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.
Best Answer
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.