2

I am trying to understand the Closed Graph Theorem and so I would like to see an example of an unbounded linear operator $\hat{T}: \mathscr{D}(\hat{T})\rightarrow Y$ where $\mathscr{D}(\hat{T})\subset X$ is a closed domain and $X$ and $Y$ are Banach spaces. In the Closed Graph theorem the domain $\mathscr{D}(\hat{T})$ and the graph $\mathscr{G}(\hat{T})$ must be closed and so I would like to see the necessity of $\mathscr{G}(\hat{T})$ being closed. The main doubt I have is if an unbounded linear operator is allowed to be defined in a closed domain.

Thanks.

glS
  • 7,963
Fabio Paolini
  • 23

2 Answers2

5

The closed graph theorem is precisely the reason that it is "hard" to construct an everywhere defined unbounded operator (say acting between Banach spaces). I heared that there is a serious theorem which states that you cannot construct unbounded $T:E \to E$ where $E$ is Banach without axiom of choice (but I don't know where you can find the proof of this fact): please note that in the solution given by Akatsuki you need axiom of choice (since it is equivalent to the existence of linear basis). So in practise I believe that every unbounded operator which you will meet will be not everywhere defined (it's domain would not be closed).

truebaran
  • 4,848
  • What is the relationship between not being everywhere defined and the domain not being closed? – Abel Jun 22 '22 at 17:10
  • 1
    Since every closed subspace of a Banach space is again a Banach space and in general we are interested in operators acting between Banach spaces, there is no gain in considering operators $T$ viewed as operators between $X$ and $Y$ but defined only on $X_0$ which is closed in $X$ (and not being the whole $X$). What is interesting is to investigate operators which are densely defined – truebaran Jun 24 '22 at 15:29
2

Define $T(e_k)=kf$, where $\{e_k\}$ are linear independent in $X$ with $|e_k|=1$, and $f$ is a fixed nonzero vector in $Y$. For other $e$ that cannot represented by $\{e_k\}$, define $T(e)=0$. Then $T$ is defined on the whole space $X$, and it is obviously unbounded.

Akatsuki
  • 3,390
  • Akatsuki, suppose the element $x\in X$ like $x=\frac{\sqrt{6}}{\pi}\sum \frac{e_k}{k}$. Its norm is $|x|^2 = \frac{6}{\pi} \sum \frac{1}{k^2}=1$, but $T x = \frac{\sqrt{6}}{\pi}\sum \frac{T e_k}{k} = \frac{\sqrt{6}}{\pi}\sum f$, so I suppose $x\notin\mathscr{D}(T)$. Is it not true? If I am right, then $\mathscr{D}(T)$ is not closed since we have a cauchy sequence $x_n=\frac{\sqrt{6}}{\pi}\sum_{k=1}^{n} \frac{e_k}{k}$ that goes to $x$, I mean $x$ is an accumulation point of $\mathscr{D}(T)$ – Fabio Paolini Jan 23 '16 at 02:58
  • @FabioPaolini Here we should not regard $x$ as a linear combination of ${e_k}$ because when we say linear combination, we always mean finite linear combination.(Just think about the linear structure, do not consider any topology, hence no convergence) So we actually have $Tx=0$. – Akatsuki Jan 23 '16 at 12:31
  • @ Akatsuki, you say that the vector $x$ I have constructed should be thought as an example of an element like the $e$ of your answer, that cannot be represented by ${e_k}$ ? – Fabio Paolini Jan 24 '16 at 00:21
  • @FabioPaolini Yes. – Akatsuki Jan 24 '16 at 07:04
  • could you explain it in more details? I think it is hard to understand, because if ${e_k}$ is a linear independent base of a subset $X'\subset X$ and ${e_k}+e$ spam $X$ as a linear independet base then the vector $ae_k+\sum_k\alpha_k e_k$ (where $a$ and $\alpha_k$ are numbers) should be $0$ just if $\alpha_k$ and $a$ are zero. If $T$ is a linear operator then the relation $Te=T \sum_k (\alpha_k e_k)=\sum_k(\alpha_k Te_k)$ should be valid, at least if it is valid to interchange $T$ with $\sum$. If it is not true then it seems that the behaviour of $T$ depends on the base adopted. – Fabio Paolini Jan 24 '16 at 13:49
  • what can be said if we take the example of the differential operator $T:\mathscr{D}(T)\rightarrow X$ where $X=C[0,1]$, $\mathscr{D}(T) = C^1[0,1] $ with $Tx = \dot{x}$. It is a closed operator but not bounded. We can take a limit $x_n\rightarrow x$ where $x_n\in\mathscr{D}(T)$ but $x\notin\mathscr{D}(T)$. Can we apply the same reasoning here? I mean, define $Te = 0$ always $e\notin C^1[0,1]$? Since $T$ is unbounded, will it loose its closed property? Because if it is closed and defined in a closed domain it should be bounded. In some sense that was my original doubt. – Fabio Paolini Jan 25 '16 at 02:16
  • @FabioPaolini I made a little mistake in my last comment (so I have deleted it). What I want to say is you cannot have $Te=\Sigma T e_k$. In fact it is just the interchange of $T$ with $\Sigma$ that is invalid. – Akatsuki Jan 26 '16 at 08:15
  • And as for your question of differential operator $T$, it will definitely not be closed since the closed property implies bounded property, which you already know that $T$ does not have. You can also check it directly: let $f_n$ be some elements in $C[0,1]$ but not in $C^1[0,1]$, and let $f_n$ tend to some $f$ in $C[0,1]$ where $f$ is in $C^1[0,1]$ with $f' \neq 0$. Then $T(f_n)=0$ for all $n$ but $T(f)\ne 0$. Hence $T$ is not closed. – Akatsuki Jan 26 '16 at 08:16
  • @Akatsuki your defined linear operator is not well defined because Take $x=\sum a_n e_n$ and $y=\sum b_n e_n$ such that $x\ne y$ but $a_n=b_n \forall n\ge 10$ Then by your def T(x)=T(y)=0 implies T(x-y)=0 but it should't be. – Biplab Jan 12 '23 at 05:50