How to rigorously differentiate the convolution of a distribution and a $L^2$ function?

Question

I want to prove the following: (Here, $W^{2,2}$ is a Sobolev space as defined in Evans, 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}

---

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*}

---

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?

Answer 1

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.)

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.