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- Problem Statement
- Show no eigenvalues exist for the operator ##Af(x) = xf(x)## where ##A:C[0,1]\to C[0,1]##.

- Relevant Equations
- Nothing comes to mind.

Eigenvalues ##\lambda## for some operator ##A## satisfy ##A f(x) = \lambda f(x)##. Then

$$

Af(x) = \lambda f(x) \implies\\

xf(x) = \lambda f(x)\implies\\

(\lambda-x)f(x) = 0.$$

How do I then show that no eigenvalues exist? Seems obvious one doesn't exists since ##\lambda-x \neq 0## for all ##x\in [0,1]##.

$$

Af(x) = \lambda f(x) \implies\\

xf(x) = \lambda f(x)\implies\\

(\lambda-x)f(x) = 0.$$

How do I then show that no eigenvalues exist? Seems obvious one doesn't exists since ##\lambda-x \neq 0## for all ##x\in [0,1]##.