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. 



Injectivity of a class of integral operators
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


$$ \eqalign{\int\_0^1 f(xy) g(x)\; dx &= \frac{1}{3} \int\_0^1 f(xy)\; dx + \frac{4}{3} \int\_0^{1/2}f(xy)\; dx\cr
&= \frac{1}{3} \int\_0^1 f(xy)\; dx - \frac{1}{3} \int\_0^1 f(xy)\; dx = 0}$$


