# $X$ and $Y$ Banach Spaces, $T \in B(X,Y)$, $Y = \operatorname{im}T \oplus M$, for $M \subseteq Y$, then $\operatorname{im}T$ is closed in $Y$

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?

• 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. – Michael Hardy Apr 29 '17 at 17:55
• I have a feeling this will go through because of the closed graph theorem – sntx Apr 30 '17 at 7:03
• 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. – Daniel Fischer May 1 '17 at 12:39
• 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. – Dragonite May 1 '17 at 18:57

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 $Y$.