Does Closure of Y Guarantee Continuity of Projection in Norm Space X?

[jostpuur]

Let X be a norm space, and X=Y+Z so that $Y\cap Z=\{0\}$. Let P:X->Z be the projection $y+z\mapsto z$, when $y\in Y$ and $z\in Z$.

I see, that if P is continuous, then Y must be closed, because $Y=P^{-1}(\{0\})$.

Is the converse true? If Y is closed, does it make the projection continuous?

If yes, fine. If not, would finite dimensionality of Z make P continuous then?

 

[morphism]

Assuming Y and Z are linear subspaces, the converse is false in general. The root of the problem is that Z may not be a closed subspace, even when Y is. For instance, take $Y=c_0$ and $X=\ell_\infty$. It is a well-known theorem that $c_0$ is not complemented in $\ell_\infty$, i.e. that there does not exist a closed subspace Z such that $X=Y \oplus Z$. But using a standard Zorn's lemma argument, we can produce a linear subspace Z of X such that $X=Y \oplus Z$. This implies that I-P is not continuous, since ker(I-P)=Z is not closed. As a result, P is not continuous either.

Edit:
If Z is finite-dimensional, then P is always going to be continuous. I'll let you figure out why.

 

[jostpuur]

Edit:
If Z is finite-dimensional, then P is always going to be continuous. I'll let you figure out why.

Frankly, I think you left a pretty damn big task for me! Fortunately I had got some explanations for this from elsewhere too. I think I understood how this is done now, by considering some basis $\{e_1,\ldots,e_n\}$ of Z, linear functionals $f_i:Z\to\mathbb{C}$, $\alpha_1 e_1+\cdots\alpha_n e_n\mapsto \alpha_i$, and Hahn-Banach theorem. To be fully precise, I was told how to choose Y so that it becomes closed. Now I realized how the same procedure gives also the continuous projection.

 

[jostpuur]

If Z is finite-dimensional, then P is always going to be continuous.

Are you sure this is fully correct? I think it is possible to choose P:X->Z so that it is continuous, but isn't it also possible to choose it so that it is not continuous?

Edit: I see. You meant, that when Z is finite dimensional and Y is fixed and closed, then $y+z\mapsto z$ is always going to be continuous? It could be true, but I'm not sure how to show it.

 

[morphism]

Edit: I see. You meant, that when Z is finite dimensional and Y is fixed and closed, then $y+z\mapsto z$ is always going to be continuous? It could be true, but I'm not sure how to show it.

That is what I meant. This follows from the fact that finite dimensional subspaces are closed.

But since, as you observed, finite dimensional subspaces are always complemented, we don't really need to start with the assumption that Y is closed.

 

[jostpuur]

That is what I meant. This follows from the fact that finite dimensional subspaces are closed.

But since, as you observed, finite dimensional subspaces are always complemented, we don't really need to start with the assumption that Y is closed.

Do you also mean that this is true: "If Y and Z are closed, then projection X->Z is continuous."?

 

[morphism]

Hold on... All along I've been assuming that X is a Banach space, but I've just reread your original post and noticed that X is only normed. In this case, it's not necessarily true that the projection X->Z is continuous, even if Z and its complement are closed, but I can't come up with an easy example.