Inverse convolution of a distribution.

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.

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