Let $X$, $Y$ be Banach spaces. Let $T:X\rightarrow Y$ 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 $T:X\rightarrow T(X)$ is a bounded linear bijection between Banach spaces, so it is at least an isomorphism onto its image by bounded inverse theorem.

  • 3
    $\begingroup$ 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 '11 at 23:10

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.

  • $\begingroup$ In fact, if the injectivity of the operator can replaced by the Fredholmness. $\endgroup$ – DLIN Sep 7 '18 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 $Y=T(X)$.

  • $\begingroup$ 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 '17 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 '17 at 17:55

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