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Theorem 2.10, on page 37

Let $E$ be a Banach space. Assume that G and L are two closed linear sub spaces such that $G+L$ is closed. Then there exists a constant $C \geqslant 0$ such that every $z \in G+L$ admits a decomposition of the form $z=x+y$ with $x \in G, y \in L, \vert\vert x \vert\vert \leqslant C \vert\vert z \vert\vert$ and $\vert\vert y \vert\vert \leqslant C \vert\vert z \vert\vert$.

The proof uses the open mapping theorem, consider the map: $$T:G \times L \to G+L$$ The norm of $G \times L$ defined as $\vert\vert [x,y] \vert\vert = \vert\vert x \vert\vert + \vert\vert y \vert\vert $. The theorem guarantees that the image of $\{[x,y] : \vert\vert x \vert\vert + \vert\vert y \vert\vert\}$ conclude $\{z : \vert\vert z \vert\vert < c\}$. By homogeneity, $$z = x+y,\qquad \vert\vert x \vert\vert + \vert\vert y \vert\vert \leqslant 1/c \cdot \vert\vert z \vert\vert$$

I would like to ask the final step. Why $\vert\vert x \vert\vert + \vert\vert y \vert\vert \leqslant 1/c \cdot \vert\vert z \vert\vert$ ? I think if $z$ is quite small, it cannot prove this result.

Math
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MichaelS
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1 Answers1

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Open Mapping Theorem tells you that there exists $\delta >0$ such that for any $z \in G+L$ with $\|z\| <\delta $ we ca n write $z=x+y$ with $\|x\|+\|y\|<1$. Now take any $z \neq 0$ in $G+L$. Let $z'=\frac {\delta z} {2 \|z\|}$. Then $\|z'\|<\delta$. So $z'=x+y$ with $x \in G, y \in L$ and $\|x\|+\|y\|<1$. Now let $x'=\frac {2\|z||} {\delta}x$ and $y'=\frac {2\|z||} {\delta}y$. Then $z=x'+y'$ and $\|x'\|+\|y'\| \leq \frac 1 c \|z\|$ where $c=\frac {\delta} 2$.

Kavi Rama Murthy
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