Let $H$ be a $\mathbb R$-Hilbert space. If $(\mathcal D(A),A_i)$ is a symmetric linear operator on $H$, write $A_1\le A_2$ if $$\langle A_1x,x\rangle_H\le\langle A_2x,x\rangle_H\;\;\;\text{for all }x\in\mathcal D(A)\tag1.$$ Let $(H_\lambda)_{\lambda\ge0}$ with
- $H_\lambda$ is a closed subspace of $H$ for all $\lambda\ge0$
- $(H_\lambda)_{\lambda\ge0}$ is nondecreasing and right-continuous, i.e. $$\bigcap_{\mu>\lambda}H_\mu=H_\lambda\tag2$$
Now, let $\pi_\lambda$ denote the orthogonal projection of $H$ onto $H_\lambda$ for $\lambda\ge0$. How can we show that $$[0,\infty)\ni\lambda\mapsto\pi_\lambda\tag3$$ is nondecreasing (in the sense of $(1)$) and $$[0,\infty)\ni\lambda\mapsto\pi_\lambda x\tag4$$ is right-continuous for all $x\in H$ (are we even able to show that $(3)$ is right-continuous?)?
Intuitively, it seems to be trivial, but how can we prove it?
Best Answer
We first show that for all $t\ge s$, $$ \pi_t\pi_s=\pi_s\pi_t = \pi_{s}. $$ We can observe that $H_s\le H_t$, which implies that $$ \text{Im}\ \pi_s =H_s \le H_t = \ker(1-\pi_t), $$ and $$ \text{Im}\ (1-\pi_t) = H_t^\perp \le H_s^\perp = \ker \pi_s. $$ This implies $(1-\pi_t)\pi_s = \pi_s(1-\pi_t) = 0$, i.e. $\pi_t\pi_s=\pi_s\pi_t = \pi_s$.
It is a direct consequence that for $t\ge s$, $$ \langle x,\pi_s x\rangle =\|\pi_sx\|^2=\|\pi_s\pi_tx\|^2\le \|\pi_tx\|^2 =\langle x,\pi_tx\rangle, $$ since $\|\pi_s\|\le 1$ for every $s\ge 0$.
It remains to prove right continuity of $t\mapsto \pi_t x$ for every $x$. Let $x\in H$ and $s\ge 0$ be arbitrarily given. We can find that for $t\ge s$, $$ f(t) =\|\pi_{t}x-\pi_s x\|^2=\langle x,\pi_{t}x-\pi_s x\rangle \ge 0 $$ is a decreasing function of $t\in[s,\infty)$. Thus there is a finite limit $\lim_{t\downarrow s}f(t)$, which implies for each and every decreasing sequence $t_n \downarrow s$, $$ \|\pi_{t_n}x-\pi_{t_m}x\|^2 =\langle x,\pi_{t_n}x-\pi_{m} x\rangle =f(t_n)-f(t_m)\xrightarrow{n,m\to\infty} 0, $$ i.e. $(\pi_{t_n}x)_{n\ge 1}$ forms a Cauchy sequence. If we denote $y=\lim_{n\to\infty}\pi_{t_n}x$, then we can find that $$ \|\pi_{t_n}x-\pi_s x\|^2=\|\pi_{t_n}x-\pi_s \pi_{t_n}x\|^2\xrightarrow{n\to\infty} \|y-\pi_s y\|^2\ge 0. $$ Finally, from the fact that $\pi_{t_n}x \in H_t$, $t>s$ for all sufficiently large $n$'s, we can deduce that $$ y=\lim_{n\to\infty}\pi_{t_n}x \in H_t,\quad \forall t>s, $$ hence $$ y\in \bigcap_{t>s}H_t =H_s. $$ This proves $$ \|\pi_{t_n}x-\pi_s x\|^2\xrightarrow{n\to\infty} \|y-\pi_s y\|^2=0 $$ for all decreasing sequence $(t_n)_{n\ge 1}$ with $\lim_{n\to\infty} t_n =s$. Since $s$ was arbitrary, the path $t\mapsto \pi_t x$ is right continuous for every $x$.