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.


  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.


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

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

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    $\begingroup$ 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 !! $\endgroup$ – alvarezpaiva Jul 19 at 12:44
  • $\begingroup$ That eases my mind that when there is injectivity it is not for general "geometric" reasons. Thank you very much! $\endgroup$ – alvarezpaiva Jul 19 at 17:31

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