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Notation. Let ${\mathcal{D}'}_+(\mathbb{R})$ be the set of distributions on $\mathbb{R}$ supported on $[0,+\infty[$.

One easily derives the:

Proposition. Let $T,S\in{\mathcal{D}'}_+(\mathbb{R})$, then $T$ and $S$ are convolvable and $T\ast S\in{\mathcal{D}'}_+(\mathbb{R})$.

Let me recall what is a convolutive inverse.

Definition. Let $T\in{\mathcal{D}'}_+(\mathbb{R})$, $T$ is invertible in ${\mathcal{D}'}_+(\mathbb{R})$ if and only if there exists $S\in{\mathcal{D}'_+}(\mathbb{R})$ such that: $$T\ast S=\delta_0.$$

One easily show that an element of ${\mathcal{D}'}_+(\mathbb{R})$ has at most one convolutive inverse.

Let $\lambda\in\mathbb{C}$ and $E_{\lambda}$ be the inverse convolution of ${\delta_0}'-\lambda\delta_0$. I was able to compute $E_\lambda$ for $\lambda\neq 1$, if found: $$E_{\lambda}=\frac{H}{1-\lambda}.$$ Where $H$ is the Heaviside step function. However, I have hard time computing $E_1$, since ${\delta_0}'-\delta_0$ is a tempered distribution, I tried finding its convolution inverse in $\mathcal{S}'(\mathbb{R})$. Using Fourier transform, one has: $$(-i\xi-1)\widehat{E_1}=1.$$ Hence, $$\widehat{E_1}=-\frac{1}{1+i\xi}.$$ I am stuck computing the inverse Fourier transform of $\displaystyle\xi\mapsto-\frac{1}{1+i\xi}$. I thought of using residues theorem but first is this method computing $E_1$ even correct? Is there any better one? Any help will be appreciated.

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You miscalculated, but otherwise your approach looks fine, except for one potential problem that I'll comment on below. Let me suggest an alternative method: We're looking for an $E$ so that $\delta'*E-\lambda\delta*E=\delta$. Proceeding formally, this becomes $E'-\lambda E=\delta$, and now we "solve" this (still formally, let's not worry about anything at this point) by variation of constants. This gives $$ E(x) = \chi_{(0,\infty)}(x) e^{\lambda x} . $$ Now it's an easy matter to check with the actual rigorous definition of convolution of two distributions one of which has compact support that this $E$ works.

Your approach will work too if $\textrm{Re}\,\lambda\le 0$; in the other case, we have the potential problem that $E$ is not a tempered distribution (though we might still get the right answer from a formal calculation, I haven't checked this).

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  • $\begingroup$ Indeed, I miscalculated! Your method is clear enough, thanks! I assume $\chi_{(0,+\infty)}$ is the Heaviside step function, isn't it? $\endgroup$ – C. Falcon Dec 5 '15 at 15:33
  • $\begingroup$ Yes, it's $H$ ($\chi$ as in characteristic function). $\endgroup$ – user138530 Dec 5 '15 at 19:21

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