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.**

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

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

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.

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.