$X$ is a normed vector space and $T:X\to X$ is a function that has a closed graph, does $T$ map closed sets to closed sets?

Question

Here is the question:

Suppose that $X$ is a normed vector space and $T:X\to X$ is a function that has a closed graph. Is it true that $T$ maps closed sets to closed sets? Is it true if $T$ is linear?

So I think I am over thinking this one, but I am having some issues getting started with this. I am a bit unsure if I should be working directly with the definition of a graph that I have been given:

When $X$ and $Y$ are normed linear spaces and $T:X\to Y$ is a linear map, the graph of $T$ is $\{(x,Tx):x\in X\}$. It should be noted that the graph of $T$ is a subset of $X\times Y$.

Or if this can be worked out from the Closed Graph Theorem:

If $X$ and $Y$ are Banach spaces and $T:X\to Y$ is linear, then $T$ is bounded if and only if $graph(T)$ is closed in $X\times Y$.

I am thrown off though, as I do not have Banach spaces to work with, and in the first case $T$ is not linear, and therefore the assumptions of the Closed Graph Theorem are not met...

Any guidance as to where to begin, would be appreciated.

$\Bbb R$ is a normed vector space; for the general question consider $T(x)=e^x$. — Brian M. Scott, Oct 23 '14 at 21:16
So in your example, $T:\mathbb{R}\to\mathbb{R}$ and $T(x)=e^x$. If I take a closed subset of the reals, $[a,b]$, then this gives me the closed subset $[e^a,e^b]$... but I don't see what this proves? @BrianM.Scott — user3784030, Oct 23 '14 at 21:47
Sorry if this seems extremely basic, but how do we know that $\mathbb{R}$ is a closed subset? @BrianM.Scott — user3784030, Oct 23 '14 at 21:54
Ah, okay - and to follow up, take a sequence $x_n\in\mathbb{R}$ such that $T(x_n)\to 0$, however $0$ is not an element of $T(x)$, therefore $T$ does not map closed sets to closed sets - would this be correct? @BrianM.Scott — user3784030, Oct 23 '14 at 22:08
What about the case that $T$ is linear? I am thinking that under this assumption, it would be true that $T$ would map closed sets to closed sets... @BrianM.Scott — user3784030, Oct 23 '14 at 22:27
Not necessarily: @T.A.E.’s answer gives a counterexample, though it’s a fairly sophisticated one. — Brian M. Scott, Oct 23 '14 at 22:37
Would you have any suggestions on a less complex counterexample? @BrianM.Scott — user3784030, Oct 23 '14 at 23:32
I’m afraid not. I suspect that any counterexample needs to be quite non-trivial, but it’s a bit outside my usual range of interests, so I don’t have a good feel for it. — Brian M. Scott, Oct 23 '14 at 23:36

score 2 · Answer 1 · answered Oct 23 '14 at 21:55

Let $X=C[0,1]$ and define $T : C[0,1]\rightarrow C[0,1]$ by $Tf = \int_{0}^{x}f(t)\,dt$. $X$ is a Banach space, and $T$ has a closed graph because $T$ is a bounded linear operator. And $C[0,1]$ is closed, but $T C[0,1]$ is not closed in $C[0,1]$.

To see that $TC[0,1]$ is not closed, first notice that $T$ is injective because $Tf=0$ implies $f=0$ by the Fundamental Theorem of Calculus. So, if the range of $T$ were closed, then it would have a bounded inverse by the Closed Graph Theorem, which would give the existence of a constant $m > 0$ such that $\|Tf\| \ge m\|f\|$ for all $f \in C[0,1]$. To see that such a constant $m > 0$ cannot exist, define $$ f_{n}(t) = \left\{\begin{array}{cc} 1-nt, & 0 \le t \le 1/n \\ 0, & 1/n \le t \le 1 \end{array} \right.. $$ Then $\|f_{n}\|=1$ for all $n=1,2,3,\cdots$, and $\|Tf_{n}\|=1/(2n)$.

Direct proof: Consider f_n(t)=1 for [0,1/2], 1-nt for (0,1/2+1/n] and 0 for (1/2+1/n,1]. Tf_n->g, where g(t)=min(t,1/2) in C[0,1]. By the Fundamental Theorem of Calculus, if g is in TC[0,1], dg/dt exists and is in C[0,1]. However, dg/dt doesn't exist at t=1/2. Thus, g is not in TC[0,1]. — YouJiacheng, Jan 10 '24 at 20:42

Robert Israel · Answer 2 · 2014-10-23T22:54:38.613

1

Consider $T:\mathbb R^2 \to \mathbb R^2$ defined by $T(x,y) = (x + \sqrt{2} y,0)$, and the closed set $\mathbb Z^2$.

edited Oct 23 '14 at 22:54

answered Oct 23 '14 at 21:27

Robert Israel

470,583

Shouldn't $T$ be from $\mathbb{R}\to\mathbb{R}$, as it is given that $T:X\to X$? @Robert – user3784030 Oct 23 '14 at 21:38
OK, make the second coordinate $0$ (I edited). $\mathbb R \to \mathbb R$ is just a bit too trivial. – Robert Israel Oct 23 '14 at 22:54
I am not quite sure how this example works... it seems to me that every infinite sequence diverges to positive or negative infinity. @Robert – user3784030 Oct 23 '14 at 23:12
$T(\mathbb Z^2)$ is a dense subgroup of $\mathbb R \times {0}$: see http://en.wikipedia.org/wiki/Equidistribution_theorem – Robert Israel Oct 24 '14 at 00:03

$X$ is a normed vector space and $T:X\to X$ is a function that has a closed graph, does $T$ map closed sets to closed sets?

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