[proofplan] The strategy is to bound the partial sums uniformly by an iterated Cauchy–Schwarz argument that exploits both halves of the almost-orthogonality hypothesis. For any $m \in \mathbb{N}$ and any finite index set $J \subset \mathbb{Z}$, we expand the operator $(S_J^* S_J)^m$ where $S_J := \sum_{j \in J} T_j$, take the operator norm, and bound it by a sum over length-$2m$ index tuples. We then estimate each tuple by alternating between the two hypotheses (factoring as $\|T_a^* T_b\|^{1/2}$ and $\|T_b T_c^*\|^{1/2}$), giving telescoping factors of $\gamma$. The $2m$-th root and the spectral radius formula (which for a positive self-adjoint bounded operator $A$ on a Hilbert space gives $\|A^m\|_{\mathcal{L}(H)} = \|A\|_{\mathcal{L}(H)}^m$, a consequence of the $C^*$-identity $\|A\|^2 = \|A^*A\|$ and induction on $m$) give a bound independent of $m$ and of the cutoff $J$. Strong convergence of $S_N$ as $N \to \infty$ then follows by a density argument and a Cauchy criterion on the increments. [/proofplan] [step:Set notation and reduce to the operator-norm bound on finite partial sums] For a finite subset $J \subset \mathbb{Z}$, set \begin{align*} S_J: H &\to H \\ x &\mapsto \sum_{j \in J} T_j x. \end{align*} This is a finite sum of bounded operators, hence bounded. Our main task is to show \begin{align*} \|S_J\|_{\mathcal{L}(H)} \le A \quad \text{for every finite } J \subset \mathbb{Z}. \end{align*} Writing $S_N = S_{\{j : |j| \le N\}}$, the bound $\|S_N\|_{\mathcal{L}(H)} \le A$ then holds uniformly in $N$, after which strong convergence (Step 5) and the limit bound $\|T\|_{\mathcal{L}(H)} \le A$ will follow from the uniform estimate. [/step] [step:Expand $(S_J^* S_J)^m$ and bound its operator norm by a sum over index tuples] Fix a positive integer $m \in \mathbb{N}$. Since $S_J^* S_J$ is a positive self-adjoint bounded operator on $H$, the operator $(S_J^* S_J)^m$ is also positive and self-adjoint. Expanding $S_J = \sum_{j \in J} T_j$ and $S_J^* = \sum_{j \in J} T_j^*$, \begin{align*} (S_J^* S_J)^m = \sum_{j_1, k_1, \dots, j_m, k_m \in J} T_{j_1}^* T_{k_1} T_{j_2}^* T_{k_2} \cdots T_{j_m}^* T_{k_m}. \end{align*} Taking operator norms and applying the triangle inequality, \begin{align*} \|(S_J^* S_J)^m\|_{\mathcal{L}(H)} \le \sum_{j_1, k_1, \dots, j_m, k_m \in J} \|T_{j_1}^* T_{k_1} T_{j_2}^* T_{k_2} \cdots T_{j_m}^* T_{k_m}\|_{\mathcal{L}(H)}. \end{align*} [/step] [step:Estimate each length-$2m$ product by alternating between the two almost-orthogonality bounds] Fix indices $j_1, k_1, \dots, j_m, k_m \in J$ and write the product as \begin{align*} P := T_{j_1}^* T_{k_1} T_{j_2}^* T_{k_2} \cdots T_{j_m}^* T_{k_m}. \end{align*} We bound $\|P\|_{\mathcal{L}(H)}$ in two complementary ways and take the geometric mean of the resulting estimates. **Estimate (a): pair adjacent factors as $T_a T_b^*$ on the inside.** Group $P$ as \begin{align*} P = T_{j_1}^* \cdot (T_{k_1} T_{j_2}^*) \cdot (T_{k_2} T_{j_3}^*) \cdots (T_{k_{m-1}} T_{j_m}^*) \cdot T_{k_m}. \end{align*} Submultiplicativity of the operator norm and the bounds $\|T_j\|_{\mathcal{L}(H)} = \|T_j^*\|_{\mathcal{L}(H)} \le \|T_j^* T_j\|_{\mathcal{L}(H)}^{1/2} \le \gamma(0)$ on the outer factors (using the $C^*$-identity $\|T\|^2 = \|T^*T\|$, valid in $\mathcal{L}(H)$ for any Hilbert space $H$) combined with $\|T_{k_i} T_{j_{i+1}}^*\|_{\mathcal{L}(H)} \le \gamma(k_i - j_{i+1})^2$ from the second hypothesis give \begin{align*} \|P\|_{\mathcal{L}(H)} \le \gamma(0) \cdot \gamma(k_1 - j_2)^2 \cdot \gamma(k_2 - j_3)^2 \cdots \gamma(k_{m-1} - j_m)^2 \cdot \gamma(0). \end{align*} **Estimate (b): pair adjacent factors as $T_a^* T_b$ on the inside.** Group $P$ as \begin{align*} P = (T_{j_1}^* T_{k_1}) \cdot (T_{j_2}^* T_{k_2}) \cdots (T_{j_m}^* T_{k_m}), \end{align*} and bound each factor by $\|T_{j_i}^* T_{k_i}\|_{\mathcal{L}(H)} \le \gamma(j_i - k_i)^2$ from the first hypothesis. This yields \begin{align*} \|P\|_{\mathcal{L}(H)} \le \gamma(j_1 - k_1)^2 \cdot \gamma(j_2 - k_2)^2 \cdots \gamma(j_m - k_m)^2. \end{align*} Since $\|P\|$ is bounded by both the right-hand sides above, and the geometric mean of two non-negative real numbers does not exceed either, we have \begin{align*} \|P\|_{\mathcal{L}(H)} &\le \big[\gamma(0) \cdot \gamma(k_1 - j_2)^2 \cdots \gamma(k_{m-1} - j_m)^2 \cdot \gamma(0)\big]^{1/2} \\ &\quad \cdot \big[\gamma(j_1 - k_1)^2 \cdots \gamma(j_m - k_m)^2\big]^{1/2} \\ &= \gamma(0) \cdot \gamma(j_1 - k_1) \cdot \gamma(k_1 - j_2) \cdot \gamma(j_2 - k_2) \cdots \gamma(k_{m-1} - j_m) \cdot \gamma(j_m - k_m). \end{align*} The right-hand side is a product of $2m - 1$ values of $\gamma$ at consecutive index differences in the alternating chain $j_1, k_1, j_2, k_2, \dots, j_m, k_m$, multiplied by the prefactor $\gamma(0)$. [/step] [step:Sum over index tuples and take the $2m$-th root to bound $\|S_J\|$] Combining Steps 2 and 3, \begin{align*} \|(S_J^* S_J)^m\|_{\mathcal{L}(H)} \le \gamma(0) \sum_{j_1, k_1, \dots, j_m, k_m \in J} \gamma(j_1 - k_1)\,\gamma(k_1 - j_2)\,\gamma(j_2 - k_2) \cdots \gamma(j_m - k_m). \end{align*} We bound the right-hand sum by carefully tracking which index ranges over $J$ versus $\mathbb{Z}$. Substitute the difference variables $d_1 := j_1 - k_1$, $d_2 := k_1 - j_2$, $d_3 := j_2 - k_2$, $\dots$, $d_{2m-1} := j_m - k_m$, keeping $j_1$ as the free index that anchors the chain. The map \begin{align*} (j_1, k_1, j_2, k_2, \dots, j_m, k_m) \mapsto (j_1, d_1, d_2, \dots, d_{2m-1}) \end{align*} is a bijection of $\mathbb{Z}^{2m}$ onto itself, and the integrand becomes $\gamma(d_1)\,\gamma(d_2)\cdots\gamma(d_{2m-1})$, independent of $j_1$. In the original sum, $j_1$ ranges over $J$ (so the sum over $j_1$ contributes a factor $|J|$), and each of $k_1, j_2, k_2, \dots, k_m$ ranges over $J$, which translates into constraints on $d_1, d_2, \dots, d_{2m-1}$. Dropping these constraints by non-negativity (each $\gamma \ge 0$, so removing constraints can only increase the sum) and replacing the $d_i$-ranges by all of $\mathbb{Z}$: \begin{align*} \sum_{j_1, k_1, \dots, j_m, k_m \in J} \gamma(j_1 - k_1)\cdots\gamma(j_m - k_m) &\le \sum_{j_1 \in J}\sum_{d_1, \dots, d_{2m-1} \in \mathbb{Z}} \gamma(d_1)\,\gamma(d_2)\cdots\gamma(d_{2m-1}) \\ &= |J| \cdot \left(\sum_{d \in \mathbb{Z}} \gamma(d)\right)^{2m-1} = |J| \cdot A^{2m-1}, \end{align*} where $|J|$ is the cardinality of $J$ and $A = \sum_{d \in \mathbb{Z}} \gamma(d)$ by hypothesis. Thus \begin{align*} \|(S_J^* S_J)^m\|_{\mathcal{L}(H)} \le \gamma(0) \cdot |J| \cdot A^{2m-1}. \end{align*} On the other hand, $S_J^* S_J$ is a positive self-adjoint bounded operator, so its operator norm equals its spectral radius and \begin{align*} \|S_J\|_{\mathcal{L}(H)}^{2m} = \|S_J^* S_J\|_{\mathcal{L}(H)}^m = \|(S_J^* S_J)^m\|_{\mathcal{L}(H)}. \end{align*} Combining, \begin{align*} \|S_J\|_{\mathcal{L}(H)}^{2m} \le \gamma(0) \cdot |J| \cdot A^{2m-1}, \end{align*} hence \begin{align*} \|S_J\|_{\mathcal{L}(H)} \le A \cdot \big(\gamma(0) \cdot |J| / A\big)^{1/(2m)}. \end{align*} The factor $(\gamma(0)\,|J|/A)^{1/(2m)} \to 1$ as $m \to \infty$, so letting $m \to \infty$ gives \begin{align*} \|S_J\|_{\mathcal{L}(H)} \le A. \end{align*} This bound holds for every finite $J \subset \mathbb{Z}$, with constant $A$ independent of $J$. [/step] [step:Extract a bounded limit operator $T$ via a Cauchy argument on the partial sums] For each $x \in H$ we show that $(S_N x)_{N \ge 1}$ is a Cauchy sequence in $H$, and define $Tx := \lim_N S_N x$. **Reduction to Cauchy on a dense set.** Define \begin{align*} H_0 := \{x \in H : (S_N x)_{N \ge 1} \text{ is a Cauchy sequence in } H\}. \end{align*} We first show $H_0$ is a closed linear subspace of $H$. Linearity is immediate from linearity of each $S_N$. For closure, fix $x \in \overline{H_0}$ and $\eta > 0$. By Step 4, $\|S_N - S_M\|_{\mathcal{L}(H)} \le \|S_N\|_{\mathcal{L}(H)} + \|S_M\|_{\mathcal{L}(H)} \le 2A$. Choose $x_0 \in H_0$ with $\|x - x_0\|_H < \eta/(4A + 1)$, so that $\|(S_N - S_M)(x - x_0)\|_H \le 2A\,\|x - x_0\|_H < \eta/2$ for all $M, N$. Since $x_0 \in H_0$, choose $N_0$ such that $\|(S_N - S_M) x_0\|_H < \eta/2$ for $M, N \ge N_0$. Then by the triangle inequality, $\|(S_N - S_M) x\|_H < \eta$ for $M, N \ge N_0$, proving $x \in H_0$. Hence $H_0$ is closed. Since $H_0$ is closed, it suffices to show that $H_0$ is dense in $H$ — equivalently, $H_0^\perp = \{0\}$. **$H_0$ contains the range of every $T_{j_0}^*$.** Fix $j_0 \in \mathbb{Z}$ and $y \in H$. We show $x := T_{j_0}^* y \in H_0$ by exhibiting absolute convergence of $\sum_k T_k x$. The second hypothesis $\|T_j T_k^*\|_{\mathcal{L}(H)}^{1/2} \le \gamma(j - k)$ applied with $j = k$ and $k = j_0$ — i.e., the index pair $(j, k) = (k, j_0)$ in the hypothesis variable names — gives \begin{align*} \|T_k T_{j_0}^*\|_{\mathcal{L}(H)}^{1/2} \le \gamma(k - j_0), \qquad\text{hence}\qquad \|T_k T_{j_0}^*\|_{\mathcal{L}(H)} \le \gamma(k - j_0)^2. \end{align*} We do not assume $\gamma$ is symmetric; the argument from $j_0 - k$ to $k - j_0$ requires evenness of $\gamma$ which is **not** in our hypotheses. We therefore use $\gamma(k - j_0)$ throughout. Combining, \begin{align*} \|T_k T_{j_0}^* y\|_H \le \|T_k T_{j_0}^*\|_{\mathcal{L}(H)}\,\|y\|_H \le \gamma(k - j_0)^2\,\|y\|_H. \end{align*} Since $\gamma$ is non-negative and summable, with $\gamma(d) \le \sum_{e \in \mathbb{Z}}\gamma(e) = A$ for every $d$, \begin{align*} \sum_{k \in \mathbb{Z}} \|T_k T_{j_0}^* y\|_H \le \|y\|_H \sum_{k \in \mathbb{Z}} \gamma(k - j_0)^2 \le \|y\|_H \cdot A \cdot \sum_{k \in \mathbb{Z}} \gamma(k - j_0) = A^2\,\|y\|_H < \infty, \end{align*} where the change of variable $d = k - j_0$ in the final sum gives $\sum_{d \in \mathbb{Z}} \gamma(d) = A$. Hence the series $\sum_k T_k T_{j_0}^* y$ converges absolutely in $H$, so its symmetric partial sums $S_N (T_{j_0}^* y) = \sum_{|k| \le N} T_k T_{j_0}^* y$ form a Cauchy sequence in $H$. Therefore $T_{j_0}^* y \in H_0$. **$H_0^\perp = \{0\}$.** Let $x \in H_0^\perp$. Then $(T_{j_0}^* y, x)_H = 0$ for all $j_0 \in \mathbb{Z}$ and $y \in H$. By the adjoint relation, $(T_{j_0}^* y, x)_H = (y, T_{j_0} x)_H$ for all $y \in H$, so $T_{j_0} x = 0$ for every $j_0 \in \mathbb{Z}$. Hence $S_N x = \sum_{|j_0| \le N} T_{j_0} x = 0$ for every $N$, so $S_N x$ is trivially Cauchy and $x \in H_0$. Combined with $x \in H_0^\perp$ and $H_0 \cap H_0^\perp = \{0\}$, this gives $x = 0$. Hence $H_0^\perp = \{0\}$ and $H_0 = H$. **Definition of $T$ and bound on its norm.** For every $x \in H$, define \begin{align*} T: H &\to H \\ x &\mapsto \lim_{N \to \infty} S_N x \qquad (\text{strong limit in } H). \end{align*} Linearity follows from linearity of each $S_N$ and continuity of vector addition. Boundedness with $\|T\|_{\mathcal{L}(H)} \le A$ follows from \begin{align*} \|T x\|_H = \lim_{N \to \infty}\|S_N x\|_H \le \limsup_{N \to \infty} \|S_N\|_{\mathcal{L}(H)}\,\|x\|_H \le A\,\|x\|_H, \end{align*} using continuity of the norm and the uniform bound $\|S_N\|_{\mathcal{L}(H)} \le A$ from Step 4. Hence $T \in \mathcal{L}(H)$ with $\|T\|_{\mathcal{L}(H)} \le A$, and $T = \sum_{j \in \mathbb{Z}} T_j$ in the strong operator topology. [/step]