No identity for convolution (without Fourier)

Question

it's well known, that the convolution has no identity in $L^1(\mathbb{R}^d)$, so there is no $g \in L^1(\mathbb{R}^d)$ so that $f*g=f$ for all $f\in L^1(\mathbb{R}^d)$. The usual proof goes over the Fourier transform and Riemann-Lebesgue theorem. Unfortunately, I cannot use the Fourier transform or Riemann-Lebesgue theorem. So I'm looking for another proof. Supposedly, it should use the absolute continuity of the Lebesgue integral. Thanks in advance for ideas and hints.

Answer 1

Let $\varphi$ be a mollifier, i.e. a $C^\infty(\mathbb{R}^d)$ function with compact support, taking values in the closed interval $[0, 1]$ such that $\int_\mathbb{R^d} \varphi = 1$. Define $\varphi_\varepsilon(x) = \frac{1}{\varepsilon^n}\varphi(\frac{x}{\varepsilon})$. It is well known that $\varphi_\varepsilon \ast g$ converges to $g$ a.e. for all $g \in L^1(\mathbb{R^d})$. Now suppose $g$ is an identity for convolution. Then, $\varphi_\varepsilon \ast g = \varphi_\varepsilon$. But $\varphi_\varepsilon(x)$ converges to $0$ for all $x \neq 0$, i.e. for a.e. $x \in \mathbb{R}^d$. Hence, $g = 0$ almost everywhere. However, $0$ is certainly not an identity for convolution. We reach a contradiction.

Edit: I guess you could give a simpler proof using absolute continuity. Take the convolution of $g$ with the characteristic function of a set $E$ with positive measure. Then, the convolution at $x$ is simply the integral of $g$ over $x - E$, which is smaller in absolute value than $1/2$ if $E$ has sufficiently small measure (use absolute continuity). Therefore, $\lVert \chi_E \ast g \rVert_{L^\infty} \le 1/2$. However, if $g$ is an identity for convolution, then $\lVert \chi_E \ast g \rVert_{L^\infty} = \lVert \chi_E \rVert_{L^\infty} = 1$. Thus, there can not be an identity for convolution.