[proofplan] We decompose $H$ into the fixed-point subspace $\mathcal{H}_1$ and its orthogonal complement, then show that the Cesàro averages act as the identity on $\mathcal{H}_1$ and vanish on $\mathcal{H}_1^\perp$. The vanishing is first proved for exact coboundaries of the form $Ug-g$, where the Cesàro sum telescopes. We then identify $\mathcal{H}_1^\perp$ with the closure of $\operatorname{Range}(U-I)$ and use the uniform bound $\|A_N\|_{\mathcal{L}(H)} \leq 1$ to pass from exact coboundaries to their norm closure. [/proofplan] [step:Define the Cesàro averages and decompose $f$ orthogonally] Let $\langle \cdot,\cdot\rangle_H$ denote the inner product on $H$, and let $\|\cdot\|_H$ denote the induced norm. For each integer $N \geq 1$, define the Cesàro average operator \begin{align*} A_N: H &\to H \\ h &\mapsto \frac{1}{N}\sum_{n=0}^{N-1} U^n h. \end{align*} Since $P: H \to \mathcal{H}_1$ is the orthogonal projection onto $\mathcal{H}_1$, every $f \in H$ decomposes as \begin{align*} f = f_1 + f_0, \end{align*} where $f_1 := Pf \in \mathcal{H}_1$ and $f_0 := f - Pf \in \mathcal{H}_1^\perp$. [/step] [step:Show that Cesàro averages fix the invariant component] Since $f_1 \in \mathcal{H}_1 = \ker(U-I)$, we have $Uf_1 = f_1$. By induction on $n \in \mathbb{N}$, this gives $U^n f_1 = f_1$ for every $n \geq 0$. Hence, for every $N \geq 1$, \begin{align*} A_N f_1 = \frac{1}{N}\sum_{n=0}^{N-1} U^n f_1 = \frac{1}{N}\sum_{n=0}^{N-1} f_1 = f_1. \end{align*} Thus $A_N f_1 = f_1 = Pf$ for the invariant component. [/step] [step:Identify the orthogonal complement with the closure of the coboundaries] We first prove that \begin{align*} \mathcal{H}_1^\perp = \overline{\operatorname{Range}(U-I)}. \end{align*} Let $U^*: H \to H$ denote the Hilbert-space adjoint of $U$. Since $U$ is an isometry, $\|Uh\|_H = \|h\|_H$ for every $h \in H$. The orthogonal complement of $\operatorname{Range}(U-I)$ is \begin{align*} \operatorname{Range}(U-I)^\perp = \ker((U-I)^*) = \ker(U^* - I), \end{align*} by the standard Hilbert-space identity $\operatorname{Range}(T)^\perp = \ker(T^*)$ applied to the bounded operator $T := U-I$. We claim that $\ker(U^*-I) = \ker(U-I)$. If $h \in \ker(U-I)$, then $Uh = h$, and for every $k \in H$, \begin{align*} \langle k, U^*h\rangle_H = \langle Uk,h\rangle_H = \langle Uk,Uh\rangle_H = \langle k,h\rangle_H, \end{align*} where the last equality uses that $U$ preserves inner products as an isometry. Hence $U^*h = h$. Conversely, if $h \in \ker(U^*-I)$, then $U^*h = h$. Therefore \begin{align*} \|Uh-h\|_H^2 &= \|Uh\|_H^2 - 2\operatorname{Re}\langle Uh,h\rangle_H + \|h\|_H^2 \\ &= \|h\|_H^2 - 2\operatorname{Re}\langle h,U^*h\rangle_H + \|h\|_H^2 \\ &= \|h\|_H^2 - 2\|h\|_H^2 + \|h\|_H^2 \\ &= 0. \end{align*} Thus $Uh=h$, so $h \in \ker(U-I)$. Hence $\ker(U^*-I)=\mathcal{H}_1$. Taking orthogonal complements and using the identity $(M^\perp)^\perp=\overline{M}$ for a linear subspace $M \subseteq H$, we obtain \begin{align*} \overline{\operatorname{Range}(U-I)} = \operatorname{Range}(U-I)^{\perp\perp} = \mathcal{H}_1^\perp. \end{align*} [guided] The goal of this step is to justify the approximation used later: every vector orthogonal to the fixed points can be approximated by vectors of the form $Ug-g$. These vectors are called coboundaries because they lie in the range of $U-I$. Let $U^*: H \to H$ be the Hilbert-space adjoint of $U$. We use the standard Hilbert-space identity \begin{align*} \operatorname{Range}(T)^\perp = \ker(T^*) \end{align*} for a bounded linear operator $T: H \to H$. Here $T := U-I$ is bounded because $U$ is an isometry and hence bounded with operator norm $1$. Therefore \begin{align*} \operatorname{Range}(U-I)^\perp = \ker((U-I)^*) = \ker(U^*-I). \end{align*} It remains to relate $\ker(U^*-I)$ to the fixed-point space $\mathcal{H}_1 = \ker(U-I)$. First suppose $h \in \ker(U-I)$. Then $Uh=h$. Since an isometry preserves inner products, for every $k \in H$ we have \begin{align*} \langle k, U^*h\rangle_H = \langle Uk,h\rangle_H = \langle Uk,Uh\rangle_H = \langle k,h\rangle_H. \end{align*} Because this equality holds for every $k \in H$, the defining uniqueness property of the Riesz representation gives $U^*h=h$. Thus $h \in \ker(U^*-I)$. Conversely, suppose $h \in \ker(U^*-I)$, so $U^*h=h$. We want to prove $Uh=h$. Compute the norm of the difference: \begin{align*} \|Uh-h\|_H^2 &= \|Uh\|_H^2 - 2\operatorname{Re}\langle Uh,h\rangle_H + \|h\|_H^2 \\ &= \|h\|_H^2 - 2\operatorname{Re}\langle h,U^*h\rangle_H + \|h\|_H^2 \\ &= \|h\|_H^2 - 2\|h\|_H^2 + \|h\|_H^2 \\ &= 0. \end{align*} The first equality is the Hilbert-space norm expansion, the second uses both the isometry property $\|Uh\|_H=\|h\|_H$ and the definition of the adjoint, and the third uses $U^*h=h$. Hence $Uh=h$, so $h \in \ker(U-I)$. We have proved $\ker(U^*-I)=\ker(U-I)=\mathcal{H}_1$. Therefore \begin{align*} \operatorname{Range}(U-I)^\perp = \mathcal{H}_1. \end{align*} Taking orthogonal complements and using $(M^\perp)^\perp=\overline{M}$ for a linear subspace $M \subseteq H$ gives \begin{align*} \overline{\operatorname{Range}(U-I)} = \operatorname{Range}(U-I)^{\perp\perp} = \mathcal{H}_1^\perp. \end{align*} [/guided] [/step] [step:Prove convergence to zero for exact coboundaries] Let $g \in H$, and define the exact coboundary $h \in H$ by \begin{align*} h := Ug-g. \end{align*} For every $N \geq 1$, the Cesàro average telescopes: \begin{align*} A_N h &= \frac{1}{N}\sum_{n=0}^{N-1} U^n(Ug-g) \\ &= \frac{1}{N}\sum_{n=0}^{N-1}(U^{n+1}g-U^n g) \\ &= \frac{1}{N}(U^N g-g). \end{align*} Since $U$ is an isometry, $\|U^N g\|_H=\|g\|_H$. The triangle inequality gives \begin{align*} \|A_N h\|_H = \frac{1}{N}\|U^N g-g\|_H \leq \frac{1}{N}(\|U^N g\|_H+\|g\|_H) = \frac{2\|g\|_H}{N}. \end{align*} Thus $\|A_N h\|_H \to 0$ as $N \to \infty$. [/step] [step:Use the closure of coboundaries and the uniform bound on Cesàro averages] For every $N \geq 1$, the operator $A_N$ has norm at most $1$. Indeed, for every $h \in H$, \begin{align*} \|A_N h\|_H &= \left\|\frac{1}{N}\sum_{n=0}^{N-1} U^n h\right\|_H \\ &\leq \frac{1}{N}\sum_{n=0}^{N-1}\|U^n h\|_H \\ &= \frac{1}{N}\sum_{n=0}^{N-1}\|h\|_H \\ &= \|h\|_H. \end{align*} Since $f_0 \in \mathcal{H}_1^\perp$ and $\mathcal{H}_1^\perp=\overline{\operatorname{Range}(U-I)}$, for every $\varepsilon>0$ there exists $g_\varepsilon \in H$ such that the coboundary \begin{align*} h_\varepsilon := Ug_\varepsilon-g_\varepsilon \end{align*} satisfies \begin{align*} \|f_0-h_\varepsilon\|_H < \varepsilon. \end{align*} Using the triangle inequality, the bound $\|A_N\|_{\mathcal{L}(H)} \leq 1$, and the exact-coboundary estimate, we obtain \begin{align*} \|A_N f_0\|_H &\leq \|A_N(f_0-h_\varepsilon)\|_H+\|A_N h_\varepsilon\|_H \\ &\leq \|f_0-h_\varepsilon\|_H+\frac{2\|g_\varepsilon\|_H}{N} \\ &< \varepsilon+\frac{2\|g_\varepsilon\|_H}{N}. \end{align*} Choose $N_\varepsilon \geq 1$ such that $2\|g_\varepsilon\|_H/N < \varepsilon$ whenever $N \geq N_\varepsilon$. Then for all $N \geq N_\varepsilon$, \begin{align*} \|A_N f_0\|_H < 2\varepsilon. \end{align*} Since $\varepsilon>0$ was arbitrary, $A_N f_0 \to 0$ in the norm of $H$. [guided] We now pass from exact coboundaries to all vectors in $\mathcal{H}_1^\perp$. The reason this is possible is that the operators $A_N$ are uniformly bounded in operator norm. First we prove the uniform bound. For every $h \in H$ and every $N \geq 1$, the triangle inequality gives \begin{align*} \|A_N h\|_H &= \left\|\frac{1}{N}\sum_{n=0}^{N-1} U^n h\right\|_H \\ &\leq \frac{1}{N}\sum_{n=0}^{N-1}\|U^n h\|_H. \end{align*} Because $U$ is an isometry, each iterate $U^n$ is also an isometry, so $\|U^n h\|_H=\|h\|_H$ for every $n \geq 0$. Therefore \begin{align*} \|A_N h\|_H \leq \frac{1}{N}\sum_{n=0}^{N-1}\|h\|_H = \|h\|_H. \end{align*} Thus $\|A_N\|_{\mathcal{L}(H)} \leq 1$ for every $N \geq 1$. Now use the closure identity proved above. Since $f_0 \in \mathcal{H}_1^\perp$ and \begin{align*} \mathcal{H}_1^\perp=\overline{\operatorname{Range}(U-I)}, \end{align*} for every $\varepsilon>0$ there exists $g_\varepsilon \in H$ such that \begin{align*} h_\varepsilon := Ug_\varepsilon-g_\varepsilon \end{align*} satisfies \begin{align*} \|f_0-h_\varepsilon\|_H < \varepsilon. \end{align*} This is the approximation step: $h_\varepsilon$ is an exact coboundary, and we already know that its Cesàro averages converge to zero. We estimate $A_N f_0$ by adding and subtracting $h_\varepsilon$: \begin{align*} \|A_N f_0\|_H &\leq \|A_N(f_0-h_\varepsilon)\|_H+\|A_N h_\varepsilon\|_H. \end{align*} The first term is controlled uniformly in $N$ by the operator norm bound: \begin{align*} \|A_N(f_0-h_\varepsilon)\|_H \leq \|f_0-h_\varepsilon\|_H < \varepsilon. \end{align*} The second term is controlled by the telescoping computation for exact coboundaries: \begin{align*} \|A_N h_\varepsilon\|_H \leq \frac{2\|g_\varepsilon\|_H}{N}. \end{align*} Combining these estimates gives \begin{align*} \|A_N f_0\|_H < \varepsilon+\frac{2\|g_\varepsilon\|_H}{N}. \end{align*} Since $g_\varepsilon$ is fixed after choosing $\varepsilon$, we may choose $N_\varepsilon \geq 1$ so that \begin{align*} \frac{2\|g_\varepsilon\|_H}{N}<\varepsilon \end{align*} for every $N \geq N_\varepsilon$. Hence \begin{align*} \|A_N f_0\|_H < 2\varepsilon \end{align*} for all sufficiently large $N$. Since $\varepsilon>0$ was arbitrary, this proves $A_N f_0 \to 0$ in the norm of $H$. [/guided] [/step] [step:Combine the invariant and orthogonal components] For every $N \geq 1$, linearity of $A_N$ gives \begin{align*} A_N f = A_N f_1 + A_N f_0. \end{align*} The invariant component satisfies $A_N f_1=f_1=Pf$, and the orthogonal component satisfies $A_N f_0 \to 0$ in $H$. Therefore \begin{align*} \|A_N f - Pf\|_H = \|A_N f_0\|_H \to 0 \end{align*} as $N \to \infty$. Equivalently, \begin{align*} \frac{1}{N}\sum_{n=0}^{N-1} U^n f \to Pf \end{align*} in the norm of $H$. This proves the theorem. [/step]