Does there exist a function in $L^1$ such that $u * f = f$ for all f in $L^1$ While studying for exams, a practice question came up which is 
Does there exist a function $u \in L^1({\mathbb{R}^d})$ such that $u * f = f$ for all $f \in L^1({\mathbb{R}^d})$?
I was thinking something with the delta dirac but realized thats not even a function but rather a distribution.
 A: Suppose we had a $u$ satisfying $u \ast f = f$ for all $f \in L^1(\Bbb{R}^d)$. Then in particular $u \ast u = u $. Then taking Fourier transforms we get $\hat{u}(x)\hat{u}(x) = \hat{u}(x)$ for all $x \in \Bbb{R}^d$. Now the Fourier transform of an $L^1$ function is always continuous and bounded. So for every $x$, either $\hat{u}(x) = 0$ or else is not-zero. If it is not zero, we can cancel both sides to see that $\hat{u}(x) = 1$. So we see by continuity that $\hat{u}(x)$ is either 0 or $1$. 
If it is zero then clearly the identity $u \ast f = f$ cannot hold for all $f$. So it has to be $1$. But this contradicts the Riemann-Lebesgue lemma which says that the Fourier transform of an $L^1$ function decays at infinity. Thus no such $u$ exists.
A: For $\epsilon = 1/2>0, \exists \delta > 0$ such that
$$
\mu(A) < \delta \Rightarrow \int_A |u(z)|dz < 1/2
$$
Let $A = (-\delta/4, \delta/4)$, then if $f = \chi_{A}$
$$
1 = f(0) = u*f (0) = \int f(y)u(0-y)dy = -\int_{-A} u(z)dz
$$
But $A=-A$, so
$$
1 = \left | - \int_A u(z)dz \right | < 1/2
$$
Note: This proves it for $d=1$, but you can just tweak this argument for other $\mathbb{R}^d$
