Let $A\in B(L^2(0,1))$ be defined by $(Af(t)=tf(t)$ show that $A\ge 0$ and find $A^\frac{1}{n}$

Question

Let $A\in B(L^2(0,1))$ be defined by $(Af(t)=tf(t)$ show that $A\ge 0$ and find $A^\frac{1}{n}$

i'm proving $A\ge 0$

\begin{align*} \langle g , Ag \rangle &= \int_0^1 g(t) \overline{Ag}(t)dt \\ & = \int_0^1 t g(t) \overline{g(t)} dt = \int_0^1 t |g(t)|^2 dt>0 \end{align*}

so $A\ge 0$

how to prove remaining part

Answer 1

Notice that if $M_g:f\mapsto f\cdot g$ is the multiplication operator by the function $g\geq 0$, then for all $n\in \mathbb{N}$ we have $$M_g^n(f)=f\cdot \underbrace{ g\cdot \dots g}_{n\text{ times}}=f\cdot g^n=M_{g^n}(f)$$ and thus we also have $$M_g^{1/n}f=f\cdot g^{1/n}=M_{g^{1/n}}(f) $$ In particular in this case we have $A=M_g$ with $g(t)=t\geq 0$ for all $t\in [0,1]$, so $$A^{1/n}=M_{t^{1/n}}:f(t)\mapsto t^{1/n}f(t)$$