0
$\begingroup$

My professor asked us for an example of a Bounded Operator with whose range is not closed, after some attempts I thought this, but I am not sure about it:

Consider $n\in \mathbb{N}$ and $\mathbb{R}^n$ with the $\max$ norm, i.e if $\mathbb{R}^n\ni x = (x_1,\dots,x_n)$ then $|| x|| = \displaystyle \max_{i=1,\dots,n}{x_i}. $ And the operator $T=\mathbb{R}^n\to \mathbb{R}^n$ such that, if we denote $||x||=a$ then $T:x\mapsto(a,\dots,a) \in \mathbb{R}^{n}$. This is all the non-negative multiples of the vector $(1,1,…,1)$.

Clearly $||T(x)||=||x||$, so $T$ is bounded, but I am not sure if the range is closed, for this I thought the following: We know that a set $A$ is closed iff $A^{c}$ is open. In this case $A= T(\mathbb{R}^n)$ so I can consider an element in the complement of the image of $T$, this is $y=(y_1,\dots,y_n)$ such that there are at least two $y_i$ that are different. Let's denote $b=||y||$ and consider an open ball $B(y,\epsilon)$, note that for every possible $\epsilon$ the element $T(y)$ is contained in $B(y,\epsilon)$, so there is no open ball in $A^c$ for $y$, so $A^c$ is not an open set, therefore $A$ is not a closed set.

Is this correct? I feel like I am missing something but I am not sure. If I am wrong, can you give me some other example of a bounded operator with no closed range? I saw a couple but most of the examples I found use the $L^p$ spaces that we haven't seen on class. Thanks.

$\endgroup$
  • $\begingroup$ Is this a linear map? You're going to have to use infinite dimensional spaces, I'm afraid. This function does have closed range, by the way. Describe the image concretely: It's all the non-negative multiples of the vector $(1,1,\dots,1)$, which is homeomorphic to the set of non-negative real numbers. $\endgroup$ – Ted Shifrin Oct 26 at 5:16
  • $\begingroup$ @TedShifrin I edited it with that aclaration on the image, thanks. But I don't get why you are afraid of infinite dimensional, I am only using $\mathbb{R}^n$. And jmm, now I am not sure if it is linear... $\endgroup$ – J.Rodriguez Oct 26 at 5:25
  • $\begingroup$ Sorry for my English. Figure out why your map isn't linear, but I'm telling you that you need infinite dimensions. Linear subspaces of finite dimension are always closed. $\endgroup$ – Ted Shifrin Oct 26 at 5:30
  • $\begingroup$ The term 'bounded operator' is generally used for continuous linear maps in FA. I am sure the question is about linear operators. $\endgroup$ – Kavi Rama Murthy Oct 26 at 5:47
1
$\begingroup$

The proof is faulty because $T$ is not linear ($T(-y)=T(y)$) and even so, it is not true that $T(y)$ is contained in $B(y,\epsilon)$. (Take $n=2$, $y=(1,0)$, then $T(y)=(1,1)$ is not in a small ball around $y$.) In fact, $B(y,\epsilon)\cap T(\mathbb{R}^n)=\emptyset$ for $\epsilon$ small enough.

The mapping $T : \ell^2\to\ell^2$, defined by $T (a_n) := (a_0, a_1/2, a_2/3,...)$, is linear and bounded, with $\|T\|=\pi/\sqrt6$. Its image is not closed in $\ell^2$.

Proof: Consider $y_n:=(1, \frac{1}{2},..., \frac{1}{n}, 0, 0,...)=T(\underbrace{1,1,\ldots,1}_n,0,\ldots)\in T(\ell^2)$. It converges to the sequence $y=(1,\frac{1}{2},...)\in\ell^2$ but $y\notin T(\ell^2)$ otherwise $y=T(x)$ implies $x=(1,1,\ldots)\notin\ell^2$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Ohh, thanks, this example I can understand, but still have a question, now I know that my proof is wrong because T is not linear, but why $B(y,\epsilon) \cap T(\mathbb{R}^n) = \varnothing$ ? In your example if I take a ball arround $y$ of radious epsilon, this contains all elements $x$ of $\mathbb{R}^n$ such that $||x||\in (1-\epsilon,1+\epsilon)$, and $||T(y)||=1$ (this is because we are not using the usual norm). Am I wrong about this? $\endgroup$ – J.Rodriguez Oct 26 at 11:56
  • $\begingroup$ $B(y,\epsilon)$ in the $\infty$-norm (max-norm) has the shape of a square of side $2\epsilon$. One can find such a small square around $(1,0)$ that does not touch the line $A$ (through $(0,0)$ and $(1,1)$). I don't understand your second assertion. $B(y,\epsilon)$ does not contain $-y$ for example even though $\|-y\|=1$. $\endgroup$ – Chrystomath Oct 26 at 12:07
  • $\begingroup$ Ohh sorry, I was misinterpreting something, my bad. Thanks for the good examples. $\endgroup$ – J.Rodriguez Oct 26 at 12:12
1
$\begingroup$

Every linear subspace of a finite dimensional space is closed so there is no hope of such an example in finite dimensions. Here is a valid example: Define $Tf(x)=\int_0^{x} f(t)dt$ on $C[0,1]$. Then $\|T\|\leq 1$ but the range is not closed. The range consist precisely of continuously differentiable functions vanishing at $0$. Take any continuous function $f$ vanishing at $0$ which is not differentiable and use Weierstrass Theorem to construct a sequence polynomials in the range of $T$ converging uniformly to $f$.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

Let X be a non reflexive Banach space and $Y$ be a reflexive Banach space. Suppose that $A \colon X \to Y$ is an injective bounded linear operator. I claim that the range of $A$, denoted by $R(A)$, can't be closed in $Y$.

Arguing by contradiction, suppose that $R(A)$ is closed and thus reflexive. Then, the operator $ A \colon X \to R(A) $ is a continuous bijection and thus (by virtue of the Open Mapping Theorem) is an isomorphism. It's not hard to check that $X$ has to be reflexive, since it's isomorphic to a reflexive Banach space, which leads to a contradiction.

| cite | improve this answer | |
$\endgroup$

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.