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 Post subject: Bounded
PostPosted: Fri Feb 05, 2016 4:02 pm 
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Joined: Mon Nov 09, 2015 1:52 pm
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Let \(\displaystyle{\left(X,||\cdot||\right)}\) be a \(\displaystyle{\rm{Banach}}\) space, \(\displaystyle{Y\,,Z}\)

two closed subspaces of \(\displaystyle{\left(X,||\cdot||\right)}\) such that \(\displaystyle{Y\cap Z=\left\{0\right\}}\)

and \(\displaystyle{Y+Z}\) is also closed subspace. Prove that there exists \(\displaystyle{C>0}\)

such that \(\displaystyle{||y||\leq C\,||y+z||\,,\forall\,y\in Y\,,\forall\,z\in Z}\) .


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 Post subject: Re: Bounded
PostPosted: Mon Mar 14, 2016 10:37 am 
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We define

\(\displaystyle{f:Y+Z\to Y\,,y+z\mapsto y}\) .

Let \(\displaystyle{y+z=y'+z'\in Y+Z}\). Then, \(\displaystyle{y-y'=z'-z}\) and

\(\displaystyle{y-y'\in Y\,,y-y'=z'-z\in Z}\), so, \(\displaystyle{y-y'\in Y\cap Z=\left\{0\right\}}\) and

\(\displaystyle{y-y'=0\iff y=y'}\) . (Also, \(\displaystyle{z=z'}\)) .

Therefore, the function \(\displaystyle{f}\) is a well defined map of the \(\displaystyle{\rm{Banach}}\)

space \(\displaystyle{Y+Z}\) to the \(\displaystyle{\rm{Banach}}\) space \(\displaystyle{Y}\) .

It's obvious that \(\displaystyle{f}\) is linear and onto \(\displaystyle{Y}\) . Futhermore,

the graph of \(\displaystyle{f}\)

\(\displaystyle{G_{f}=\left\{\left(y+z,f(y+z)\right): y\in Y\,,z\in Z\right\}\subseteq (Y+Z)\times Y\subseteq X\times X}\)

is a closed subset of \(\displaystyle{(Y+Z)\times Y}\).

According to the closed graph theorem, the function \(\displaystyle{f}\), is bounded, that is

\(\displaystyle{\left(\exists\,C>0\right)\,, ||f(y+z)||=||y||\leq C\,||y+z||\,,\forall\,y\in Y\,,\forall\,z\in Z}\) .


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