# Condition for an operator to have closed Image

I just did an exercise and I wanted to make sure that my proof was correct

Let $$X$$ and $$Y$$ be Banach spaces $$T$$ a one-to-one continuous linear operator and $$N$$ a closed subspace such that $$Y=ImT\bigoplus N$$, algebraically, show that $$Im T$$ is closed.

So first since $$Y$$ is a banach space and $$N$$ is a closed subspace of $$Y$$ we know that $$Y/N$$ will be a banach space. Now if we consider the composition $$X\rightarrow Y\rightarrow Y/N$$ of $$\pi\circ T$$ we will have that this is a continuous linear operator and that it will be bijective because $$N \cap ImT=\{0\}$$, and so we can use the open mapping theorem to conclude that we have an inverse $$T^{-1} \in L(Y/N,X)$$. Now consider an element $$y$$ in the closure of $$Im T$$ so that we have a sequence $$y_n=T(x_n)$$ of elements such that $$||y_n-y||\rightarrow 0$$. If we consider the elements $$\pi(y_n)$$ and $$\pi(y)$$ we will have that $$||[y_n]-[y]||_{Y/N}\rightarrow 0$$, because this norm is less or equal than the previous one, and using the fact that $$L^{-1}$$ is continuous we can get a cauchy sequence $$x_n$$ and so it will be converge to some $$x$$, and we will get that $$y= \lim\limits_{n\rightarrow \infty} T(x_n)=T(\lim\limits_{n\rightarrow \infty}x_n)=T(x)$$, and so we get that $$ImT$$ is closed, using the fact that $$T$$ is continuous.

Now this is my idea of the proof and I wonder if there is anything I forgot or did wrong , or if there are some alternative resolutions to the exercise. Thanks in advance.

• With there being essentially no hypotheses about $T$, it seems to be a red herring. The question really boils down to showing that a subspace with closed complement is closed itself. But that is not true. Are you assuming that $T$ is bounded? Commented Jun 13, 2020 at 11:04
• Yes I am assumning T is bounded sorry I will edit @tomasz Commented Jun 13, 2020 at 11:25

The proof seems okay to me, except for minor typos, like writing $$T^{-1}$$ instead. It would also be improved by breaking it into more paragraphs. You also did not justify the conclusion --- that in fact $$T(x)=y$$ (this is the only place where you really use the fact that $$N$$ and $$\operatorname{im}(T)$$ are disjoint). But the ideas are all sound.