So here is my problem,
I would like to prove the following,
Let $X,Y$ be Banach-Spaces and $T:X\rightarrow Y$ a linear and bounded operator.
Then $TX$ is closed if it is of finite codimension i.e $dim(Y/TX)<\infty$.
I thought about constructing some $\hat{T}:X\times Z\rightarrow Y$ such that is linear, continuous and surjective to work with the open mapping theorem. I thought maybe it is even possible to construct $\hat{T}$ s.t it is bijective. In that case I could try to map the Graph of $T$ on $TX$ to conclude that $TX$ is closed since by the continuity of $T$ it follows that the graph of $T$ is closed. But none of the upper ideas worked… i.e I could not find a apropriate candiate for $Z$.
Could someone help me? Thank you!
Best Answer
Your idea
is the right one. To make things a little easier to see, we assume that $T$ is injective (by considering $\tilde{T}\colon X/\ker T \to Y$, we lose no generality).
Since $Y/\mathcal{R}(T)$ is finite-dimensional, we can choose $Z$ to be an algebraic complement of $\mathcal{R}(T)$ in $Y$ and get a Banach space $Z$.
Then endow the product $X\times Z$ with the norm $\lVert(x,z)\rVert_{X\times Z} = \lVert x\rVert_X + \lVert z\rVert_Y$. That makes $X\times Z$ a Banach space, and
$$S\colon X\times Z \to Y; \quad (x,z) \mapsto T(x) + z$$
is easily seen to be continuous and bijective. The open mapping theorem finishes the proof.