A function in $L^1(\mathbb{R}^d)$ satisfying $f(x)+f\ast g(x)=e^{-|x|^2}.$

Question

Let $g\in L^1(\mathbb{R}^d)$ with $\|g\|_{L^1(\mathbb{R}^d)}<1$. I would like to show that there is a unique $f\in L^1(\mathbb{R}^d)$ such that $$f(x)+f\ast g(x)=e^{-|x|^2}.$$ My first instinct is to take Fourier transforms to find that, with $\hat{f}(\eta)=\int_{\mathbb{R}^d}e^{-i\eta\cdot x}f(x)\,dx$, $$\hat{f}(1+\hat{g})=\pi^{d/2}e^{-|\eta|^2/4}.$$ From my hypothesis, I know that $$\|\hat{g}\|_{L^\infty (\mathbb{R}^d)}\le \|g\|_{L^1 (\mathbb{R}^d)}<1.$$ Therefore, I may divide by $1+\hat{g}$ to obtain $$\hat{f}=\frac{\pi^{d/2}e^{-|\eta|^2/4}}{1+\hat{g}}. $$ I am unsure how to go from the conclusion to a unique $f\in L^1 (\mathbb{R}^d)$.

Answer 1

Here is another approach using Banach-Picard's fixed point theorem, which has the benefit of not relying on the form of the "constant" function (constant in regards to $f$ and $g$) of the equation:

Let $a : x \mapsto e^{-|x|^2}$, and let $T : f \in L^1 \mapsto - f*g + a \in L^1$.
Since $\|(f_1 - f_2) * g\|_{L^1} \leq \|g\|_{L^1}\|f_1 - f_2\|_{L^1}$ and $\|g\|_{L^1} < 1$, $T$ is a contraction, on $L^1$ complete, thus by Banach-Picard it admits a unique fixed point $f$, which satisfies the equation $f + f*g = a$.

Answer 2

Let $T:X\to X$ be a bounded linear operator. It is well known that u if $\|T\|<1,$ the operator $I+T$ is invertible, i.e. the equation $x+Tx=y$ has a unique solution $x$ for any fixed $y.$ Moreover the solution is given by the formula $$x=y+\sum_{n=1}^\infty (-1)^nT^ny$$

Consider the operator $Tf=g*f$ for $X=L^1.$ Since $\| f*g \|_1\le \|g\|_1\|f\|_1,$ we get $\|T\|\le \|g\|_1<1.$ Thus the operator $I+T$ is invertible.

We can essentially relax the assumption on $g.$ It suffices to have $\|\widehat{g}\|_\infty<1$ or even $\widehat{g}+1\neq 0.$ The spectrum of the operator $T$ is equal $\widehat{g}(\mathbb{R})\cup\{0\},$ i.e. is contained in the open unit disc. Therefore the operator $I+T$ is invertible.