# Is this an example of a Bounded Operator with NO closed range?

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.

• 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. – Ted Shifrin Oct 26 at 5:16
• @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... – J.Rodriguez Oct 26 at 5:25
• 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. – Ted Shifrin Oct 26 at 5:30
• The term 'bounded operator' is generally used for continuous linear maps in FA. I am sure the question is about linear operators. – Kavi Rama Murthy Oct 26 at 5:47

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 $$x=(1,1,\ldots)\notin\ell^2$$.

• 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? – J.Rodriguez Oct 26 at 11:56
• $B(y,\epsilon)$ 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 $\|-y\|=1$. – Chrystomath Oct 26 at 12:07
• Ohh sorry, I was misinterpreting something, my bad. Thanks for the good examples. – J.Rodriguez Oct 26 at 12:12

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.

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 $$f$$.