Show that if $y\in C^0[0,\infty)$ and $\int\limits_0^\infty (\varphi'(s)+c\varphi(s))y(s)ds=-a\varphi(0)$ then $y\in C^\infty$ and $y'=cy, y(0)=a$ Suppose that $y\in C^0[0,\infty)$ satisfies $$\int\limits_0^\infty (\varphi'(s)+c\varphi(s))y(s)ds=-a\varphi(0)$$ for any compact support function $\varphi\in C^1_0(\Bbb R)$. Then why is $y\in C^\infty[0,\infty)$ and a solution to $y'(t)=cy(t), y(0)=a$?
Clearly the solution of the differential equation above is $y(t)=ae^{ct}$ and it happens to satisfy the two properties of being $C^\infty[0,\infty)$ and integrating by parts $$\int\limits_0^\infty h'g+hg'=[hg]_0^\infty$$ with $h=\varphi$ and $g=f$ gives the second property. But that just proves the "easy way". 
So how can I show the other way? There has to be a well chosen $\varphi$ that makes it easier.
 A: The integral in the given equation can be extended to all of $\Bbb R$ if $y$ is extended as piecewise continuous function with $y(t)=0$ for $t<0$.
If $\phi(s)$ is a test function, then so is $\phi(s-\tau)$ for any constant $\tau$.  Inserting that and integrating $\tau$ over $[0,t]$ gives
\begin{align}
-a\int_0^t\phi(−τ)\,dτ&=\int_0^t\int_{\Bbb R} (\phi'(s−τ)+c\phi(s−τ))y(s)\,ds\,dτ
\\
-a\int_{-t}^0\phi(s)\,ds
&=\int_{\Bbb R} \left[\phi(s)-\phi(s-t)\right] y(s)\,ds+c\int _0^\infty \int_0^t\phi(s−τ) y(s)\,dτ\,ds
\\
-a\int_{\Bbb R}[H(s+t)-H(s)]\phi(s)\,ds
&=\int_{\Bbb R} \phi(s)[y(s)- y(s+t)]\,ds+c\int _0^\infty \int_0^t\phi(s) y(s+τ)\,dτ\,ds,
\end{align}
where $H$ be the Heaviside function, $H(s)=0$ for $s<0$, $H(s)=1$ for $s\ge 0$.
To hold for all $\phi$ with the stated properties, the other factor in the integrand has to be identical on both sides, so that
$$
-a[H(s)-H(s+t)]=y(s)-y(s+t)+\int_0^ty(s+τ)\,dτ,
\\\iff
y(s+t)-aH(s+t)=y(s)-aH(s)+\int_0^ty(s+τ)\,dτ
$$
For $s=-t$ this specializes to
$$
y(0)-a=y(-t)=0
$$
so that $y(0)=a$. Then for $s=0$ we get
$$
y(t)-a=y(0)-a+c\int_0^ty(τ)\,dτ,
$$
which by Picard-Lindelöf has a continuously differentiable solution that solves the IVP $y'(t)=cy(t)$ for $t>0$, $y(0)=a$.

If you go deeper into functional analysis, if $y$ is again piecewise continuous on the whole of $\Bbb R$ with $y(t)=0$ for $t<0$, then $y$ is or defines a regular distribution $T_y$ with support in $[0,\infty)$ and the equation has a direct translation as
$$
(-T_y'+cT_y)=-a\delta_0
$$
where applying the integrating factor $e_{-c}$ with $e_{-c}(s)=e^{-cs}$ gives
$$
(e_{-c}T_y)'=a\delta_0\implies e_{-c}T_y=aH+b
$$
Comparing the supports of the terms one finds directly $b=0$ for the integration constant. Then $y(t)=ae^{ct}$ for $t\ge 0$.
