# range of weighted shift operator

Consider the weighted shift operator on $\ell^2$ space defined by $T(x_0, x_1, x_2, ...) = (0, x_0, 2x_1, 3x_2, 4x_3, ...)$ with domain $$\mathcal{D}(T) = \{(x_n) \in \ell^2 : \sum_{n=0}^{\infty}|(n+1)x_n|^2 < \infty\}.$$

Then the point spectrum of $T$ and $T^*$ are $\sigma_p(T) = \emptyset$ and $\sigma_p(T^*) = \mathbb{C}$. What about the approximate spectrum? Is the range of $T - \lambda I$ closed for $\lambda \in \mathbb{C}$? I know that the range $\overline{\mathcal{R}(T - \lambda I)} \neq \ell^2$ for any complex number $\lambda$ because $\lambda \in \sigma_p(T^*).$ But what about the closedness of the range.

Let $H$ be the subspace spanned by the orthonormal set $\{1,e^{ix},e^{2ix},\cdots\}$ in $L^2[0,2\pi]$ with the inner product $$(f,g)=\frac{1}{2\pi}\int_{0}^{2\pi}f(t)\overline{g(t)}dt$$ Consider the operator $$T = -i\frac{d}{dx}(e^{ix}f(x)).$$ Then $T(e^{inx})=(n+1)e^{i(n+1)x}$ for $n=0,1,2,3,\cdots$. So this is the same as your $T$ because $(x_0,x_1,x_2,\cdots)$ is mapped to $(0,x_0,2x_1,3x_2,\cdots)$ if $f = x_0 + x_1e^{ix}+x_2e^{2ix}+\cdots$.
To solve $(T-\lambda I)f = g$, consider the equation $$-i(e^{ix}f)'-\lambda f=g \\ (e^{ix}f)'+i\lambda e^{-ix}(e^{ix}f)=ig$$ You can always solve the ODE with an integrating factor, and the only real issue that keeps the ODE solution from being an actual one is whether or not $f(0)=f(2\pi)$. The integrating factor is any non-zero multiple of $$h(x)=\exp\left\{\int_{0}^{x}i\lambda e^{-it}dt\right\}.$$ Note that $h(2\pi)=h(0)$ holds for all $\lambda$. The new ODE becomes $$\frac{d}{dx}(h(x)e^{ix}f(x))=ih(x)g(x) \\ h(x)e^{ix}f(x)-h(0)f(0) = \int_{0}^{x}ih(t)g(t)dt$$ Because $h(0)=h(2\pi)$, this gives an actual solution where $f(0)=f(2\pi)$ iff $$\int_{0}^{2\pi}h(t)g(t)dt=0.$$ So it appears to me that the range is of co-dimension 1, making it closed. This should be confirmed, and you must show that $f$ lies in the span of the required basis.