Exercise 7, § 4

a) The notations being those of Exer. 6, now prove that, in order that $M + N^{\circ}$ be closed, it is necessary and sufficient that $M + N^{\circ} = (M^{\circ} \cap N)^{\circ}$. b) Now prove that the following properties are equivalent: α) $\varepsilon(M, N) < 1$; β) $M + N^{\circ}$ is closed in E ; γ) If $\tilde{M}$ is the orthogonal supplement of $M \cap N^{\circ}$ in M and $(N^{\circ})^{\sim}$ that of $M \cap N^{\circ}$ in $N^{\circ}$, E is direct sum of $M^{\circ} \cap N$, $M \cap N^{\circ}$, $\tilde{M}$ and $(N^{\circ})^{\sim}$. Moreover, if R and S are the projectors of E onto $\tilde{M}$ and $(N^{\circ})^{\sim}$ respectively corresponding to this decomposition, one has $\|R\| = \|S\| = (1 - \varepsilon^2(M, N))^{1/2}$. (To see that α) implies β) first remark that if $x \in \tilde{M}$ and $y \in (N^{\circ})^{\sim}$ one has $|\langle x|y \rangle| \leq \varepsilon(M, N) \|x\|.\|y\|$; let then $u = x + y + t$ be the decomposition of an element $u \in M + N^{\circ}$ with $x \in M$, $y \in (N^{\circ})^{\sim}$ and $t \in M \cap N^{\circ}$, deduce that $\|x\| \leq (1 - \varepsilon^2(M, N))^{-1/2} \|u\|$, $\|y\| \leq (1 - \varepsilon^2(M, N))^{-1/2} \|u\|$ and $\|t\| \leq \|u\|$. To prove that γ) implies α), consider, for a $v \in \tilde{M}$, the decomposition $v = v_1 + v_2$, where $v_1$ is the orthogonal projection of $v$ onto N, supplement of $N \cap M^{\circ}$ in N ; one has $R.v_1 = v$, and if one had $\varepsilon(M, N) = 1$, there would exist a sequence $(v_n) \in \tilde{M}$ such that $\|v_n\|=1$ and that $\|(v_n)_1\|$ tends to 0. Now prove that the restriction $R_1$ of $R$ to $\tilde{N}$ is a bijection of $\tilde{N}$ onto $\tilde{M}$; to calculate $\|R\|$, now prove that $\|R_1^{-1}\|\leq(1-\varepsilon^2(M,N))^{1/2}.$) c) Deduce from b) that if $M+N^\circ$ is closed, the same is true of $M^\circ+N$.