Example of Closed Linear Operator whose range is not closed

Question

I had found example of Linear operator whose range is not closed.

But I am intersted in finding exmple of closed operator (which has closed graph) but do not have closed range.

Please can anyone give me hint to find such example

Thanks a lot

Answer 1

If $(a_n)\subset\mathbb C$ with $a_n\neq 0$ for all $n\in\mathbb N$ is unbounded such that also $(a_n^{-1})$ is unbounded, then the operator $T$ in $\ell^2$, defined by $(Tx)_n = a_nx_n$ with $\text{dom}T = \{x\in\ell^2 : (a_nx_n)\in\ell^2\}$ is closed and has a non-closed range. This is pretty easy to see: the operator is a multiplication operator, so it's closed and densely defined. Its inverse is given by $(T^{-1}x)_n = a_n^{-1}x_n$ with the appropriate domain. Its domain is dense (also a multiplication operator) and coincides with the range of $T$.

Answer 2

Let $Lf = \int_0^x f(t)dt$ be defined on $C[0,1]$ with the sup norm. $L : C[0,1]\rightarrow C[0,1]$ is bounded and, hence, has a closed graph. $L$ in injective because $Lf=0$ for some $f\in C[0,1]$ implies that $f=\frac{d}{dx}Lf=0$.

If $\mathcal{R}(L)$ were closed in $C[0,1]$, then the closed graph theorem would imply that $L$ would have a bounded inverse $$L^{-1} : \mathcal{R}(L)\subset C[0,1]\rightarrow C[0,1].$$ Of course the inverse must be $L^{-1}=\frac{d}{dx}$. It is easy to verify that $\frac{d}{dx}$ is not bounded on the bounded set $\{ x,x^2,x^3,\cdots\}\subset \mathcal{R}(L)$. So $\mathcal{R}(L)$ cannot be closed.

Answer 3

Any compact operator of infinite rank is an example.

Some examples:

• the inclusions $\ell^p\to \ell^q$, $L^q[0,1]\to L^p[0,1]$ with $p<q$
• integral operators on $C[0,1]$ of the form $f\mapsto (x\mapsto \int_0^x f(x)g(x)\,\mathrm{d}x)$ for some nonzero $g\in C[0,1]$,
• any map $\ell^p\to \ell^p$ defined by $x\mapsto xy$, where $y$ converges to $0$ (and has infinite support).