Let $\sigma>0$, $$\phi(x):=\frac1{\sigma\sqrt{2\pi}}e^{-\frac12\left|\frac x\sigma\right|^2}\;\;\;\text{for }x\in\mathbb R,$$ $\lambda$ denote the Lebesgue measure on $\mathcal B(\mathbb R)$ and $$\mathcal N(x,B):=\int_B\lambda({\rm d}y)\varphi(y-x)\;\;\;\text{for }(x,B)\in\mathbb R\times\mathcal B(\mathbb R).$$ If $$\iota:\mathbb R\to[0,\infty)\;,\;\;\;x\mapsto x-\lfloor x\rfloor,$$ then the wrapped normal distribution kernel on $[0,1)$ with variance $\sigma^2$ is defined by $$\mathcal W(x,\;\cdot\;):=\mathcal N(x,\;\cdot\;)\circ\iota^{-1}\;\;\;\text{for }x\in[0,1).$$ If $$\psi(x):=\sum_{k\in\mathbb Z}\phi(x+k)\;\;\;\text{for }x\in\mathbb R,$$ we can show that $$\mathcal W(x,B)=\int_B\lambda({\rm d}y)\psi(y-x)\;\;\;\text{for all }(x,B)\in[0,1)\times\mathcal B([0,1))\tag1.$$
Given a bounded Borel measurable $f:[0,1)\to[0,\infty)$, can we show that $$[0,1)\ni x\mapsto\int_{[0,\:1)}\lambda({\rm d}y)\frac{f(y)}{\psi(y-x)}\tag2$$ is constant?
Intuitively, I think that this claim could hold by the circular nature of $\psi$, but I don't know an easy way to prove or disprove this.