For every $a>0$, show that $\langle f_a, \psi\rangle= \int_{|x|>a} \frac{\psi(x)}{|x|}dx+\int_{|x|>a} \frac{\psi(x)- \psi(0)}{|x|}dx$ To prove that the inner product is a distribution it must satisfy the following property"
$$|T(\phi)|=|\langle T,\psi\rangle| \leq C_N \sum_{|\alpha| \leq N} \|\partial^\alpha \psi\|_\infty$$
Part A: For every $a>0$, show that $\langle f_a, \psi\rangle= \int_{|x|>a} \frac{\psi(x)}{|x|}dx+\int_{|x|<a} \frac{\psi(x)- \psi(0)}{|x|}dx$ is a distribution. What is the order of $f_a$?
My work:
$$|\langle f_a, \psi\rangle|= |\int_{|x|>a} \frac{\psi(x)}{|x|}dx+\int_{|x|<a} \frac{\psi(x)- \psi(0)}{|x|}dx|
=|\int_{|x|>a} \frac{\psi(x)}{|x|}dx+\int_{|x|<a} \psi'(c)dx|
=|\int_{|x|>a} \frac{\psi(x)}{|x|}dx+2a\psi'(c)|
\leq |\int_{|x|>a} \frac{\psi(x)}{|x|}dx| + |2a\psi'(c)|
\leq \int_{|x|>a} |\frac{\psi(x)}{|x|}|dx + |2a\psi'(c)|
= \int_{|x|>a} \frac{|\psi(x)|}{|x|}dx + 2a|\psi'(c)|$$
This is where I am stuck.
Part B: Prove that $f_a$ does not depend on a, and consequentely that $\langle g, \psi\rangle= \lim_{a \rightarrow 0+} \int_{|x|>a} \frac{\psi(x)}{|x|}dx$ is a distribution.
My work:
$$|\langle g, \psi\rangle |= |\lim_{a \rightarrow 0+} \int_{|x|>a} \frac{\psi(x)}{|x|}dx|
\leq \lim_{a \rightarrow 0+} \int_{|x|>a} |\frac{\psi(x)}{|x|}|dx
=\lim_{a \rightarrow 0+} \int_{|x|>a} \frac{|\psi(x)|}{|x|}dx
\leq \lim_{a \rightarrow 0+} \int_{|x|>a} \frac{\|\psi\|_\infty}{|x|}dx
=\|\psi\|_\infty \times\lim_{a \rightarrow 0+} [\int_{-\infty}^{-a} \frac{1}{-x}dx+\int_a^{\infty} \frac{1}{x}dx]
=\|\psi\|_\infty$$
Please check
 A: The support of $\psi$ must appear somewhere, otherwise $x\to 1$ is not a distribution, for example.
Part A. We bound things:
$$
\left|\int_{|x|>a} \frac{1}{x}\psi dx\right|\leq \sup_{|x|>a} \frac{1}{|x|} |\int_{|x|\geq a} \psi dx| \leq  \frac{1}{a}\|\psi\|_{\infty}|\textrm{supp} \psi|.
$$
For the second part,
$$
\left|\int_{|x|<a}\frac{\psi(x)−\psi(0)}{x} dx \right|\leq \int_{|x|<a}\frac{\left|\psi(x)−\psi(0)- x \psi^\prime(0)\right|}{|x|} dx  + |\psi^\prime(0)|\min(|\textrm{supp} \psi|,2a)
$$
and from Taylor's expansion Theorem in zero,
 $$\left|\psi(x)−\psi(0)- x \psi^\prime(0)\right|\leq\frac{x^2}{2}\|\psi^{(2)}\|_\infty,$$
so 
$$
\left|\int_{|x|<a}\frac{\psi(x)−\psi(0)}{x}dx \right| \leq \|\psi^\prime\|_{\infty}+\|\psi^{(2)}\|_\infty \int_{|x|\leq a} \frac{1}{2}|x| dx =(\|\psi^{(2)}\|_\infty \frac{a}{2}+ \|\psi^\prime\|_{\infty})|\min(|\textrm{supp} \psi|,2a)
, 
$$ 
Thus $N=2$.  and $$C_2=\max\left(\frac1a, 1,\frac{a}{2}\right)$$
Part B 
Suppose $0<b<a$. As it is written, 
$$\langle f_a,\psi\rangle-\langle f_b,\psi \rangle=-\int_{a>|x|>b} \frac{\psi}{|x|}dx + \int_{a>|x|>b} \frac{\psi(x)-\psi(0)}{|x|}dx =-2\psi(0)\ln \frac{b}{a}.  $$
This clearly depends on $a$ and $b$. However, if instead the question was :
$$
\langle f_a, \psi \rangle = \int_{|x|>a} \frac{\psi}{x} + \int_{|x|<a} \frac{\psi(x)-\psi(0)}{x} dx
$$
Then, Part A is unchanged  (in fact I answered it in that corrected form). but $<f_a -f_b,\psi>=0$, because it is 
$$
<f_a -f_b,\psi>=\int_{a>|x|>b} \frac{-\psi(0)}{x}dx=0
$$
(an odd function on a symmetric domain..). 
Now, for any $\psi$,
$$
\left|\int_{|x|<a}\frac{\psi(x)−\psi(0)}{x} dx \right|\leq 
(\|\psi^{(2)}\|_\infty \frac{a}{2}+ \|\psi^\prime\|_{\infty})|\min(|\textrm{supp} \psi|,2a) \to 0
$$
as $a\to 0$, so
$$<f_1,\psi>=\lim_{a\to0}<f_a,\psi>=\lim_{a\to0} \int_{|x|<a} \frac{\psi(x)}{x}dx
$$
and since $f_1$ is a distribution, so is the limit. 
