Find the pseudoinverse $T^+$ of $$T: L^2(a,b) \to L^2(a,b), \ f \mapsto g f,$$ where $0 \ne g \in \mathcal{C}(a,b)$. First find $\mathcal{N}(T)$ and $\mathcal{R}(T)$ (which denote kernel and range, respectively) and their orthogonal complements. Then find $\widetilde{T}^{-1}$, where $\widetilde{T}: \mathcal{N}(T)^{\perp} \to \mathcal{R}(T)$ is the restriction of $T$ to $\mathcal{N}(T)^{\perp}$.
First I noticed that $T$ is unbounded, as $g f$ need not be in $L^2(a,b)$, i.e. pick $(a,b) := (0,1)$, $f := g := x^{-1/4}$. Then $$ \| f \|_{L^2(0,1)} = \| g \|_{L^2(0,1)} = \int_0^1 x^{-1/2} dx = 2 $$ but $$ \| f g \|_{L^2(0,1)} = \int_{0}^{1} |x^{-1/2}|^2 dx = \int_0^1 x^{-1} dx = \infty. $$ Is this correct? I am struggling with this as normally with unbounded operators we are given a domain $\text{dom}(T)$.
Let $G := \{ x \in (a,b): g(x) \ne 0\}$ I think that $\mathcal{N}(T) = \{ f \in L^2(a,b): f|_{G} = 0\}$. Then we would have \begin{align} \mathcal{N}(T)^{\perp} & = \{ h \in L^2(a,b): \int_a^b h(x) j(x) dx = 0 \ \forall j \in \mathcal{N}(T)\} \\ & = \{ h \in L^2(a,b): h(x) j(x) = 0 \ \forall j \in \mathcal{N}(T)\} \\ & = \{ h \in L^2(a,b): h|_{G^{\complement}} = 0\}. \end{align} I only found $\mathcal{R}(T) = \{ f g: f \in L^2(a,b)\}$ and therefore \begin{align} \mathcal{R}(T)^{\perp} & = \{ h \in L^2(a,b): \int_{a}^{b} h(x) f(x) g(x) dx = 0: f g \in L^2(a,b) \} \\ & = \{ h \in L^2(a,b): h|_{G} = 0\} = \mathcal{N}(T). \end{align} Is this correct?
We would therefore have $$ \widetilde{T}: \{ f \in L^2(a,b): f|_{{G^{\complement}}} = 0\} \to \{ f g: f \in L^2(a,b)\}, \ f \mapsto f g. $$ Thus $\widetilde{T}^{-1}$ is given by $$ \widetilde{T}^{-1}: \{ f g: f \in L^2(a,b)\} \to \{ f \in L^2(a,b): f|_{{G^{\complement}}} = 0\}, f \mapsto f / g $$ and therefore $$ T^+: \mathcal{R}(T) \oplus \mathcal{N}(T) \to \mathcal{N}^{\perp}, f \mapsto f / g. $$