Proof of the nonexistence of an identity $\phi$ involving convolution The Banach space $L^1(\mathbb{R}^n)$ is an algebra with a product (convolution) which is both commutative and associative. But this algebra does not have a multiplicative identity.
An attempt to show the nonexistence:
If $\phi$ exists, consider $\chi_{B(0,r)}*\phi=\chi_{B(0,r)}$ where $B(0,r)$ is the ball with radius $r$ centered at the origin, and $\chi_A$ is the characteristic function on set $A$. Then there exists $x\in B(0,r)$ such that $(\chi_{B(0,r)}*\phi)(x)=1$.
Now, why does it follow that $\int_{B(0,2r)}|\phi(y)|dy\ge1$? Does this follow from the inequality $||f*g||_1\le||f||_1||g||_1$?
Thank you!
 A: Let me suggest a different path:
$\chi_{B(0,r)} \in L^1(\mathbb{R}^n) \cap L^\infty(\mathbb{R}^n)$, therefore $\phi\ast \chi_{B(0,r)}$ is a continuous function for every $\phi \in L^1(\mathbb{R}^n)$ But a continuous function cannot be equal almost everywhere to $\chi_{B(0,r)}$. Hence there is no identity for convolution in $L^1(\mathbb{R}^n)$.
For the route you took, we have by assumption $(\chi_{B(0,r)}\ast\phi)(x) = 1$ for almost all $x \in B(0,r)$. That means for such an $x$
$$\begin{align}
1 &= \int_{\mathbb{R}^n} \chi_{B(0,r)}(y) \phi(x-y)\,dy\\
&= \int_{B(0,r)} \phi(x-y)\,dy\\
&= \int_{B(x,r)} \phi(z)\,dz\\
&\leqslant \int_{B(x,r)} \lvert \phi(z)\rvert\,dz\\
&\leqslant \int_{B(0,2r)} \lvert \phi(z)\rvert\,dz,
\end{align}$$
where the last inequality follows since $B(x,r) \subset B(0,2r)$ for $x\in B(0,r)$.
Letting $r \searrow 0$, we obtain a contradiction to the integrability of $\phi$, since
$$\lim_{r\searrow 0} \int_{B(0,2r)} \lvert f(z)\rvert\,dz = 0$$
for every $f \in L^1(\mathbb{R}^n)$ by the dominated convergence theorem.
