27
$\begingroup$

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.

$\endgroup$
  • 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
22
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ In fact, if the injectivity of the operator can replaced by the Fredholmness. $\endgroup$ – DLIN Sep 7 '18 at 12:15
12
$\begingroup$

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 $Y=T(X)$.

$\endgroup$
  • $\begingroup$ 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

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.