Multiplication Operator on $L^2[(0,1])$

Question

Let $$T: L^2([0,1]) \mapsto L^2([0,1]),\quad T(f) = tf(t) $$

I have read that for $z\in \mathbb{C}$ follows $T - z\text{id}$ is invertable iff $z \notin [0,1]$. We didnt have spectral theory yet so i am struggling to show it. Any help is appreciated

Answer 1

You have $[(T-z I)f](t)=(t-z)f(t)$.

If $z\not\in[0,1]$, then $g(t)=\frac1{t-z}$ is continuous on $[0,1]$, so in $L^2$. This shows that $Sf=gf$ defines an inverse for $T-z I$.

Conversely, if $z\in[0,1]$, then $T-z I$ is not surjective. For instance, the constants are not in its range. For if you had $(t-z)f(t)=1$ for some $f$, then $f(t)=\frac1{t-z}$, which is not in $L^2[0,1]$