I am trying to show that for an appropriate choice of constants $c_{\delta}$ which diverge as $\delta \to 0$ a distribution $W\in \mathcal{S}'(\mathbb{R})$ can be defined by: $$ W(\phi)=\lim_{\delta \to 0}\,\int_{|x|>\delta} \bigg(\frac{1}{|x|}\phi(x)-c_{\delta}\phi(0)\bigg)\,dx \,\,\,\,\, \phi \in \mathcal{S}(\mathbb{R})$$
I am new to distributions and I am finding them a bit confusing, we first have to show that this defines a linear functional on $\mathcal{S}(\mathbb{R})$ and clearly the only thing to prove here is that the limit converges. Then do we prove that the above satisfies the necessary and sufficient boundedness/continuity condition to lie in $\mathcal{S}'(\mathbb{R})$ or can we prove the latter first to save time without knowing the integral makes sense?
So we have to show that we can choose constants $\delta$ so that the integral converges independently of $\delta$? I have been looking at $$\bigg|\frac{\phi(x)-(c_{\delta}|x|)\phi(0)}{x}\bigg|$$
and have been trying to use the differentiability of $\phi$ but I'm not sure this will work.
The starting point is that the limit $$\lim_{\delta \to 0}\,\int_{\delta<|x|<1} \frac{1}{|x|}(\phi(x)- \phi(0))\,dx\tag1$$ is finite, because the integrand is bounded by the Mean Value Theorem : $|\phi(x)- \phi(0)|\le |x|\sup |\phi'|$.
This tells you what $c_\delta$ should be: it's $-2\log\delta$.
To complete the proof that this is a distribution, you need continuity with respect to $\phi$. This will follow from the estimate $$W(\phi)\le C(\sup|\phi'| + \sup (|x|+1) |\phi(x)|)$$ because the suprema on the left bound (1) together: one helps for $|x|<1$, the other for $|x|\ge 1$.