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 Post subject: Bounded Posted: Fri Feb 05, 2016 4:02 pm
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Joined: Mon Nov 09, 2015 1:52 pm
Posts: 426
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 $\displaystyle{||y||\leq C\,||y+z||\,,\forall\,y\in Y\,,\forall\,z\in Z}$ .

Top   Post subject: Re: Bounded Posted: Mon Mar 14, 2016 10:37 am
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Joined: Mon Nov 09, 2015 1:52 pm
Posts: 426
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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