[proofplan] The proof has four stages. First, we show eigenvalues are real and eigenvectors are orthogonal (directly from self-adjointness). Second, we prove $\|T\|$ is attained as an eigenvalue using the variational characterisation and compactness. Third, we construct the full eigensequence by induction, restricting $T$ to the orthogonal complement of eigenspaces found so far. Fourth, we prove the expansion formula by showing the remainder $T - T_n$ has operator norm $|\mu_{n+1}| \to 0$. [/proofplan] [step:Show eigenvalues are real and eigenvectors for distinct eigenvalues are orthogonal] [claim:Reality of Eigenvalues] If $Tx = \mu x$ with $x \neq 0$, then $\mu \in \mathbb{R}$. [/claim] [proof] $\mu \|x\|^2 = (\mu x, x)_H = (Tx, x)_H = (x, Tx)_H = \bar{\mu} \|x\|^2$. Since $\|x\|^2 > 0$, $\mu = \bar{\mu}$. [/proof] [claim:Orthogonality of Eigenvectors] If $Tx = \mu x$ and $Ty = \nu y$ with $\mu \neq \nu$, then $(x, y)_H = 0$. [/claim] [proof] $\mu(x, y)_H = (Tx, y)_H = (x, Ty)_H = \nu(x, y)_H$. Since $\mu \neq \nu$, $(x, y)_H = 0$. [/proof] [/step] [step:Show the operator norm is attained as an eigenvalue] [claim:Extremal Eigenvalue] If $T \neq 0$, then $\|T\| = \sup_{\|x\|=1} |(Tx, x)_H|$ and this supremum is attained by an eigenvector. [/claim] [proof] For self-adjoint $T$, $\|T\| = \sup_{\|x\|=1} |(Tx, x)_H|$ (by the polarisation identity). Assume $\mu_1 := \sup_{\|x\|=1}(Tx, x)_H = \|T\|$ (otherwise replace $T$ by $-T$). Let $\{x_k\}$ be a maximising sequence in the unit sphere. By compactness, a subsequence satisfies $Tx_k \to z$. Computing: \begin{align*} \|Tx_k - \mu_1 x_k\|^2 = \|Tx_k\|^2 - 2\mu_1(Tx_k, x_k)_H + \mu_1^2 \leq \mu_1^2 - 2\mu_1^2 + \mu_1^2 = 0 \end{align*} in the limit (using $\|Tx_k\|^2 \leq \mu_1^2$ and $(Tx_k, x_k)_H \to \mu_1$). So $Tx_k - \mu_1 x_k \to 0$, giving $x_k \to z/\mu_1$, and $Te_1 = \mu_1 e_1$ for $e_1 = z/(\mu_1\|z/\mu_1\|)$. [/proof] [/step] [step:Construct the full eigensequence by induction on orthogonal complements] Define $H_1 = \{e_1\}^\perp$. Since $T$ is self-adjoint, $T(H_1) \subseteq H_1$ (for $x \in H_1$: $(Tx, e_1)_H = (x, Te_1)_H = \mu_1(x, e_1)_H = 0$). The restriction $T|_{H_1}$ is compact and self-adjoint. Apply the previous step to get $\mu_2, e_2$; proceed inductively on $H_k = \{e_1, \ldots, e_k\}^\perp$. The eigenvalues satisfy $|\mu_{k+1}| = \|T|_{H_k}\| \leq |\mu_k|$. In the infinite case, $|\mu_k| \to 0$ because the orthonormal sequence $\{e_k\}$ with $Te_k = \mu_k e_k$ must have $\{Te_k\}$ containing a convergent subsequence (compactness), and pairwise distances $\|Te_j - Te_k\|^2 = \mu_j^2 + \mu_k^2$ force $\mu_k \to 0$. [/step] [step:Verify finite multiplicity of each nonzero eigenvalue] [claim:Finite Multiplicity] For each $\mu \neq 0$, $\dim \ker(T - \mu I) < \infty$. [/claim] [proof] An infinite-dimensional eigenspace would contain an orthonormal sequence $\{f_k\}$ with $Tf_k = \mu f_k$ and $\|Tf_j - Tf_k\|^2 = 2|\mu|^2 > 0$, contradicting compactness. [/proof] [/step] [step:Prove norm convergence of the spectral expansion] [claim:Norm Convergence of the Spectral Expansion] $Tx = \sum_k \mu_k (x, e_k)_H \, e_k$ for every $x \in H$. [/claim] [proof] Define $T_n x = \sum_{k=1}^n \mu_k (x, e_k)_H \, e_k$. Write $x = \sum_{k=1}^n (x, e_k)_H e_k + r_n$ with $r_n \in H_n$. Then $(T - T_n)x = Tr_n$, so \begin{align*} \|(T - T_n)x\| = \|Tr_n\| \leq \|T|_{H_n}\| \cdot \|r_n\| \leq |\mu_{n+1}| \cdot \|x\|. \end{align*} Since $|\mu_{n+1}| \to 0$, $\|T - T_n\| \to 0$ in operator norm. [/proof] The decomposition $H = \ker(T) \oplus \overline{\operatorname{span}\{e_k\}}$ follows: if $x \perp e_k$ for all $k$, then $Tx = \lim T_n x = 0$. [/step]