Multiplication Operator on L^2(R) that has no eigenvalues on the unit circle

Question

Statement: We define the multiplication operator $T_n$ on $L^2(\mathbb{R})$ by $(T_nf)(x)=e^{\frac{iq(x)}{n}}f(x), \ x\in\mathbb{R}, \ f\in L^2{(\mathbb{R})}$, where $q:\mathbb{R}\rightarrow [0,1]$ is a strict monotone function. I have to prove, that $T_n$ has no eigenvalues on the unit circle. Attempt: We need to show, that $\not\exists\ \lambda\in \mathbb{C}$ with $|\lambda|=1$ with $T_nf=\lambda f$. Equivalently, we show, that $f(x)(\lambda-e^{\frac{iq(x)}{n}})\neq0$ for $f(x)\neq 0, \ x\in \mathbb{R}, \ \lambda\in S^1$. I know, that $T_n$ is unitary and $\lambda$ has the form $\lambda=e^{ik}$ for some $k\in\mathbb{R}$. How do I show: $\lambda-e^{\frac{iq(x)}{n}}\neq 0?$

Answer 1

Note that it's quite possible (unavoidable, I think) that there will be at least one $x$ where we can find $\lambda$ such that $\lambda-e^{\frac{iq(x)}{n}}=0$. But on $L^2$, for $\lambda$ to be an eigenvector, that equation needs to hold for all $x\in\mathbb{R}$ (or at least the measure of all $x$ where it doesn't hold must be 0), not just a single value.

So suppose for some $x_0$, $f(x_0)(\lambda-e^{\frac{iq(x_0)}{n}})=0$ and $f(x_0)\neq 0$. Thus we have that $\lambda=e^{\frac{iq(x_0)}{n}}$. Now take any other $x_1\neq x_0$ such that $f(x_1)\neq 0$. What happens?