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

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 $$A(a)$$).


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?

• do u know how $A'$ is defined? Aug 7 at 21:04
• You still have the usual (existence and uniqueness) theory available also for ODEs in this more general interpretation; this is discussed in many books, for example Coddington-Levinson. Your formula is then the variation-of-constants formula for the solutions of an inhomogeneous linear ODE. Aug 7 at 22:00
• @ChristianRemling Thank you! I will check out the Coddington-Levinson book that you mentioned in the next days Aug 7 at 23:24
• @mathworker21 (I do not know of a sensible way to define the convolution $T*f$ if $T$ is any tempered distribution and $f$ is any $L^2$-function.) Aug 7 at 23:32
• @mathworker21 ahh I apologize for getting a bit pissed then 😅 Aug 8 at 21:51

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.