# Closed unbounded operator with domain not closed

I am looking for an example for further understanding of the Closed Graph Theorem:

Let $X,Y$ be Banach spaces and $T:X\to Y$ closed (i.e. the graph of $T$ is closed in $X\times Y$). Then if $\mathcal{D}(T)$ is closed in $X$, $T$ is bounded.

I am looking for an unbounded operator whose graph $\mathcal{G}(T)$ is closed in $X\times Y$ and whose domain $\mathcal{D}(T)$ is not closed in $X$, to clearify the necessity of $\mathcal{D}(T)$ being closed.

This question arose due to the definition of the norm of a graph $\lVert (x,Tx)\rVert:=\lVert x\rVert+\lVert Tx\rVert$, where I thought the following statement would be true: [$\mathcal{G}(T)$ closed $\Rightarrow\mathcal{D}(T)$ closed] which in general is false.

• open ≠ not closed :-) – Martin Mar 28 '13 at 14:31
• It's not a linear operator, but an example that can help to understand this phenomenon is the function $f(x) = 1/x$ considered as a function on $\mathbb{R}$ whose domain is $\mathbb{R} \setminus \{0\}$. The graph is closed but the domain is not. – Nate Eldredge Mar 28 '13 at 14:49

## 1 Answer

One of the simplest examples is the following:

Let $D(T) = C^1[0,1]$ be the space of of continuously differentiable functions (one-sided derivative at the end points) and let $X = C^0[0,1]$ be equipped with the norm $\lVert f \rVert_\infty = \sup_{x \in [0,1]} \lvert f(x)\rvert$.

Then the derivative $T = \frac d{dx}$ is an operator $D(T) \subset C^0[0,1] \to C^0[0,1]$ and $T$ is easily checked to have closed graph. Remember: if $f_n \to f$ pointwise and $f_{n}' \to g$ uniformly then $g$ is the derivative of $f$ by an application of the fundamental theorem of calculus.

However, $D(T)$ is not closed (it is dense but not all of $C^0[0,1]$) and $T$ is not bounded (consider $x^n$).