# Injectivity of a class of integral operators

Given a probability measure $$\mu$$ on the interval $$[0,1]$$, the linear operator $$T_\mu \! f(y) := \int_0^1 f(yx) \, d\mu(x)$$ takes the space of continuous functions $$f: [0, \infty) \rightarrow \mathbb{R}$$ such that $$f(0) = 0$$ to itself.

Assuming that $$\mu$$ is not the delta function at the origin, is this operator injective?

For some measures this is really easy to show. For instance, if $$d\mu(x) = g(x)|dx|$$ with $$g$$ a homogeneous function.

Remarks.

1. $$T_\mu$$ acts like an invertible diagonal matrix on polynomials of a given degree.

2. If a function $$g$$ is in the kernel of $$T_\mu$$, then so are the functions $$x \mapsto g(\lambda x)$$ $$(0 \leq \lambda < \infty)$$.

3. Although the only requirement on $$\mu$$ is that it is not the delta at zero, in the applications I have in mind $$\mu$$ is positive almost everywhere.

4. In the original question (addressed in Robert Israel's answer) the domain of the functions was $$[0,1]$$ instead of $$[0,\infty)$$. If there is an advantage in dealing with the Banach space of continuous function on $$[0,1]$$ (I thought there might be), then just add the hypothesis that $$\mu$$ is positive almost everywhere.

At first I thought this was going to be easy, but I've been blocked for a while. It may still be easy though.

## 1 Answer

If the support of $$\mu$$ is contained in $$[0, b]$$ for some $$b \in (0,1)$$, then $$T_\mu f = 0$$ for any $$f$$ that is $$0$$ on $$[0,b]$$, so it is not injective.

EDIT: Another example, where the support of $$\mu$$ is all of $$[0,1]$$: $$d\mu(x) = g(x)\; dx$$ where $$g(x) = 5/3$$ for $$0 \le x < 1/2$$, $$1/3$$ for $$1/2 \le x \le 1$$. Let $$f(x) = x \sin(\pi \log_2(x))$$, and note that $$f(2x) = - 2 f(x)$$. Then

\eqalign{\int_0^1 f(xy) g(x)\; dx &= \frac{1}{3} \int_0^1 f(xy)\; dx + \frac{4}{3} \int_0^{1/2}f(xy)\; dx\cr &= \frac{1}{3} \int_0^1 f(xy)\; dx - \frac{1}{3} \int_0^1 f(xy)\; dx = 0}

• Yes, I missed that when I reformulated the problem from $[0,\infty)$ to $[0,1]$ to make it "simpler". I'll edit the OP. Thanks !! – alvarezpaiva Jul 19 at 12:44
• That eases my mind that when there is injectivity it is not for general "geometric" reasons. Thank you very much! – alvarezpaiva Jul 19 at 17:31