Let $\mathcal{H}$ a complex Hilbert space (with inner product $(\cdot, \cdot)$ conjugate linear in the second argument). Denote the set of bounded operators on $\mathcal{H}$ by $L(\mathcal{H})$.
Let $S,T \in L(\mathcal{H})$. Say $S$ is positive if $\forall \psi \in \mathcal{H}: ( S\psi, \psi) \geq 0$. We denote this by $S \geq 0$. Say $S \geq T \iff S - T \geq 0$.
1. Uniformly Bounded Monotone Sequence of Self-Adjoint Operators are Strongly Convergent
Proposition: $\leq$ is a partial order on $L(\mathcal{H})$.
Proof: Reflexivity is apparent. Antisymmetry is straightforward. Suppose $A \leq B, B \leq A$. Then $((A-B) \psi, \psi) \geq 0$ and $((B-A) \psi, \psi) = - ((A-B)\psi,\psi) \geq 0$. Hence $((A-B)\psi,\psi) = 0$ for all $\psi$, hence $A-B = 0$. Now we show transitivity. Suppose $A \leq B, B \leq C$. Let $\psi \in \mathcal{H}$. Then $((C-A)\psi,\psi) = ((C-B) \psi, \psi) + ((B-A)\psi,\psi) \geq 0$, so $C \geq A$.
Theorem: Let $T_n$ a sequence of bounded, self-adjoint operators s.t. $T_n \leq T_{n+1}$ for all $n$ and s.t. there exists some $C>0$ s.t. $\|T_n\| \leq C$ for all $n$. Then $T_n \to T$ strongly for some self-adjoint, bounded $T$.
Proof:
Claim 1: $\forall \psi \in \mathcal{H}: (T_n \psi, \psi)$ is a convergent sequence in $\mathbb{R}$.
Proof of claim 1: Self-adjointness gives that $(T_n \psi, \psi) \in \mathbb{R}$. Note also that $|(T_n \psi, \psi)| \leq \|T_n \psi\| \|\psi\| \leq \|T_n\| \|\psi\|^2 \leq C \|\psi\|^2$, so $(T_n \psi, \psi)$ forms a bounded sequence in $\mathbb{R}$. Every bounded monotone sequence is convergent by a result in elementary analysis, so it is sufficient to show this sequence is monotone. $(T_{n+1} \psi, \psi) - (T_n \psi, \psi) = ((T_{n+1} - T_n) \psi, \psi) \geq 0$, by monotonicity. So the sequence is monotone, hence convergent, so the claim holds.
Claim 2: $\forall \psi, \phi \in \mathcal{H}: (T_n \psi, \phi)$ is a convergent sequence in $\mathbb{C}$.
Proof of claim 2: Direct computation verifies the so-called Polarization Identity: $4(T_n x,y) = (T_n(x+y), x+y) - (T_n(x-y), x-y) + i (T_n(x+iy), x+iy) - i (T_n(x-iy), x-iy)$. Thus convergence on the diagonals (i.e. on terms of the form $(T_n \psi, \psi)$) guarantees convergence in general, thus proving the claim.
So we have that $(T_n x, y) \to F_{x,y} \in \mathbb{C}$ for each $x,y \in \mathcal{H}$ and that the limit is unique. So we define a function $F : \mathcal{H}^2 \to \mathbb{C}$ by $F(x,y) = F_{x,y}$. Consider the function $\phi_z: \mathcal{H} \to \mathbb{C}$ given by $\phi_z(x) = \overline{F(z,x)}$. It is easy to check that this define a linear functional. Also $|\phi_z(x)| = |F(z,x)| = \lim |(T_nz,x)| \leq C \|T_n z\| \|x\| \leq C \|z\| \|x\|$, where we use the Cauchy-Schwartz inequality and the uniform boundedness of the $T_n$. Hence for each $z$, $\phi_z$ is a bounded linear functional, so the Reisz representation theorem gives that there is some unique map $T: \mathcal{H} \to \mathcal{H}$ s.t. $(x, Tz) = \phi_z(x) = \overline{F(z,x)}$. Hence $(Tz,x) = F(z,x)$. A straightforward computation then verifies that the map $T$ is linear (exploiting the uniqueness condition on the Reisz representation theorem).
Now we want to show that $T$ is self adjoint and bounded. $\|Tz\|^2 =|(Tz, Tz)| = |F(z, Tz)| = \lim |(T_nz, Tz)| \leq C \|z\| \|Tz\|$. So $\|Tz\| \leq C \|z\|$, hence $\|T\| \leq C$ and $T$ is bounded. $(Tx, z) = F(x,z) = \lim (T_n x, z) = \lim (x, T_nz) = \overline{\lim (T_n z,x)} = \overline{F(z,x)} = (x,Tz)$, so $T$ is self-adjoint.
Then note that $T_n \to T$ in the weak operator topology, by construction. As $(T_n x, y) \to (Tx, y) $ for all $x,y \in \mathcal{H}$. We want to show that $T_n \to T$ strongly.
Claim 3: $T- T_n \geq 0$ for all $n$.
Proof of claim 3: $((T - T_n)x ,x) = (Tx, x) - (T_nx,x) = \lim_{m \to\infty} (T_m x, x) - (T_nx, x) = \lim_{m\to\infty} ((T_m-T_n)x,x) \geq 0$, since this sequence is eventually $\geq 0$ as for any $m \geq n$, $T_m - T_n$ is positive by the monotonicity assumption.
We now utilize the functional calculus for bounded self-adjoint operators (e.g. see B.C. Hall, Quantum Theory for Mathematicians, Chapters 7 and 8). We want to be able to take square roots; for this to work the way we want, we need that the spectrum is positive, leading to the following claim:
Claim 4: Let $A$ a bounded, self-adjoint, positive operator, then the spectrum of $A$, $\sigma(A)$, is a subset of $[0,\infty)$.
Proof of claim 4: Suppose $\lambda \in \sigma(A)$. Note that we know that $\lambda \in \mathbb{R}$ as $A$ is self-adjoint (e.g. see B.C. Hall, Quantum Theory for Mathematicians, Proposition 7.7). Suppose $\lambda < 0$. Also by Hall, proposition 7.7, we have a sequence $x_n \in \mathcal{H}$ s.t.\ $\lim \frac{\|(A - \lambda I)\psi_n\|}{\|\psi_n\|} = 0$. Note that we can only have $\|\psi_n\| = 0$ for finitely many $\psi$. WLOG then suppose none of the $\psi_n =0$. Then let $\phi_n = \psi_n/\|\psi_n\|$. Then $\frac{\|(A - \lambda I) \phi_n\|}{\|\phi_n\|} = \frac{\|(A- \lambda I)\psi_n\|}{\|\psi_n\|}$, so we can WLOG suppose $\|\psi_n\| = 1$ for all $n$. Then we have $\|(A - \lambda I)\psi_n\| \to 0$ for some unit vectors $\psi_n$. Then $((A- \lambda I)x, x) = (Ax,x) - \lambda\|x\|^2 \geq - \lambda \|x\|^2 = |\lambda|\|x\|^2$ for any $x \in \mathcal{H}$, using that $A$ is positive and that $\lambda <0$. Thus $((A-\lambda I) x, x) = |((A-\lambda I)x,x)| \leq \|(A-\lambda I)x\|\|x\|$. Putting it together for $x = \psi_n$ yields $\|(A- \lambda I) \psi_n\| \geq |\lambda|\|\psi_n\|^2/\|\psi_n\| = |\lambda| > 0$, a contradiction. So the claim is shown.
So now we use the continuous functional calculus for bounded, self-adjoint operators (Hall, Proposition 8.4), which provides a map from $C(\sigma(A), \mathbb{R}) \to L(\mathcal{H})$ where $\sigma(A)$ is the spectrum of $A$. In particular, for $A$ positive, $\sigma(A) \subseteq [0,\infty)$ and so $x \mapsto \sqrt{x}$ is a well-defined continuous function in $C(\sigma(A), \mathbb{R})$. Then directly using the properties given by 8.4, we have get that positive self-adjoint bounded operators have positive self-adjoint bounded square roots.
Define $S_n := T - T_n$. Claim 3 shows that $S_n \geq 0$. Clearly as the sum of bounded operators $S_n$ is bounded. Then $S_n^* = T^* - T_n^* = T - T_n = S_n$, so $S_n$ is self-adjoint. Thus the proceeding discussion yields a positive bounded self-adjoint square root $R_n$.
Claim 5: If $A$ is self-adjoint and $\|A\| \leq M$, then $A \leq M \cdot I$.
Proof of Claim 5: Note that $\overline{(Ax, x)} = (x,Ax) = (Ax,x)$, so $(Ax,x) \in \mathbb{R}$ for all $x \in \mathcal{H}$. Then $((M\cdot I - A)x, x) = M \|x\|^2 - (Ax, x) \geq M \|x\|^2 - |(Ax,x)| \geq M \|x\|^2 - \|Ax\|\|x\| \geq M \|x\|^2 - M \|x\|^2 = 0$ for all $x \in \mathcal{H}$.
Then we have that $\|S_n\| \leq \|T\| + \|T_n\| \leq 2C$, hence $S_n \leq 2C \cdot I$. Then we have that $\|S_nx\|^2 = (S_nx,S_nx) = (S_n^2x,x) = (R_n S_n R_n x,x) = (S_n R_n x, R_n x)$. Then note that $2C \cdot I - S_n \geq 0$, hence $0 \leq ((2C \cdot I - S_n) R_n x , R_n x) = 2C (R_n x, R_n x) - (S_n R_n x, R_nx) = 2C(R_n^2 x, x) - (S_n R_n x, R_n x) = 2C(S_n x, x) - (S_n R_n x, R_n x)$. Hence $\|S_nx\|^2 = (S_n R_n x, R_n x) \leq 2 C (S_n x,x)$. Thus $0 \leq \|S_nx\|^2 \leq 2C (S_n x,x)$, and $(S_nx,x) = ((T_n - T)x, x) \to 0$ as $T_n \to T$ weakly, hence $\|S_nx\|^2 \to 0$, so $\|S_n x\| \to 0$. Thus $T_n x \to Tx$ in the norm topology for all $x$, i.e. $T_n \to T$ strongly.
QED.
So we have the monotone convergence result for bounded operators. We now want to apply that to orthogonal projection, some simple facts about which we now recall.
2. Countable Sum of Orthogonal Projections with Pairwise Orthogonal Ranges is the Orthogonal Projection onto the Closed Span of the Ranges
Let $M \subseteq \mathcal{H}$ a closed vector subspace. Then denote $M^\perp := \{x \in \mathcal{H} \mid \forall y \in M:(x,y) = 0\}$. $H = M \oplus M^\perp$. For each $x \in \mathcal{H}$ we can then write $x = y +z$ where $y \in M, z \in M^\perp$ and this decomposition is unique. The map sending $x \mapsto y$ in the above decomposition is a bounded linear operator, denote it $P_M$, and has the following properties: (i) $P_M^2 = P_M$ and (ii) $P_M^* = P_M$. Note that $Range(P_M) = M$. We call $P_M$ the orthogonal projection onto $M$. It is an elementary result that if we have a bounded operator $P$ s.t.\ $P^2 = P, P^* = P$, then $P$ is the orthogonal projection onto its range.
Proposition: If $P_1, P_2$ are orthogonal projections with orthogonal ranges, then $P_1 + P_2$ is the orthogonal projection onto the span of their ranges.
Proof: $(P_1 + P_2)^2 = P_1^2 + P_2^2 + P_1P_2 +P_2P_1 = P_1 + P_2 +P_1P_2 + P_2P_1$, so we want show that $P_1P_2 = P_2P_1 =0$. Let's do the $P_1P_2$ case the other follows similarly. Let $x \in \mathcal{H}$. Denote the range of $P_1$ by $M_1$ and the range of $P_2$ by $M_2$. Then $M_1$, $M_2$ are closed vector subspaces. So $\mathcal{H} = M_1 \oplus M_1^\perp$. By assumption $M_2 \subseteq M_1^\perp$. We can then decompose $M_1^\perp = M_2 \oplus M_2^\perp \cap M_1^\perp$. Hence $\mathcal{H} = M_1 \oplus M_2 \oplus M_1^\perp \cap M_2^\perp$. Then $x = a+b +c$ where we decompose as in the direct sum. Then $P_2 x = b$ and $P_1P_2 x= P_1 b =0$. Thus $P_1P_2 = 0$. Similarly $P_2P_1 =0$. So $P_1 + P_2$ is idempotent. As the sum of self adjoint operators, $P_1 + P_2$ is self adjoint. So $P_1 + P_2$ is an orthogonal projection onto it range. So we just need to show that range of $P_1 + P_2$ is the span of $M_1$ and $M_2$. First it's clear that $Range(P_1 + P_2) \subseteq M_1 + M_2$ as $(P_1+P_2)x = P_1x +P_2x$ where $P_1x \in M_1$ and $P_2x \in M_2$. Since $Range(P_1+P_2)$ is a vector subspace, it is sufficient to show that $M_1,M_2 \subseteq Range(P_1 + P_2)$. We'll do the $M_1$ case, the $M_2$ case follow similarly. Let $x \in M_1$. Then $x \in M_2^\perp$, hence $P_2 x = 0$. Thus $(P_1 + P_2)x = P_1 x = x$. So $x \in Range(P_1 + P_2)$. So $M_1 \subseteq Range(P_1+P_2)$. Thus the claim follows.
Proposition: If $P_1,...,P_n$ are a finite collection of orthogonal projections with pairwise orthogonal ranges, then $P_1 + \cdots + P_n$ is an orthogonal projection onto the span of the ranges of the $P_j$.
Proof: Suppose the result has been shown for $P_1 + \cdots + P_j$ for $j < n$.
Claim: $Range(P_1 + \cdots + P_j) \perp Range(P_{j+1})$.
Proof of claim: We know that $Range(P_1+\cdots+ P_j) = Range(P_1)+\cdots + Range(P_j)$ by the inductive hypothesis. So let $x \in Range(P_1+\cdots+ P_j)$. Then $x = x_1 + \cdots + x_j$ where $x_i \in Range(P_i)$. Then let $y \in Range(P_{j+1})$. By pairwise orthogonality, $(y,x_i) =0 $ for each $i$. Hence $(y,x) = 0$, and so the claim holds.
Then we can use the result for the sum of two projections to get the result for $P_1 + \cdots + P_{j+1}$, completing the inductive step.
Proposition: If $\{P_j\}_1^\infty$ is a sequence of orthogonal projections with pairwise orthogonal ranges, then $S_n := \sum_{j=1}^n P_i$ defines a sequence of orthogonal projections s.t. $S_1 \leq S_2 \leq \cdots$.
Proof: The previous proposition gives that the $S_j$ are orthogonal projections, so we just have to show the monotonicity. Note that $S_{j+1} - S_j = P_{j+1}$, so really we just have to show that orthogonal projections are positive operators. But this is clear: let $P$ an orthogonal projection with range $M$. Write $x = y+z$ where $y \in M, z \in M^\perp$. Then $(Px, x) = (y,y+z) = (y,y) + (y,z) = (y,y) = \|y\|^2 \geq 0$. So orthogonal projections are positive operators and so the monotonicity condition holds.
Proposition: The strong limit of orthogonal projections is an orthogonal projection.
Proof: Suppose $\{P_j\}_1^\infty$ is a sequence of orthogonal projections s.t.\ $P_j \to P$ strongly. We want to show that $P$ is an orthogonal projection, for which it is sufficient to show that $P^2 = P = P^*$. Let $x,y \in \mathcal{H}$ arbitrary. Then $(P^*x, y) = (x,Py) = \lim_j (x,P_jy) = \lim_j (P_jx, y) = (Px,y)$. So $((P^*-P)x,y) = 0$ for all $x,y \in \mathcal{H}$. So $P = P^*$. Then $(P^2 x, y) = (Px,Py) = \lim_j (P_j x, P_j y) = \lim_j (P_j^2 x, y) = \lim_j (P_j x, y) = (Px,y)$. So as above $P^2 = P$. Hence the proposition holds.
Theorem: If $\{P_j\}_1^\infty$ is a sequence of orthogonal projections with pairwise orthogonal ranges, then $\sum_1^\infty P_j = P$ with convergence in the strong operator topology where $P$ is the orthogonal projection onto the closed span of the ranges of the $P_j$.
Proof: Let $S_j$ denote the partial sums as in the previous proposition. Note that the $S_j$ are projections and that projections have norm at most 1: if $P$ is an orthogonal projection with range $M$, then for any $x \in \mathcal{H}$, we have $x = x_1 + x_2$ where $x \in M, x_2 \in M^\perp$. Then $\|Px\| = \|x_1\|$ and $\|x\|^2 = \|x_1\|^2 + \|x_2\|^2$ (by the Pythagorean theorem), $\geq \|x_1\|^2$, sp $\|x_1\| \leq \|x\|$, so $\|Px\| \leq \|x\|$. Thus $\|P\| \leq 1$. Thus by the previous proposition, we have monotonicity and we just showed uniform boundedness of the $S_j$, hence by the previous theorem on monotone convergence of self-adjoint operators, we have the $S_j \to P$ for some bounded self-adjoint $P$ with convergence in the strong operator topology. Then by the previous proposition, we have that $P$ is an orthogonal projection. Thus to show the theorem, it is sufficient to show that the range of $P$ is the closed span of the ranges of $P_j$, call this closed span $M$.
Let $x \in \mathcal{H}$. Then $Px = \lim_j S_j x$. Then each $S_j x \in M$ as is easy to see. Then by closedness $Px \in M$. Hence $Range(P) \subseteq M$. Note also that $Range(P)$ is a closed vector subspace, so to show that that $M$, the closed span of the $Range(P_j)$ is in $Range(P)$, it is sufficient to show that $Range(P_j) \subseteq M$ for each $j$. Thus this is clear, since for any $x \in Range(P_j)$, $P_i x = x$ if $i =j$ and $P_i x = 0$ if $ i\neq j$ (by pairwise orthogonality of the ranges). Hence for $i > j$, $S_i x = x$. Thus $Px = \lim_i S_i x =x$, hence $Range(P_j) \subseteq Range(P)$. Thus the theorem holds.
