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let $M$ and $N$ be closed linear subspace of a Banach space $X$ such that $M \ \cap N=\{0\}.$

prove that $M\oplus N$ is closed in $X$ if and only if there is a constant $K>0\ $ such that $||x-y||\geq K \ \forall \ x\in M,y \ \in N$ with $||x||=||y||=1.$

any suggestion how to approach

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Sufficiency is quite straightforward to prove. First of all note that the inequality gives you that for any $x \in M$ and $y \in N\setminus \{0\}$ we have $\|x\| K \leq \|x+ \frac{\|x\|}{\|y\|} y \|$. But by the triangle inequality and the reverse triangle inequality, we have $$ \|x+ \frac{\|x\|}{\|y\|} y \| \leq \|x+y\|+ \| y - \frac{\|x\|}{\|y\|} y \| \leq 2 \| x+y \|. $$

Take a sequence $(z_n)_n$ in $M \oplus N$ that converges to some $z \in X$. Then write it as $z_n = x_n +y_n$ with $x_n \in M$ and $y_n \in N$. Then it's easy to prove that $(x_n)_n$ is a Cauchy sequence by using the above property.

The converse is proven by looking at the map $$ T: M\oplus N \to M: (x,y) \mapsto x. $$ Because both $M\oplus N$ and $M$ are closed, they must be Banach spaces. The closed graph theorem will provide that $T$ is continuous. So there is a constant $A>0$ such that $\|x\| =\|T(x,y)\| \leq A\|x+y \|$.

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  • $\begingroup$ why is $||x ||K \leq \ || x+y||$, with $|| y||=1$? $\endgroup$ – Eklavya Oct 22 '17 at 13:57
  • $\begingroup$ @Eklavya I edited my answer, what I wrote before probably wasn't entirely correct. $\endgroup$ – Demophilus Oct 22 '17 at 16:19
  • $\begingroup$ from where 2 come from in expression $2|| x+y||$?? $\endgroup$ – Eklavya Oct 22 '17 at 16:37
  • $\begingroup$ It comes from the reverse triangle inequality: $\| y - \frac{\|x\|}{\|y\|} y \| = \lvert \|y \| - \|x\| \rvert \leq \| x+y\|$. $\endgroup$ – Demophilus Oct 22 '17 at 16:54
  • $\begingroup$ By closed graph theorem, we only have $\lVert T(x,y)\rVert\leq A(\lVert x\rVert+\lVert y\rVert).$ What's the reason for the last line? $\endgroup$ – C.Ding Oct 23 '17 at 0:39

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