I want to prove the following: (Here, $W^{2,2}$ is a Sobolev space as defined in [Evans](https://bookstore.ams.org/gsm-19-r), chapter 5; $S$ is a Schwartz space; and if $A$ is a distribution and $a$ a function, then $\langle A, a\rangle$ means $A(a)$).



> 
> **Theorem.** Let $\newcommand{\C}{\mathbb C}\newcommand{\R}{\mathbb R}C\in]0,\infty[$. For every $f\in L^2 (\mathbb R;\C)$ there exists a $g\in (L^2 \cap W^{2,2}\_{\text{loc}})(\R;\C)$ such that (in the weak sense)
> \begin{equation}
> -g'' + C^2 g = f
> \end{equation}
> and an explicit solution is given by
> \begin{equation}
> g(t) = \frac{1}{2C} \int\_{\R} e^{-C|t-\tau|} f (\tau)\,\mathrm d\tau.
> \end{equation}
> 
> 
> 




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**My attempt.** (Skip to the bottom for my question)


Consider the tempered distribution
\begin{equation\*}\begin{split}
 \mathscr Z: S(\R;\C)&\to\C, \\
 \phi&\mapsto\int\_\R e^{-{C\lvert t\rvert}} \phi(t)\,\mathrm dt = \int\_0^\infty e^{-Ct}\phi(t)\,\mathrm dt+\int\_{-\infty}^0 e^{Ct} \phi(t)\,\mathrm dt.
\end{split}\end{equation\*}
Then we have, for all $\phi\in S(\R;\C)$,
\begin{equation\*}
\langle\mathscr Z',\phi\rangle=-\langle\mathscr Z,\phi'\rangle=\int\_0^\infty e^{-Ct}\phi'(t)\,\mathrm dt+\int\_{-\infty}^0 e^{Ct} \phi'(t)\,\mathrm dt.
\end{equation\*}
Integrating both terms by parts, where the exponential term gets differentiated and $\phi'$ gets integrated, we get
\begin{equation\*}\begin{split}
 -\langle\mathscr Z',\phi\rangle &= \left[e^{-Ct}\phi(t)\right]^\infty\_0+C\int\_0^\infty e^{-Ct}\phi(t)\,\mathrm dt+\left[e^{Ct}\phi(t)\right]^0\_{-\infty}-C\int\_{-\infty}^0 e^{Ct}\phi(t)\,\mathrm dt \\
 &= C\int\_0^\infty e^{-Ct}\phi(t)\,\mathrm dt-C\int\_{-\infty}^0 e^{Ct}\phi(t)\,\mathrm dt.
\end{split}\end{equation\*}
Integrating both terms by parts in the same way, we get (where $\delta\_0$ is the Dirac distribution at $0$)
\begin{equation\*}\begin{split}
 \langle\mathscr Z'',\phi\rangle &= \langle-\mathscr Z',\phi'\rangle \\
 &= C\left[e^{-Ct}\phi(t)\right]^\infty\_0+C^2\int\_0^\infty e^{-Ct}\phi(t)\,\mathrm dt - \left(C \left[e^{Ct}\phi(t)\right]\_{-\infty}^0-C^2\int\_{-\infty}^0 e^{Ct}\phi(t)\,\mathrm dt\right) \\
 &= C^2 \langle\mathscr Z,\phi\rangle-2C\phi(0) = C^2 \langle\mathscr Z,\phi\rangle-2 C\langle{\delta\_0,\phi}\rangle.
\end{split}\end{equation\*}




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Now I would like to finish by writing


$$g = \frac{\mathscr Z\* f}{2C}$$ and therefore


$$-g''+C^2 g = \frac{-(\mathscr Z'' \* f)+C^2 (\mathscr Z\* f)}{2C} = \frac{-((C^2\mathscr Z-2C\delta\_0)\*f)+C^2 (\mathscr Z\* f)}{2C}=\delta\_0\*f = f.$$


**My question:** However, to do this, I formally use
$$ (A\*a)'=(A'\*a)$$
when $A$ is a distribution. Is there some result that justifies this? And furthermore, does this result also imply that $g$, defined as the convolution of $f$ with $\mathscr Z$, can be written as a $W^{2,2}\_{\text{loc}}$ function?



How to rigorously differentiate the convolution of a distribution and a $L^2$ function?
The last step is formally justified by **15.8, differentiation property** of José Sebastião e Silva's "Integrals and orders of growth of distributions." (The paper is currently available [here](http://jss100.campus.ciencias.ulisboa.pt/Publicacoes/Artigos-de-Investigacao/Inv-JSS/Integrals%20and%20orders%20of%20growth%20of%20distributions.pdf).)


More precisely: $\mathscr K$ is defined by a continuous function. Also, by Lemma 8.2 of Haim Brezis' *Functional Analysis, Sobolev Spaces and Partial Differential Equations* (2010), we have, in the weak sense, since $f\in L^2\subset L^1\_{\text{loc}}$, $F'=f$ where $F(x):=\int\_0^x f$ is continuous.


Therefore, one can indeed apply 15.8 mentioned above.


