Analysis Problem of the Day 69

Today’s Problem appeared as Problem 3 on the UCLA Fall 2023 Analysis Qual:

Problem 3. Let $U,V$ be closed subspaces of a real Hilbert space such that $\sup_{\|u\|=\|v\|=1} \langle u,v \rangle <1,$ and let $W=U+V.$

a) Show that $W=U \oplus V.$

b) Show that $W$ is a closed subspace.

c) Show that there exists a bounded linear map $T:W \to U$ such that $w-Tw \in V$ for all $w \in W.$

Solution: a) Note that the assumptions of the problem imply that $|\langle u,v \rangle| \leq c \|u\|\|v\|$ for some $0<c<1$ and all $u \in U, v \in V.$ It suffices to show that $U \cap V = \{0\}.$ Indeed, if the intersection is nonempty, there exists a unit vector $w \in U \cap V.$ Then,

$1=\|w\|^2 = \langle w,w \rangle \leq c <1,$

since $w \in U$ and $w \in V,$ which is a contradiction. Thus, $U \cap V = \{0\},$ so that $W = U \oplus V.$

b) I claim that for all $\epsilon >0$ there exists $\delta>0$ such that $\|u\|,\|v\|<\delta$ whenever $\|u-v\|<\epsilon, u \in U, v \in V.$ This yields the claim, since if $u_n-v_n \to y, u_n \in U, v_n \in V,$ then $u_n-v_n$ is Cauchy, and $\|u_n-v_n\|<\delta$ for large enough $n$ implies $\|u_n\|,\|v_n\|<\epsilon,$ i.e. $u_n,v_n$ are Cauchy sequences. Then, since $U,V$ are closed, $u_n \to u \in U, v_n \to v \in V,$ and therefore $y=u-v \in W,$ i.e. $W$ is closed (it is clear that $W$ is a subspace). Now, to show the claim, note that applying Cauchy-Schwarz yields

$\|u-v\|^2 =\|u\|^2+\|v\|^2-2 \langle u,v \rangle \implies \|u\|^2+\|v\|^2 \leq \|u-v\|^2+2c\|u\|\|v\| \leq \|u-v\|^2+c(\|u\|^2+\|v\|^2),$

which implies

$\|u\|^2+\|v\|^2 \leq \frac{1}{1-c}\|u-v\|^2,$

i.e. $\|u-v\|<\delta \implies \|u\|,\|v\| \leq \sqrt{\frac{1}{1-c}}\delta,$ as desired. Thus, $W$ is a closed subspace of the real Hilbert space.

c) Note that for $w=u+v, Tw = u$ satisfies $w-Tw = v \in V$ for all $w \in W.$ It is clear that $T(w_1+w_2)=u_1+u_2=Tw_1+Tw_2$ and $T(cw)=cTw,$ so the map $T$ is linear. It remains to show that $T: W \to U$ is bounded, for which we appeal to the closed graph theorem. Indeed, if $w_n \to w, Tw_n=u_n \to y, w_n = u_n+v_n, w= u+v,$ by (b) it follows that $u_n \to u,$ and by uniqueness of limits one has $y=u=Tw.$ Thus, by the closed graph theorem, $T$ is a bounded map (note that $T$ is just the orthogonal projection from $W$ onto $V.$ )

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