How to show: $A_y$ has no eigenvectors if $y$ is not constant on any subinterval of $[0,1]$

Question

Let $y\in C[0,1]$ and $A_y : C[0,1]\rightarrow C[0,1]: x\mapsto xy$

How to show:

$A_y$ has no eigenvectors if $y$ is not constant on any subinterval of $[0,1]$.

Could you please help.

Answer 1

The phrase "$\, y$ is not constant on any subinterval of $[0,1]\,$" has to be interpreted as follows: Given any nondegenerate subinterval $J\subset[0,1]$ there are two points $t_1$, $t_2\in J$ with $y(t_1)\ne y(t_2)$.

Assume that $A_y$ has an eigenvector $x\ne0$. Then there is an open interval $J\subset[0,1]$ with $x(t)\ne0$ for all $t\in J$, as well as a number $\lambda$ such that $$\bigl(A_yx(t)=\bigr)\quad y(t)\>x(t)=\lambda x(t)\qquad \forall \ t\in[0,1]\ .$$ In particular $$\bigl(y(t)-\lambda\bigr)\>x(t)=0\qquad\forall \ t\in J\ ,$$ so that $y(t)=\lambda$ for all $t\in J$. It follows that the function $y$ in question is constant on the interval $J$.

Answer 2

Hint: If $x$ is a non-zero function from $C[0,1]$, then it is non-zero on some subinterval. Yet $yx=\lambda x$ for some scalar $\lambda$.