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Problem Statement

Let $(X,\|\cdot \|)$ be a Banach space and $Y$, $Z$ closed subspaces of $X$. If $Y+Z$ is closed, then show that there exists $C>0$ such that for all $x \in Y+Z$, there exist $y \in Y$, $z \in Z$ such that $x = y+z$ and $\|y\| + \|z\| \leq C\|x\|$.

Attempt

If $Y \cap Z = \{0\}$, then by this post, we can use the open mapping theorem and we are done. So my idea is to reduce to this case somehow.

I try to consider the bijective continuous linear map $Y\oplus Z/\{(w,-w)|w \in Y \cap Z\} \to Y+Z$, and the open mapping theorem tells me the inverse is also a continuous linear operator. Hence there exists $C>0$ such that $\inf_{w \in Y \cap Z}\{\|y+w\|+\|z-w\|\} \leq C\|y+z\|$ for all $y \in Y$, $z \in Z$. But then I don't know how to go from here as I don't know whether this infimum can be achieved or not.

Thanks for your help.

zzz
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1 Answers1

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You are almost there. Note that by the definition of infimum, given that $\|x\|>0$, we can find $w\in Y\cap Z$ such that $$ \|y+w\|+\|z-w\| \le 2C\|x\|. $$ By letting $y'=y+w\in Y$ and $z'=z-w\in Z$, the desired claim follows.

A neater formulation is as follows. Let $Y\times Z$ be given the graph norm, i.e. $\|(y,z)\|_{Y\times Z}=\|y\|+\|z\|$. Define a surjective linear map $$ T:Y\times Z\ni (y,z)\mapsto y+z\in Y+Z. $$ If we consider a factored map $$ \hat{T}:(Y\times Z)/\ker T\ni (y,z)+\ker T \mapsto y+z \in Y+Z, $$ then $\hat T$ is a continuous bijection, and hence has a bounded inverse by inverse mapping theorem, i.e. there exists a $C>0$ such that $$ \|\hat{T}^{-1}(x)\|_{Y\times Z/\ker T}\le C\|x\|,\quad \forall x\in Y+Z. $$ Now we can find $(y,z)\in Y\times Z$ such that $$ \|y\|+\|z\|=\|(y,z)\|_{Y\times Z}\le 2C\|x\| $$ because $$\|\hat{T}^{-1}(x)\|_{Y\times Z/\ker T}=\inf \{\|(y,z)\|_{Y\times Z}:T(y,z)=y+z=x\}.$$

Myunghyun Song
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