Let $X=[0,1)$ the 1-d torus. Given a bounded positive function $w\colon X\to\mathbb{R}$ with unit integral (I mean $w\geq 0$, $w\in L^\infty(X)$ and $\int_X w\; dx=1$), define
\begin{align*} T_{w} \colon L^\infty(X) &\to L^\infty(X)\\ f &\mapsto T_wf(x)=\int_0^1 f(y)w(x-y)dy. \end{align*}
I need to determine conditions for $w$ under which the transformation $T_w$ is injective. I know a counterexample for $w$ continuous with compact support, so those conditions are not enough.
I have read a lot of material about convolutions and integral transformation, but it is mostly about $L^2(\mathbb{R})$, and it haven't been easy to adapt.
For example, I know that if $w=\frac{1}{(b-a)}\chi_{[a,b]} $ $(0\leq a\leq b\leq 1)$, $T_w$ is not injective because if we take $$f(x) = \begin{cases} 1 &\mbox{if } x < \frac{b-a}{2} \\
-1 & \mbox{if } x\geq\frac{b-a}{2} \end{cases} \pmod{b-a}.$$ then $T_wf=0$. But if we extend $T_w$ to $L^2(\mathbb{R})$ it happens to be injective (Another proof of the iniectivity of a linear operator).
Working on this problem, I stumbled upon this type of equation $$\int_0^1 f(y)w(x-y)dy=0\quad\forall x\in X$$ from which I would like to conclude that $f\equiv 0$. Any reference would be helpful.