Show that every test function $\varphi$ $\in$ $C^\infty_c(\mathbb{R})$ can be written as $\varphi(x) = x \psi(x) + c\varphi_0(x)$ where $\varphi_0(x)$ is a fixed test function (with $\varphi_0(0) \neq 0)$, $\psi$ $\in$ $C^\infty_c(\mathbb{R})$, and $c$ is a constant.
I'm not really sure where to begin on this. Any help would be appreciated.
On
Hint: Observe \begin{align} f(x):= \varphi(x)-\frac{\varphi(0)}{\varphi_0(0)}\varphi_0(x)=\varphi-\alpha\varphi_0 \end{align} equals $0$ when $x=0$, i.e. $f(0) = 0$. Next, observe \begin{align} f(x) = \int^x_0f'(s)\ ds = xf'(x)-\int^x_0sf''(s)\ ds = x\left(f'(x)-\frac{1}{x}\int^x_0 sf''(s)\ ds \right). \end{align} (Note that I used the fact $f(0) =0$.) Hence it suffices to show that \begin{align} G(x) = \begin{cases} \displaystyle \frac{1}{x}\int^x_0 sf''(s)\ ds & \text{ if } x \neq 0\\ 0 & \text{ if } x = 0 \end{cases} \end{align} is in $C^\infty_c(\mathbb{R})$.
More hint: for $x \neq 0$, we that \begin{align} \lim_{h\rightarrow 0}\frac{G(x+h)-G(x)}{h} =\ f''(x) -\frac{1}{x^2}\int^x_0sf''(s)\ ds= \frac{1}{2}f''(x)+\frac{1}{2x^2}\int^x_0 s^2f'''(s)\ ds \end{align}
$\newcommand{\vp}{\varphi}$ Your $\vp$s and $\psi$s are all over the place alas. I will assume you want to write $\psi$ as $x\vp+c\vp_0$.
Putting $x=0$ we need $\psi(0)=c\vp(0)$, so we need $c=\psi(0)/\vp_0(0)$. Then we must have $$\vp(x)=\frac{\psi(x)-c\vp_0(x)}{x}.$$ So we need to check this is $C^\infty$ at $0$ (when we assign a suitable value to $\vp(0)$) as it is clearly compactly supported.