9
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ open ≠ not closed :-) $\endgroup$ – Martin Mar 28 '13 at 14:31
  • 1
    $\begingroup$ 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. $\endgroup$ – Nate Eldredge Mar 28 '13 at 14:49
11
$\begingroup$

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 $x^n$).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.