Let $T: X \longrightarrow Y$ be a continuous linear map between two Banach spaces.

When is $\operatorname{Ran}(T)$ a closed subspace?

What theorems are there?

Thanks :)

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    $\begingroup$ It is not a “fact”, because it is not always true. The open mapping theorem is one handy theorem for this. Understanding quotients of Banach spaces likewise (in particular, $X/\operatorname{ker}T$). $\endgroup$ – Harald Hanche-Olsen Feb 4 '13 at 22:51
  • $\begingroup$ You may want to refer to the Wikipedia article on Closed Range Theorem. $\endgroup$ – Haskell Curry Feb 4 '13 at 22:53
  • $\begingroup$ Take a look here. $\endgroup$ – JohnD Feb 4 '13 at 22:55
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    $\begingroup$ The inclusion map from $\ell_1$ into $\ell_2$ is an example of a bounded operator whose range is not closed (else, as its range contains the standard unit vectors, we would have $\ell_1=\ell_2$. $\endgroup$ – David Mitra Feb 4 '13 at 23:06
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    $\begingroup$ This does not say to prove something, it says to study it. So that means you have to figure out what to prove yourself. $\endgroup$ – GEdgar Feb 5 '13 at 15:51

A case in which it is true that $R(T)$ is closed is if $T$ is bounded from below, i.e., there exists $c > 0$ such that $\|Tx\| \geq c\|x\|$ for all $x \in X$. To prove this, notice first that since $$Tx = 0 \implies 0 = \|Tx\| \geq c\|x\| \geq 0 \implies x= 0$$ $T$ must be injective. Now let $x_n \in R(T)$ be such that $x_n \to x$. Then we have that $$ \|x_n - x_m\| = \|TT^{-1}x_n - TT^{-1}x_m\| \geq c\|T^{-1}x_n - T^{-1}x_m\|$$ so that $T^{-1}x_n$ is a Cauchy sequence which must converge to some $z \in X$. But we have $$T(z) = T(\lim_{n\to\infty} T^{-1}x_n) = \lim_{n\to\infty} TT^{-1}x_n = \lim_{n\to\infty} x_n = x$$ which means that $x \in R(T)$ and thus $R(T)$ is closed.

In fact, if $T$ is assumed injective then this is a characterization of $T$ having closed range. For if $R(T)$ is closed in $Y$ then $R(T)$ is a Banach space under $\|\cdot\|_Y$ in its own right and the open mapping theorem shows that $T^{-1}:R(T) \to X$ is continuous which means there exists $C > 0$ such that $$\|T^{-1}x\| \le C\|x\|$$ or $$\|Tx\| \geq \frac{1}{C}\|x\|$$ for all $x \in X$.

  • $\begingroup$ How do you know that $T$ is an inversable operator ? $\endgroup$ – Iuli Feb 5 '13 at 21:02
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    $\begingroup$ @Iuli The proof above showed that $Tx = 0$ implies that $x = 0$. This is enough, because then $Tx = Ty$ implies $T(x-y) = 0$ which implies $x-y = 0$ so $x=y$. This means $T$ is injective, so invertible when considered as an operator $T:X \to R(T)$. $\endgroup$ – user38355 Feb 5 '13 at 21:49
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    $\begingroup$ We do not know if $T$ is bounded. Therefore your prove could fail at the step $T( \lim_{n\rightarrow \infty} T^{-1}x_n) = \lim_{n\rightarrow \infty} TT^{-1}x_n)$ However, taking $T^{-1}$ outside the limit, instead of bringing $T$ in fixes the problem. $\endgroup$ – nippon Jan 20 '15 at 15:26
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    $\begingroup$ @nippon Well, the hypothesis in the OP is $T \in B(X,Y)$. This means, by definition, that $T$ is bounded. $\endgroup$ – user38355 Jan 21 '15 at 17:14
  • $\begingroup$ @brom That is true, however if we generalize to general $T$, the prove still holds with my remark. Otherwise it would fail there. $\endgroup$ – nippon Jan 25 '15 at 21:36

Consider $X=Y:=\ell^\infty$, the normed space of bounded sequences. Let $T(x)(n):=\frac{x(n)}{n^2}$; this gives a linear continuous operator. Let $y\in T(X)$, then $\{n^2y(n)\}$ is bounded, and the converse holds. So $$T(X)=\{y,\sup_n|n^2y(n)|<\infty\}.$$ Let $x^{(n)}:=\sum_{j=1}^nj^{-1}e(j)$; it converges in $\ell^\infty$ to the sequence $x=\{n^{-1}\}$. Furthermore, $x^{(n)}\in T(X)$. But $x\notin T(X)$.


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