I need to find the spectrum of an operator.

Question

I need to prove that the spectrum of operator A in $L_{2}[0,1]$ is [-1;1] where $Ax(t) = \sin(\frac{1}{t})x(t)$ if $ t > 0$ and$ Ax(0) = 0$ Elementary - norm of $A$ is less or equal to $1$. Hence, all $\lambda$ from spectrum is such that $|\lambda| \leq$ 1. $g(t)$ := $\sin(\frac{1}{t})$ if $t > 0$ and $0$ if $4t = 0$. If there exists $y(t)$ such that $(A -\lambda I)x(t) = y(t)$ then if $y_{0} = 1$ almost everywhere then $x(t) = \frac{1}{g(t) - \lambda}$ $\notin L_{2}[0,1]$ if $\lambda \in [-1;1]$. So, this means that $1 \notin$ Im(A-$\lambda I$) which means that Im(A - $\lambda I$) $\neq$ $L_{2}[0,1]$ and $\lambda$ is in spectrum. But i can't seem to proof the fact that i used: $x(t) = \frac{1}{g(t) - \lambda}$ $\notin L_{2}[0,1]$ if $\lambda \in [-1;1]$ And that's where i need help. Why is $x(t) = \frac{1}{g(t) - \lambda} \notin L_{2}[0,1]$ where $\lambda $ in $[-1;1]$

Answer 1

Hint: First make the substitution $u=\frac 1 t$ in $\int |x(t)|^{2}dt$. Write $\lambda$ as $\sin y$. Use the fact that $\sin u -\sin y=2\sin (\frac {u-y} 2) \cos (\frac {u+y} 2)$. In an interval around $y$ the function $\frac 1 {\sin (\frac {u-y} 2)}$ behaves like $\frac 2 {u-y}$ and the square of this is not integrable around $y$.