How does one prove the following proposition for convolution $*$?
There is for $p\in[1,\infty)$ there is no $g\in L^p$ such that $f*g=f, \forall f\in L^p$ where $*$ is the convolution.
It will be great if one can show it first for $p=1$ then for $p>1$.
I can prove the case for $p=1$ easily. Fourier transforming, denoted by $\hat f$ of a function $f$, the above equation I obtain $\,\hat g=1$, contradicting $\hat f(\xi)\rightarrow 0$ as $|\xi|\rightarrow\infty$ from the Riemann-Lebesgue lemma. So such $g$ does not exist.
One can use Fourier transform for $p\in (1,2]$ if the function satisfies additional regularity conditions. But how does one prove the proposition without addition conditions and for arbitrary $p\in (2,\infty)$ with and without Fourier transform? Would the Young's inequality for convolution help?
Let $B\subset \mathbb{R}^n$ be a ball of radius $r \in (0,+\infty)$, and $f = \chi_B$ its characteristic function. Since the ball has finite measure, we have $f \in L^1 \cap L^{\infty}$, so in particular $f \in L^p$ and $f \in L^q$, where $\frac{1}{p} + \frac{1}{q} = 1$. Since $f \in L^q$, the convolution $f \ast g$ is continuous for all $g \in L^p$. But $f$ has no continuous representative, so $f \ast g \neq f$, and $g$ is not a unit for the convolution.