5
$\begingroup$

Let $X$ and $Y$ be Banach spaces. If $T \in B(X,Y)$, and $Y = \operatorname{im}T \oplus M$ for some closed linear subspace $M$ of $Y$, then $\operatorname{im}(T)$ is closed in $Y$.

I am unsure if this statement is true or not. Nonetheless, I am having difficulty proving it, any suggestions or counterexample?

$\endgroup$
  • $\begingroup$ When you write \text{im}T then you see $\text{im}T$ without proper spacing. In $a\operatorname{im} T$ or $a\operatorname{im}(T)$, coded as a\operatorname{im}T and a\operatorname{im}(T) you see spacing before and after $\operatorname{im},$ and you see that the amount of space depends on the context, so it need not be adjusted manually. In actual $\LaTeX$ (as opposed to MathJax, which is the software used here, you can put the following command before the \begin{document} command: \newcommand{\im}{\operatorname{im}} Then in the document, just write \im T. $\endgroup$ – Michael Hardy Apr 29 '17 at 17:55
  • $\begingroup$ I have a feeling this will go through because of the closed graph theorem $\endgroup$ – sntx Apr 30 '17 at 7:03
  • $\begingroup$ Step 0: Explain why we may assume that $T$ is injective. Step 1: Find a useful continuous bijection $X\times M \to Y$. Step 2: Use the open mapping theorem. $\endgroup$ – Daniel Fischer May 1 '17 at 12:39
  • $\begingroup$ Daniel Fischer, I am still unsure how to find such a useful continuous bijection. In light of the open mapping theorem, I think I see it is true though. $\endgroup$ – Dragonite May 1 '17 at 18:57
4
+50
$\begingroup$

It's true.

We can replace $X$ with $X/\ker T$ and $T$ with the induced map $\tilde{T} \colon X/\ker T \to Y$ if necessary, since $\operatorname{im} \tilde{T} = \operatorname{im} T$, and $X/\ker T$ is also a Banach space. Thus we may assume that $T$ is injective.

As a closed subspace of the Banach space $Y$, $M$ is itself a Banach space, and therefore $X \times M$ is also a Banach space if we endow it with one of the usual norms on a product of two normed spaces, e.g. $\lVert (x,m)\rVert = \lVert x\rVert_X + \lVert m\rVert_Y$.

We then define a continuous bijection $S \colon X \times M \to Y$ via

$$S(x,m) = Tx + m.$$

By the open mapping theorem, $S$ is open, and thus a homeomorphism. Since $X \times \{0\}$ is a closed subspace of $X\times M$, it follows that

$$\operatorname{im} T = S(X\times \{0\})$$

is a closed subspace of $Y$.

$\endgroup$

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.