Q:

Norm of the sum of orthogonal projections

Let $H_\lambda$ be a closed subspace of a $\mathbb R$-Hilbert space $H$ for $\lambda\ge0$ and assume that $(H_\lambda)_{\lambda\ge0}$ is nondecreasing and right-continuous, i.e. $$\bigcap_{\mu>\lambda}H_\mu=H_\lambda\;\;\;\text{for all }\lambda\ge0.$$ Let $\pi_\lambda$ denote the projection of $H$ onto $H_\lambda$ for $\lambda\ge0$.
Now, let $k\in\mathbb N$, $0<\lambda_1<\cdots<\lambda_k$ and $\alpha_0,\ldots,\alpha_k\in\mathbb R$. Moreover, let $x\in H$ and $$y:=\alpha_0\pi_{\lambda_1}x+\sum_{i=2}^k\alpha_i\left(\pi_{\lambda_i}-\pi_{\lambda_{i-1}}\right)x.$$ Are we able to show that $$\left\|y\right\|_H^2=\alpha_0^2\left\|\pi_{\lambda_1}x\right\|_H^2+\sum_{i=2}^k\alpha_i^2\left\|\left(\pi_{\lambda_i}-\pi_{\lambda_{i-1}}\right)x\right\|_H^2?$$
It's clear to me that $\pi_{\lambda_i}-\pi_{\lambda_{i-1}}$ is again an orthogonal projection for all $i\in\left\{2,\ldots,k\right\}$.

A:

There's a lot that can be simplified away from this question. Given that you're only considering finitely many of your nested family of closed subspaces, we might as well just assume we have closed subspaces
$$H_1 \subseteq H_2 \subseteq \ldots \subseteq H_k,$$
with projection $\pi_i$ onto $H_i$.
Also, you don't seem to use $\alpha_1$, so I'm going to rename $\alpha_0$ to $\alpha_1$.
Recall that orthogonal projections are self-adjoint and idempotent. Given $i < j$, since $H_i \subseteq H_j$, we have that $\pi_j \pi_i = \pi_i$. Since the projections are self-adjoint, we also get $\pi_i \pi_j = (\pi_j \pi_i)^* = \pi_i$.
Hence,
$$\pi_i \pi _j = \pi_j \pi_i = \pi_i$$
whenever $i < j$. In fact, it holds even when $i = j$.
Using Pythagoras's theorem, it suffices to show that
$$\pi_1 x, (\pi_2 - \pi_1)x, (\pi_3 - \pi_2)x, \ldots, (\pi_k - \pi_{k - 1})x$$
are pairwise orthogonal. If $2 \le i < j$, then
\begin{align*}
\langle (\pi_i - \pi_{i - 1})x, (\pi_j - \pi_{j - 1})x\rangle &= \langle(\pi_j - \pi_{j - 1})(\pi_i - \pi_{i - 1})x, x \rangle\\
&= \langle(\pi_j\pi_i + \pi_{j - 1} \pi_{i - 1} - \pi_j \pi_{i - 1} - \pi_{j - 1} \pi_i)x, x\rangle \\
&= \langle(\pi_i + \pi_{i - 1} - \pi_{i - 1} - \pi_i)x, x\rangle \\
&= 0.
\end{align*}
Also, given $i \ge 2$, we have
$$\langle (\pi_i - \pi_{i - 1}) x, \pi_1 x \rangle = \langle \pi_1(\pi_i - \pi_{i - 1}) x, x \rangle = \langle (\pi_1 - \pi_1) x, x \rangle = 0.$$
The vectors are all orthogonal, and the result you want holds by Pythagoras's theorem.

