Let the continuous spectrum of a densely defined linear operator $L$ over a Separable Hilbert space, be defined as the set of all $\lambda \in \mathbb C$ such that:
(i) $L-\lambda$ is injective,
(ii) The closure of the range of $L-\lambda$ is the whole Hilbert space,
(iii) The inverse of $L-\lambda$ (with range of $L-\lambda$ as the domain) is not bounded.
(This definition is taken from the book "Principles and Techniques of Applied Mathematics" by Bernard Friedman.)
Then, prove or provide counter-example to the following statement: If $\lambda$ is in the continuous spectrum of $L$, then the range of $L-\lambda$ can not be closed.
If the above statement is not true, is it true under the additional assumption that $L$ is a bounded linear operator?
If $\lambda$ belongs to the continuous spectrum then the image of $\lambda I-T$ is a dense proper subset of the space (see 1). Thus it cannot be closed by definition.
What follows is an example where $0$ belongs to the essential spectrum, the image is equal to the whole space and both $T$ and $T^{-1}$ are unbounded. Let $\mathcal{B}$ be a Hamel basis in $\mathcal{H}$ and $\mathcal{B}_0=\{b_n\}_{n=1}^\infty$ its countable subset. Define an operator $T:\mathcal{H}\to \mathcal{H}$ by the formula $$Tb=\begin{cases} {1\over n}b_{n} & b=b_n\\ b & b\notin \mathcal{B}_0\end{cases} $$ and extend linearly to the whole space. The inverse operator $S$ is defined by the formula $$Sb=\begin{cases} nb_{n} & b=b_n\\ b & b\notin \mathcal{B}_0\end{cases} $$ Consider $\lambda =0.$ The operator $T$ is injective and $T(\mathcal{H})=\mathcal{H}$ The range of $T$ is thus closed. The inverse operator $S$ is unbounded (as well as $T$ is unbounded).