Can $f*g = f+g$ for $f$ and $g$ compactly supported? Let $f$ and $g$ be  continuous, compactly-supported functions $\mathbb{R} \to \mathbb{C}$. Can it happen that $f*g  = f+g$? Here, $f*g$ denotes the convolution
$$(f*g)(s) = \int_\mathbb{R} f(t) g(s-t) \ dt.$$
Edit: Discount the solution $f=g=0$. 

My thoughts: My feeling is that the answer is "no". We are asking for a certain identity to hold in the dense subalgebra $C_c(\mathbb{R})$ of the Banach algebra $L^1(\mathbb{R})$. If there is a pair such that this happens, then without loss of generality, $\|f\|_1 < 1$. But then, the series
$$h = f + f*f + f*f*f + \ldots$$
converges in $L^1(\mathbb{R})$. It is not hard to see that $g = -h$ has $f*g = f+g$. I think there can only be one $g$ satisfying $f+g = f*g$, so this is the one. But, my sense is that the interated convolutions in the sum are making the support spread out. Could somebody please clarify this for me?
 A: For a continuous function $f$, write $M(f)=\max \operatorname{supp}f$ and $m(f)=\min \operatorname{supp}f$. Suppose $f*g=f+g$. On one hand, 
$$M(f+g)\le \max(M(f),M(g)),\quad m(f+g)\ge \min(m(f),m(g)) \tag{1}$$
On the other hand, the  Titchmarsh convolution theorem says that $$M(f*g)=M(f)+M(g),\quad m(f*g)=m(f)+m(g)\tag{2}$$ 
Since $a+b=\max(a,b)+\min(a,b)$ for any reals $a,b$, it follows that 
$$\min(M(f),M(g))\le 0,\quad \max(m(f),m(g))\ge 0 \tag{3}$$ 
So, the supports have disjoint interiors. Since the support of $f+g$ is the union of supports, equality holds in both places in (1), and consequently in (3).
Without loss of generality, $M(f)=0$ and $m(g)=0$. Fix $\epsilon>0$ such that $m(f)+\epsilon<0$ and $\int_{0}^{\epsilon} |g(t)|\,dt\le 1/2$. Let $\mu = \max_{[m(f),m(f)+\epsilon]}|f|$, which is positive. Pick $x\in [m(f),m(f)+\epsilon]$ such that $|f(x)|=\mu$. We have
$$ 
\mu=|f(x)|=|(f+g)(x)| = |(f*g)(x)| = \left|\int_0^\epsilon  f(x-t)g(t)\,dt \right| \le 
\mu  \int_{0}^{\epsilon} |g(t)|\,dt \le \frac{\mu}{2}
 $$
A contradiction.
