# A convolution inequality

1. Jun 1, 2012

### amirmath

Dear friends,

I am interesting to find some functions g satisfying the following convolution inequality

(g$\ast$v)(t)$\leq$v(t)

for any positive function v$\in$L$^{1}$[0,T] and * denotes the convolution between g and v.

2. Jun 4, 2012

### Undecided Guy

The way you've worded the statement, it's not possible. Suppose that $v \in L^1[0, T]$ satisfies 0 < (g*v)(0) < v(0). Let v'(t) = v(t) for all t other than 0 and v'(0) = .5(g*v)(0). Then v = v' in the sense of L1, but (g*v)(0) > v'(0).