varphi^{(k)} \|_\infty$: more precisely, for every compact $K$ we need to find a constant $C_K$ and a number $n_K$ such that for any test function $\varphi$ with support in $K$ the following must hold: $$\left|\left\langle f_a,\varphi\right\rangle\right|\le C_K\cdot (\|\varphi\|_\infty+\|\varphi' \|_\infty+\dots+\|\varphi^{(n_K)} \|_\infty).$$
Let the compact be included in $[-R,R]$ and $R>a$. We write $$\left|\left\langle f_a,\varphi\right\rangle\right| = \left|\int_{-\infty}^{-a} \! \frac{\varphi(x)}{|x|} \, \mathrm{d}x+ \int_a^{\infty} \! \frac{\varphi(x)}{|x|} \, \mathrm{d}x + \int_{-a}^{a} \! \frac{\varphi(x)-\varphi(0)}{|x|} \, \mathrm{d}x\right|$$ $$\le \|\varphi\|_\infty\left(\int_{-R}^{-a} \! \frac{\mathrm{d}x}{|x|} + \int_{a}^{R} \! \frac{\mathrm{d}x}{|x|}\right)+ \left|\int_{-a}^{a} \! \frac{\varphi(x)-\varphi(0)}{|x|} \, \mathrm{d}x\right|$$
The factor $\left(\int_{-R}^{-a} \! \frac{\mathrm{d}x}{|x|} + \int_{a}^{R} \! \frac{\mathrm{d}x}{|x|}\right)$ does not depend on $\varphi$, so everything is ok.
Now to the second term, $ \int_{-a}^{a} \! \frac{\varphi(x)-\varphi(0)}{|x|} \, \mathrm{d}x $:$$ \int_{-a}^{a} \! \frac{\varphi(x)-\varphi(0)}{|x|} \, \mathrm{d}x= \int_{0}^{a} \! \frac{\varphi(x)-\varphi(0)}{x} \, \mathrm{d}x-\int_{-a}^{0} \! \frac{\varphi(x)-\varphi(0)}{x} \, \mathrm{d}x.$$ By Taylor it equals $$ \int_{0}^{a} \! (\varphi' (0) +\varphi''(\xi_1)x/2 ) \mathrm{d}x-\int_{-a}^{0} (\varphi' (0) +\varphi''(\xi_2)x/2 ) \mathrm{d}x $$ with $\xi_1\in(-a,0)$ and $\xi_2\in(0,a)$. Thus
$$\left|\int_{-a}^{a} \! \frac{\varphi(x)-\varphi(0)}{|x|} \, \mathrm{d}x\right|\le \frac 12 \|\varphi''\|_\infty\int_{-a}^a|x|dx$$ Hence, by choosing $$C_K = \max \left(\left(\int_{-R}^{-a} \! \frac{\mathrm{d}x}{|x|} + \int_{a}^{R} \! \frac{\mathrm{d}x}{|x|}\right),\frac 12 \int_{-a}^a|x|dx\right)$$ and $n_K=2$ we can conclude that indeed $f_a$ is a distribution.
The proof for the case $a\ge R$ is done likewise.