Let $X$, $Y$ be Banach spaces. Let $T \colon X \to Y$ be a bounded linear operator. Under what circumstances is the image of $T$ closed in $Y$ (except finite-dimensional image).

In particular, I wonder under which assumptions $T \colon X \to T(X)$ is a bounded linear bijection between Banach spaces, so it is at least an isomorphism onto its image by bounded inverse theorem.

  • 5
    $\begingroup$ Probably the most useful criterion is en.wikipedia.org/wiki/Closed_range_theorem but that may not be exactly what you're looking for. It also is a nice exercise to show that an operator whose image has finite codimension has closed range (this has some use in Fredholm theory). $\endgroup$
    – t.b.
    Mar 9, 2011 at 23:10

2 Answers 2


An answer to your last question is that a bounded linear map $T$ between Banach spaces is injective with closed range if and only if it is bounded below, meaning that there is a constant $c>0$ such that for all $x$ in the domain, $\|Tx\|\geq c\|x\|$. You can read more about this in Chapter 2 of An invitation to operator theory by Abramovich and Aliprantis.

  • 1
    $\begingroup$ In fact, if the injectivity of the operator can replaced by the Fredholmness. $\endgroup$
    – DLIN
    Sep 7, 2018 at 12:15

Thrm 1: Suppose $X$ is a Banach space, $Y$ is a normed vector space, and $T:X\to Y$ is a bounded linear operator. Then the range of $T$ is closed in $Y$ if $T$ is open.

Proof: Suppose $\mathrm{ran}(T)$ is not closed in $Y$. Let $\delta>0$ be given. The goal is to show that there exists $x\in X$ such that $\|T(x)\|/\|x\|<\delta$. Since $\delta$ is arbitrary this will demonstrate that $T$ is not open.

Since $\mathrm{ran}(T)$ is not closed there is a sequence $\{y_n\}$ in $\mathrm{ran}(T)$ and point $y\in Y\setminus\mathrm{ran}(T)$ such that $y_n\to y$. This means that there are corresponding $x_n\in X$ such that $y_n=T(x_n)$. Since $T$ is continuous it cannot be that $\{x_n\}$ is a convergent sequence or $y$ would be in the range of $T$.

Since $\{x_n\}$ does not converge it is not Cauchy. So there exists an $\epsilon>0$ such that $\forall N\in\mathbb{N} \ \ \exists n,m \ge N $ s.t. $\|x_n-x_m\|>\epsilon$. On the other hand, since $y_n\to y$, there is an $M\in\mathbb{N}$ such that $\forall k\ge M \ \ \ \|T(x_k)-y\|<\delta\frac{\epsilon}{2}$. By choosing $N=M$, there exist $n,m\ge N$ such that $\|x_n-x_m\|>\epsilon$, $\ \|T(x_n)-y\|<\delta\frac{\epsilon}{2}$, and $\|T(x_m)-y\|<\delta\frac{\epsilon}{2}$. By the Triangle Inequality, $\|T(x_n)-T(x_m)\|=\|T(x_n-x_m)\|<\delta \ \epsilon$.

Let $x=(x_n-x_m) \in X$. Then \begin{align*} \frac{\|T(x)\|}{\|x\|} &< \frac{\delta \ \epsilon}{\epsilon} \\ &= \delta. \end{align*}

Thrm 2: Suppose $X$ and $Y$ are both Banach spaces, and $T:X\to Y$ is a bounded linear operator. Then $\mathrm{ran}(T)$ is closed in $Y$ if and only if $T$ is open.

Proof: If $\mathrm{ran}(T)$ is closed then $Y$ is a surjective map onto this subspace so by the Open Mapping Theorem $T$ is open.

The other direction is just the first Theorem.

EDIT: The so called "Thrm2" is false, as per the counterexample provided in comments. The "proof" makes the false assumption that $Y=T(X)$.

  • 2
    $\begingroup$ The second "Thrm" is blatantly false. Take $T:X\rightarrow X$ defined as $T(x)=0$ where $X$ is any Banach space. Then the range of $T$ is $\{0\}$, a closed set, yet the image of the open set $X$ is $\{0\}$, a set which is not open. Clearly then $T$ cannot be an open map. $\endgroup$
    – user293794
    Sep 7, 2017 at 23:20
  • $\begingroup$ @user293794 Yes, good catch. Its clear from the proof that I was thinking about $T:X\to T(X)$, as in the second part of the question. But I can't remember enough about what I was thinking when I wrote this to be able to fix this error. $\endgroup$
    – cantorhead
    Sep 9, 2017 at 17:55
  • 2
    $\begingroup$ Maybe I'm missing something, but if a bounded linear operator $T : X \to Y$ is an open map, then it is necessarily surjective. Namely, $X$ is open so $T(X)$ is an open subspace of $Y$, but this implies $T(X)= Y$. So Thrm $1$ is true, but in a degenerate way. $\endgroup$ May 11, 2020 at 21:03
  • $\begingroup$ @mechanodroid I think you are correct. Open maps are surjective, thus the image is $Y$ which is clearly closed. Would nyojne know why finding for each $\delta$ $x$ s.t $\|x\|=1$ and $\|T(x)\|<\delta$ contradicts openness of the map? $\endgroup$
    – Sorfosh
    Aug 8, 2020 at 22:15
  • 1
    $\begingroup$ @Sorfosh It doesn't. The property is clearly satisfied by open maps $T$ with $\|T\| < 1$. $\endgroup$ Aug 9, 2020 at 10:06

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .