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Let $X$ be a Banach space, $Y$ and $Z$ be its linear subspaces such that $X = Y+Z$, $\ Y\cap Z = \{0\}$ and the unit spheres in $Y$ and $Z$ are separated, i.e.

$$ \exists r>0 \ \ \lVert y-z\rVert \geq r \quad \forall y\in Y \ \ \forall z\in Z \ \text{ s.t. } \lVert y\rVert = \lVert z\rVert = 1 $$

The problem is to prove that $Y$ and $Z$ is closed in $X$.

My idea was to estimate norms of $y\in Y$ and $z\in Z$ with norm of their sum, because by Banach's isomorphism theorem it is enough to show that $$ \exists C>0 \ \ \ \lVert y\rVert + \lVert z\rVert \leq C\lVert y+z\rVert \quad \forall y\in Y \ \forall z\in Z $$ what implies $X \cong Y\oplus Z$.

Could anybody help me?

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    $\begingroup$ Suppose $(y_n)$ is Cauchy in $Y$ and $y_n\rightarrow y+z\in X$ with $y\in Y$ and $0\ne z\in Z$. What can you say about the sequence $\bigl( {y_n-y\over\Vert y_n-y\Vert}\bigr)$? $\endgroup$ – David Mitra Jan 1 '15 at 16:31
  • $\begingroup$ Thank you! I got it! This sequence is eventually separated from $z/\lVert y_n-y\rVert$ by $r/2>0$ what contradicts with $\frac{y_n-y-z}{\lVert y_n-y\rVert}\rightarrow 0$. $\endgroup$ – quartermind Jan 1 '15 at 17:32
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The solution rises from David Mitra's comment.

Suppose there is a sequence $(y_n)$ in $Y$ such that $y_n \rightarrow y+z\in X$ with $y \in Y$ and $0 \neq z\in Z$. Then we have $\lVert\frac{y_n-y}{\lVert y_n-y\rVert}-\frac{z}{\lVert z\rVert}\rVert \geq r > 0$ and, by continuity of norm, for sufficiently large $n$ $\lVert y_n-y-z \rVert = \lVert y_n-y\rVert\cdot\lVert\frac{y_n-y}{\lVert y_n-y\rVert}-\frac{z}{\lVert y_n-y\rVert}\rVert \geq \lVert z\rVert r-\varepsilon > 0$, what is contradiction.

Therefore $Y$ is closed. By the same argument, $Z$ is closed too.

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