Prove there is no $\delta \in C_c(\mathbb{R})$ such that for all $f \in C_c(\mathbb{R}) \ f= \delta \ast f$.
I think I have a solution to this problem, but am a bit unsure about it, as the author of this problem offers a more complicated solution. So I'd be grateful if anyone could check my work.
Consider $f_n(t)= 1-nt$ when $0 \leq t \leq \frac{1}{n}$, $f_n(t)= 1+nt$ when $-\frac{1}{n} \leq t \leq 0$ and $f_n(t)=0$ otherwise. Clearly this defines a sequence of compactly supported continuous functions. This sequence is uniformly bounded by $1$ and for each $n \in \mathbb{N}$ $supp(f_n)= [-\frac{1}{n}, \frac{1}{n}]$. Hence $1= f_n(0)= \int_{-\frac{1}{n}}^{\frac{1}{n}} \delta(-y) f_n(y) dy \leq \frac{2}{n} ||\delta||_{\infty}$. So any such $\delta$ has to be unbounded, so can't belong to $C_c$.
( In case it is of interest: The author considers the same sequence but with spike $(0,n)$ rather than spike $(0,1)$.)