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!


1 Answer 1


Your idea

I thought about constructing some $\hat{T}\colon X\times Z\to Y$ such that is linear, continuous and surjective

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.

  • $\begingroup$ Can you maybe explain why the existence of such a $Z$ follows directly from the fact that the qoutient is finite dimensional?? $\endgroup$
    – Thorben
    Apr 7, 2014 at 9:44
  • $\begingroup$ You always have an algebraic complement. If that is finite-dimensional, it is closed, which is what we need. We want a Banach space $Z$ such that $X\times Z \to Y$ is bijective, so $Z$ should be (algebraically isomorphic to) a complement of $T(X)$. To be Banach, it must be closed. $\endgroup$ Apr 7, 2014 at 9:49
  • $\begingroup$ Ok, thank you very much!! $\endgroup$
    – Thorben
    Apr 7, 2014 at 9:50

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