Do bounded below linear operators have closed range?

Question

Suppose $T : X \to Y$ is a linear operator between Banach spaces which is bounded below, say, $C\|x\| \leq \|Tx\|$ for all $x\in X$. Is it the case that the range of T is closed?

It's standard to show this is the case if we assume $T$ is bounded. However, in the unbounded/discontinuous case, it doesn't seem like this is true. However, I couldn't find a counterexample - I suspect it might require the axiom of choice. Does anyone know whether or not this is possible?

Answer 1

Any unbounded operator $T$ with the property $\|Tx\|\ge c\|x\|$ will do. Assume by contradiction the range $T(X)$ is closed, i.e. it is a Banach space. The operator $T^{-1}:T(X)\to X$ is bounded ($\|T^{-1}\|\le c^{-1}$) and its range is equal $X.$ Therefore by the Banach inverse mapping theorem the operator $T$ is bounded, a contradiction. So the problem reduces to finding an unbounded operator with the property $\|Tx\|\ge c\|x\|.$ For this sake take any unbounded operator $S:X\to X$ defined on the entire space (see), and the operator $T:X\to X\times X,$ where $\|(x_1,x_2)\|=\|x_1\|+\|x_2\|,$ defined by $(x,Sx).$ Then $\|Tx\|\ge \|x\|.$

Answer 2

Let $X$ be an infinite-dimensional Banach space, $X_0 \subset X$ be a dense co-dimension one subspace (which exists by the Axiom of Choice - just choose the kernel of a discontinuous linear functional, say). Let $v \in X \setminus X_0$ with norm $1$, $Y = X \oplus \mathbb{C}$. Then define $T: X \to Y$ by $T(x + \lambda v) = (x, \lambda)$ where $x \in X_0$ and $\lambda \in \mathbb{C}$. One easily verifies that $\|x + \lambda v\| \leq \|x\| + |\lambda| = \|(x, \lambda)\|$, so $\|w\| \leq \|Tw\|$ for all $w \in X$. However, the range of $T$ is $X_0 \oplus \mathbb{C}$, which is dense but of co-dimension one in $Y$, and therefore is not closed.

Answer 3

Do you mean your unbounded operator $T$ is defined on all of $X$? In this case, as the exhaustive answers above show, $T$ can not have closed range. However, usually when we talk about unbounded operators we assume that $T$ is only defined on some subspace $\mathcal{D}(T) \subset X$. In this case, we have that if $T$ is bounded below and closed then $T$ has closed range. This is in accordance with the aswers above because an unbounded operator that is defined on the whole space can never be closed by the closed graph theorem.

Answer 4

Let $T:X\to Y$ be a linear operator between Banach spaces (meaning $T$ is defined on the whole Banach space $X$).

• If $T$ is bounded below and has closed range, then $T$ is bounded.

  One way to prove it is to observe that

  1. $T$ bounded-below implies $T$ is injective. Injectivity is necessary to ensure we have a well-defined inverse $T^{-1}:\operatorname{im}(T)\to \operatorname{dom}(T)$.
  2. If $T$ has closed range, then $T^{-1}$ is a linear operator defined everywhere on a Banach space.
  3. If $T$ is bounded below then $T^{-1}$ is bounded.
  4. Thus $T^{-1}$ is an everywhere-defined bounded operator between Banach spaces, and thus by the open mapping theorem its inverse, which is $T$, must be bounded.

  Another possible approach is to leverage instead the closed graph theorem:

  1. The theorem tells us that $T$ is closed iff it is bounded, so we can focus on proving the assumptions imply $T$ is closed, which is easier. Consider then a converging sequence $X\ni x_n\to x\in X$ such that $Tx_n\to y$ for some $y\in Y$.
  2. Because the image of $T$ is closed, we must have $y\in \operatorname{im}(T)$, i.e. $y=Tx'$ for some $x'\in X$.
  3. Because $T$ is bounded below, $Tx_n\to Tx'$ implies $x_n\to x'$. Thus by uniqueness of limits, $x=x'$.

• Thus an unbounded operator that is bounded below has non-closed range.

  Note that unbounded operators can be bounded below. An example if given in this other answer, and more generally the closed range theorem gives conditions for this to happen.

  Note also that while unbounded linear operators defined on a Banach space can exist, their construction involves the axiom of choice, and most unbounded operators encountered in practice are instead defined on a dense subset rather than on a whole Banach space. See this post for more details on this topic.

Furthermore, under the assumption that $T$ is bounded, we have:

• A bounded operator that is bounded below has closed range.

  Indeed if $T$ is bounded, $x_n\to x\implies Tx_n\to Tx$. Thus for the range to not be closed there must be non-converging sequences $x^{(n)}$ such that $Tx^{(n)}$ converges (to a point outside the range). This is only possible if $T$ is not bounded below.

  Note that the other implication doesn't hold: closed range does not imply bounded below. Consider for example $T:\ell^2\to\ell^2$ with $T(x_1,x_2,x_3,...)=(x_1,x_2,0,...)$. This has closed range but it's not injective, and thus not bounded below.

  Another proof can be found in this answer.

• A bounded operator can have non-closed range.

  An example is $T:\ell^\infty\to\ell^\infty$, $Tx=(x_n/n)_n$ (taken from here). If $x^{(n)}=\sum_{k=1}^n \sqrt k e_k\in\ell^\infty$, we have $Tx^{(n)}\to y\equiv (1/\sqrt k)_{k\in\mathbb{N} }\in\ell^\infty$, but $y\notin \operatorname{im}(T)$.

  Another similar example is $T:\ell^2\to\ell^2$, $Tx=(x_n/n)_n$. Then the sequence $x^{(n)}=\sum_{k=1}^n e_k$ has $Tx^{(n)}\to (1/k)_{k\in\mathbb{N} }\notin \operatorname{im}(T)$.

  More generally, as mentioned [in this answer][9], the multiplication operator $T:\ell^2\to\ell^2$, $Tx=(a_n x_n)_{n\in\mathbb{N} }$ is

  1. Bounded iff $a\in\ell^\infty$, i.e. $\sup|a_n|<\infty$.
  2. Injective iff $a_k\neq0$ for all $k\in\mathbb{N}$.
  3. Bounded below iff $a_k\to 0$. Or more generally if $\inf |a_n|=0$.

• If $T$ is bounded, then bounded below implies injective. For the other direction, we require closed range: if a bounded operator is injective and has closed range, it is bunded below. See also When is the image of a linear operator closed? and When is the image of a linear operator (between Banach spaces) closed? about this.

  Note that the requirement of $T$ having closed range is important, as you can have bounded injective operators that have non-closed range and are not bounded below. The bounded operator $T:\ell^\infty\to\ell^\infty$, $Tx=(x_n/n)_n$ is again such an example.