Spectral Theory of Compact Operators

Example: The Right Shift Operator on \(\ell^2\)

Define the right shift \(S : \ell^2 \to \ell^2\) by \(S(x_1, x_2, x_3, \ldots) = (0, x_1, x_2, x_3, \ldots)\).

This operator has no eigenvalues. If \(S\mathbf{x} = \lambda\mathbf{x}\), then \(0 = \lambda x_1\), \(x_1 = \lambda x_2\), \(x_2 = \lambda x_3\), and so on. For \(\lambda = 0\), we get \(\mathbf{x} = 0\). For \(\lambda \neq 0\), the first equation gives \(x_1 = 0\), and the remaining equations then force \(x_n = 0\) for every \(n\). So \(\sigma_p(S) = \emptyset\).

Yet the spectrum is nonempty. A standard computation, which we take on faith here, gives \(\sigma(S) = \{\lambda \in \mathbb{C} : |\lambda| \leq 1\}\), the entire closed unit disk. Specifically, the open disk \(\{|\lambda| \lt 1\}\) forms the residual spectrum, and the boundary \(\{|\lambda| = 1\}\) forms the continuous spectrum. The residual claim, at least, will become a short exercise once the adjoint and the exchange of range and kernel are established in the next section. This phenomenon has no finite-dimensional counterpart.

Proof:

The existence of \(T^*\) follows from the Riesz Representation Theorem: for each fixed \(y\), the map \(x \mapsto \langle Tx, y \rangle\) is a continuous linear functional on \(\mathcal{H}\). By Riesz, there exists a unique vector, denoted \(T^* y\), such that \(\langle Tx, y \rangle = \langle x, T^* y \rangle\). The map \(y \mapsto T^* y\) is linear (it is the composition of two conjugate-linear maps: \(y \mapsto \varphi_y\) where \(\varphi_y(x) = \langle Tx, y\rangle\), and the Riesz sharp map \(\varphi \mapsto \varphi^\sharp\), which is conjugate-linear).

For the norm equality, \(\|T^*y\| = \|\varphi_y\|_{\mathcal{H}^*} \leq \|T\|\|y\|\) gives \(\|T^*\| \leq \|T\|\). To see that \((T^*)^* = T\), note that \(\langle T^*y, x\rangle = \overline{\langle x, T^*y\rangle} = \overline{\langle Tx, y\rangle} = \langle y, Tx\rangle\), so \(T\) satisfies the same defining relation that characterizes the adjoint of \(T^*\). Uniqueness in the theorem, applied to \(T^*\) in place of \(T\), then gives \((T^*)^* = T\). The reverse inequality follows from that identity combined with the same inequality applied to \(T^*\).

Example: Adjoints of Familiar Operators

Matrices.
For \(A \in \mathbb{C}^{n \times n}\) acting on \(\mathbb{C}^n\), the adjoint is the conjugate transpose: \(A^* = \overline{A}^\top\). Over \(\mathbb{R}\), this is simply \(A^\top\).

Integral operators.
Let \(K\) be a square-integrable kernel on \([a,b] \times [a,b]\), a condition studied in detail later on this page. If \(T : L^2([a,b]) \to L^2([a,b])\) is defined by \[ (Tf)(x) = \int_a^b K(x, t)\, f(t)\, dt, \] then the adjoint is \[ (T^* g)(x) = \int_a^b \overline{K(t, x)}\, g(t)\, dt. \] That is, the kernel is "transposed and conjugated": \(K^*(x, t) = \overline{K(t, x)}\). The verification is a direct computation with Fubini's theorem, which the square-integrability of \(K\) licenses.

Proof:

Fix \(y \in \mathcal{H}\). By the definition of the orthogonal complement, \(y \in \operatorname{Range}(S)^\perp\) means \(\langle y, Sx \rangle = 0\) for every \(x \in \mathcal{H}\). Hermitian symmetry and the defining relation of the adjoint give \[ \langle y, Sx \rangle = \overline{\langle Sx, y \rangle} = \overline{\langle x, S^* y \rangle}, \] so the condition is equivalent to \(\langle x, S^* y \rangle = 0\) for every \(x \in \mathcal{H}\). Choosing \(x = S^* y\) yields \(\|S^* y\|^2 = 0\), that is \(S^* y = 0\). Conversely, if \(S^* y = 0\) then \(\langle x, S^* y \rangle = 0\) for every \(x\), hence \(\langle y, Sx \rangle = 0\) for every \(x\). This proves \(\operatorname{Range}(S)^\perp = \ker(S^*)\).

For the second statement, \(\operatorname{Range}(S)\) is a subspace because \(S\) is linear, and it is closed by hypothesis. The Projection Theorem therefore applies to it and gives \[ \mathcal{H} = \operatorname{Range}(S) \oplus \operatorname{Range}(S)^\perp = \operatorname{Range}(S) \oplus \ker(S^*) = \operatorname{Range}(S) \oplus \{0\}, \] so \(\operatorname{Range}(S) = \mathcal{H}\).

Proof:

(a)
If \(Tx = \lambda x\) with \(x \neq 0\), then \(\lambda \langle x, x \rangle = \langle Tx, x \rangle = \langle x, Tx \rangle = \overline{\lambda} \langle x, x \rangle\). Since \(\langle x, x \rangle \neq 0\), we get \(\lambda = \overline{\lambda}\), so \(\lambda \in \mathbb{R}\).

(b)
If \(Tx = \lambda x\) and \(Ty = \mu y\) with \(\lambda \neq \mu\), then \[ \lambda \langle x, y \rangle = \langle Tx, y \rangle = \langle x, Ty \rangle = \overline{\mu} \langle x, y \rangle = \mu \langle x, y \rangle, \] where the last step uses \(\mu \in \mathbb{R}\) from part (a). Since \(\lambda \neq \mu\), we conclude \(\langle x, y \rangle = 0\).

(c)
For \(\lambda = \alpha + i\beta \in \mathbb{C}\) with \(\beta \neq 0\), self-adjointness gives the identity \[ \|(T - \lambda I)x\|^2 = \|(T - \alpha I)x\|^2 + \beta^2 \|x\|^2 \geq \beta^2 \|x\|^2, \] which is verified by expanding the left-hand side. Since \(\alpha\) is real, \(T - \alpha I\) is self-adjoint, so \(\langle (T - \alpha I)x, x\rangle\) equals its own conjugate and is therefore real. Writing \((T - \lambda I)x = (T - \alpha I)x - i\beta x\), \[ \|(T - \lambda I)x\|^2 = \|(T - \alpha I)x\|^2 + \beta^2\|x\|^2 - 2\operatorname{Re}\bigl\langle (T - \alpha I)x,\, i\beta x \bigr\rangle , \] and conjugate-linearity in the second slot turns the cross term into \(-i\beta \langle (T - \alpha I)x, x\rangle\), which is purely imaginary and so has zero real part. Hence \(T - \lambda I\) is bounded below: taking square roots, \(\|(T - \lambda I)x\| \geq |\beta|\,\|x\|\) for every \(x\). Injectivity is immediate, since \(\|(T-\lambda I)x\| = 0\) then forces \(\|x\| = 0\). The range is closed as well. Let \(z\) be a boundary point of \(\operatorname{Range}(T - \lambda I)\), so that \(\operatorname{dist}(z, \operatorname{Range}(T - \lambda I)) = 0\) and we may pick \(x_n\) with \(\|(T - \lambda I)x_n - z\| \lt 1/n\). The triangle inequality and the lower bound then give, for all \(m, n\), \[ \|x_n - x_m\| \leq \frac{1}{|\beta|}\,\|(T - \lambda I)x_n - z\| + \frac{1}{|\beta|}\,\|z - (T - \lambda I)x_m\| \lt \frac{1}{|\beta|}\left(\frac{1}{n} + \frac{1}{m}\right), \] so \(\{x_n\}\) is Cauchy and converges to some \(x \in \mathcal{H}\) by completeness. Since \(T - \lambda I\) is continuous, \((T - \lambda I)x_n \to (T - \lambda I)x\), while by construction \((T - \lambda I)x_n \to z\). The two limits must agree because limits in a metric space are unique, so \(z = (T - \lambda I)x\) lies in the range. The imaginary part of \(\overline{\lambda}\) is \(-\beta \neq 0\), so the same estimate applied to \(T - \overline{\lambda}I\) shows that it too is injective, giving \(\ker(T - \overline{\lambda}I) = \{0\}\). Since \(T\) is self-adjoint, the adjoint of \(S := T - \lambda I\) is \(S^* = T^* - \overline{\lambda}I = T - \overline{\lambda}I\), the conjugate appearing because the inner product is conjugate-linear in its second slot. So this says \(\ker(S^*) = \{0\}\). With the range of \(S\) already known to be closed, the exchange of range and kernel under the adjoint turns injectivity of \(S^*\) into surjectivity of \(S\): \[ \operatorname{Range}(T - \lambda I) = \mathcal{H}. \] Thus \(T - \lambda I\) is bijective. The bound \(\|(T-\lambda I)x\| \geq |\beta|\|x\|\) from the initial inequality forces \(\|(T-\lambda I)^{-1}\| \leq 1/|\beta|\), so \(\lambda \notin \sigma(T)\), giving \(\sigma(T) \subseteq \mathbb{R}\).

(d)
Let \(M = \sup_{\|x\|=1} |\langle Tx, x\rangle|\). The bound \(M \leq \|T\|\) follows from Cauchy-Schwarz. For the reverse, take unit vectors \(h, g \in \mathcal{H}\) and expand \(\langle T(h \pm g), h \pm g\rangle\) using self-adjointness. The cross terms satisfy \[ \begin{align*} \langle Th, g\rangle + \langle Tg, h\rangle &= \langle Th, g\rangle + \overline{\langle Th, g\rangle} \\\\ &= 2\operatorname{Re}\langle Th, g\rangle \end{align*} \] (using self-adjointness \(\langle Tg, h\rangle = \langle g, Th\rangle\) followed by the Hermitian symmetry of the inner product), so \[ \langle T(h \pm g), h \pm g\rangle = \langle Th, h\rangle \pm 2\operatorname{Re}\langle Th, g\rangle + \langle Tg, g\rangle. \] Subtracting the two equations gives \[ 4\operatorname{Re}\langle Th, g\rangle = \langle T(h+g), h+g\rangle - \langle T(h-g), h-g\rangle. \] By homogeneity, the definition of \(M\) gives \(|\langle Tf, f\rangle| \leq M\|f\|^2\) for every \(f \in \mathcal{H}\). Applying this bound and the parallelogram law, \[ 4\operatorname{Re}\langle Th, g\rangle \leq M\bigl(\|h+g\|^2 + \|h-g\|^2\bigr) = 2M\bigl(\|h\|^2 + \|g\|^2\bigr) = 4M. \] If \(\langle Th, g\rangle = 0\) the bound \(|\langle Th, g\rangle| \leq M\) is trivial. Otherwise, over \(\mathbb{C}\), replacing \(h\) with \(e^{-i\theta}h\) for \(\theta = \arg\langle Th, g\rangle\) converts the real part into the modulus (over \(\mathbb{R}\) the inner product is already real and replacing \(h\) with \(\pm h\) does the same). Either way \(|\langle Th, g\rangle| \leq M\) for all unit \(h, g\). Taking the supremum over unit \(g\) yields \(\|Th\| \leq M\), because \[ \|y\| = \sup_{\|g\| = 1} |\langle y, g \rangle| \quad \text{for every } y \in \mathcal{H}. \] That identity is immediate from Cauchy-Schwarz, which bounds \(|\langle y, g \rangle|\) by \(\|y\|\) for every unit \(g\): when \(y \neq 0\) the admissible choice \(g = y / \|y\|\) attains \(\langle y, g \rangle = \|y\|\), and when \(y = 0\) both sides vanish. Taking the supremum over unit \(h\) then yields \(\|T\| \leq M\).

Proof sketch (Hilbert space case):

(\(\Leftarrow\)) Suppose \(T_n \to T\) in operator norm with each \(T_n\) finite-rank (hence compact). Given a bounded sequence \(\{x_k\}\) with \(\|x_k\| \leq B\), iteratively extract subsequences. The sequence \(\{T_1 x_{k}\}\) has a convergent subsequence indexed by \(k_1(j)\). Within that, \(\{T_2 x_{k_1(j)}\}\) has a convergent subsequence indexed by \(k_2(j)\), and so on. The diagonal \(y_j := x_{k_j(j)}\) then has the property that \(\{T_n y_j\}_{j \geq n}\) converges for each fixed \(n\). For any \(\varepsilon \gt 0\), choose \(n\) with \(\|T - T_n\| \lt \varepsilon/(3B)\). Then for \(j, j'\) large enough that \(\|T_n y_j - T_n y_{j'}\| \lt \varepsilon/3\), \[ \begin{align*} \|T y_j - T y_{j'}\| &\leq \|(T - T_n)y_j\| + \|T_n y_j - T_n y_{j'}\| + \|(T_n - T)y_{j'}\| \\\\ &\lt \tfrac{\varepsilon}{3} + \tfrac{\varepsilon}{3} + \tfrac{\varepsilon}{3} = \varepsilon. \end{align*} \] Hence \(\{Ty_j\}\) is Cauchy in \(\mathcal{H}\), and converges by the Cauchy criterion for completeness. Thus \(T\) is compact.

(\(\Rightarrow\)) Separability of \(\overline{\operatorname{Range}(T)}\) (which follows from compactness, since \(\overline{\operatorname{Range}(T)} = \overline{\bigcup_n T(\overline{B}_n)}\) is the closure of a countable union of relatively compact, hence separable, sets) lets us pick a finite or countable orthonormal basis \(\{e_n\}\) for it. Let \(P_n\) be orthogonal projection onto \(\operatorname{span}\{e_1, \ldots, e_n\}\), and set \(T_n = P_n T\). Each \(T_n\) is finite-rank.

Since \(T(\overline{B}_\mathcal{H})\) has compact closure, the equivalence of compactness applied to that closure makes it a totally bounded subset of \(\mathcal{H}\). Fix \(\varepsilon \gt 0\) and cover the set by finitely many balls of radius \(\varepsilon/4\). Discarding the balls that miss the set and replacing each remaining ball by the \(\varepsilon/2\)-ball around one of its points from \(T(\overline{B}_\mathcal{H})\) yields a finite cover by \(\varepsilon/2\)-balls whose centers \(y_1, \ldots, y_k\) lie in \(T(\overline{B}_\mathcal{H})\). Each \(y_j\) therefore lies in \(\operatorname{Range}(T)\), hence in the closed span of \(\{e_n\}\). Choose \(N\) so that \(\|y_j - P_n y_j\| \lt \varepsilon/2\) for all \(j\) and \(n \geq N\) (possible because \(P_n y_j \to y_j\) for each fixed \(y_j\) in the closed span of \(\{e_n\}\)). For unit \(x \in \overline{B}_\mathcal{H}\), pick \(j\) with \(\|Tx - y_j\| \lt \varepsilon/2\). Then \[ \begin{align*} \|(T - T_n) x\| = \|(I - P_n) Tx\| &\leq \|(I - P_n)(Tx - y_j)\| + \|(I - P_n) y_j\| \\\\ &\leq \|Tx - y_j\| + \|y_j - P_n y_j\| \\\\ &\lt \varepsilon, \end{align*} \] using \(\|I - P_n\| \leq 1\). Taking sup over unit \(x\) gives \(\|T - T_n\| \leq \varepsilon\) for \(n \geq N\).

Remark. This characterization relies on the Hilbert space structure through the orthogonal projections \(P_n\). A Banach space in which every compact operator arriving from an arbitrary Banach space is a norm limit of finite-rank operators is said to have the approximation property. Famously, there exist separable Banach spaces without it.

Proof sketch (compactness and norm bound):

Norm bound. For \(f \in L^2([a,b])\), Cauchy-Schwarz applied pointwise in \(x\) gives \[ |(Tf)(x)|^2 = \left|\int_a^b K(x,t) f(t)\, dt\right|^2 \leq \left(\int_a^b |K(x,t)|^2\, dt\right) \|f\|_2^2. \] Integrating in \(x\) yields \(\|Tf\|_2^2 \leq \|T\|_{\text{HS}}^2 \|f\|_2^2\), that is, \(\|T\| \leq \|T\|_{\text{HS}}\).

Compactness. Fix an orthonormal basis \(\{\varphi_i\}\) of \(L^2([a,b])\). Then \(\{\varphi_i(x)\varphi_j(t)\}\) is an orthonormal basis of \(L^2([a,b]^2)\). Orthonormality follows from Fubini's theorem, which factors the double integral into a product of two inner products. Completeness holds because a kernel orthogonal to every product \(\varphi_i(x)\varphi_j(t)\) has, for almost every \(x\), a slice \(K(x, \cdot)\) orthogonal to every \(\varphi_j\), and therefore vanishes almost everywhere. Expand \(K = \sum_{i,j} c_{ij}\, \varphi_i(x)\varphi_j(t)\) and let \(K_N\) be the truncation to \(i, j \leq N\). The corresponding operator \(T_N\) is finite-rank, and \(\|T - T_N\| \leq \|T - T_N\|_{\text{HS}} \to 0\) by Parseval. By the finite-rank approximation theorem above, \(T\) is compact.

Proof:

(a) Linearity (closure under addition and scalar multiplication) is straightforward from the subsequence-extraction characterization. Closedness in operator norm was proved as the (\(\Leftarrow\)) direction of Compact Operators as Limits of Finite-Rank Operators for the Hilbert case. The same diagonal argument works for general normed spaces once "finite-rank" is replaced by "compact" throughout, since that argument never uses finite-rank approximability. A sequence \(\{T_n\}\) of compact operators with \(T_n \to T\) in norm and a bounded \(\{x_k\}\) admit, by iterated subsequence extraction, a diagonal \(\{y_j\}\) on which every \(T_n\) converges. The \(3\varepsilon\) argument then gives \(\{Ty_j\}\) Cauchy, hence convergent in the Banach space \(\mathcal{Y}\) by the Cauchy criterion for completeness.

(b) Let \(\{x_n\}\) be a bounded sequence in \(\mathcal{X}\). For \(TS\): \(\{Sx_n\}\) is bounded (\(\|Sx_n\| \leq \|S\|\|x_n\|\)), so by compactness of \(T\), \((TS)x_n = T(Sx_n)\) has a convergent subsequence. For \(ST\): by compactness of \(T\), \(\{Tx_n\}\) has a convergent subsequence \(Tx_{n_k} \to y\). By continuity of \(S\), \(S(Tx_{n_k}) \to Sy\), so \((ST)x_n\) has a convergent subsequence. Hence both \(ST\) and \(TS\) are compact.

(c) On a Hilbert space the result follows from the finite-rank approximation already established. Write \(T\) as a norm limit of finite-rank operators \(T_n\), as in Compact Operators as Limits of Finite-Rank Operators. Each \(T_n^*\) is again finite-rank. Choose an orthonormal set \(v_1, \ldots, v_m\) spanning \(\operatorname{Range}(T_n)\), so that \(T_n x = \sum_i \langle T_n x, v_i\rangle v_i\) for every \(x\). Moving \(T_n\) across the inner product gives \(T_n^* y = \sum_i \langle y, v_i\rangle\, T_n^* v_i\), so \(\operatorname{Range}(T_n^*)\) lies in the span of \(T_n^* v_1, \ldots, T_n^* v_m\). Since the adjoint is norm-preserving, \(\|T^* - T_n^*\| = \|(T - T_n)^*\| = \|T - T_n\| \to 0\), so \(T^*\) is itself a norm limit of finite-rank operators and therefore compact.

(d) If \(I\) were compact, the image of the unit ball under \(I\), which is the unit ball itself, would be relatively compact. But in infinite dimensions, the closed unit ball is not compact. This is exactly the Riesz characterization of finite-dimensionality.

Proof Sketch:

Step 1: Find the first eigenvector.
If \(T = 0\), the theorem holds trivially with \(N = 0\) and any orthonormal basis of \(\ker(T) = \mathcal{H}\), so assume \(T \neq 0\). Since \(T\) is self-adjoint, \(\|T\| = \sup_{\|x\|=1} |\langle Tx, x \rangle|\). Because every Hilbert space is reflexive, we have weak compactness of the unit ball, and Eberlein-Šmulian turns it into the sequential statement that every bounded sequence has a weakly convergent subsequence. Together with compactness of \(T\), this produces \(x_* \in \overline{B}_\mathcal{H}\) achieving \(|\langle Tx_*, x_* \rangle| = \|T\|\). The mechanism deserves its own paragraph.

Why the supremum is attained.
Compactness of \(T\) upgrades weak convergence to strong convergence of images along bounded sequences. Suppose \(\{x_n\}\) is bounded, \(x_n \rightharpoonup x\), and some subsequence of \(\{Tx_n\}\) stays at distance at least \(\delta \gt 0\) from \(Tx\). Compactness extracts from it a further subsequence with \(Tx_{n_k} \to y\) strongly. Boundedness of \(T\) makes it weakly continuous, because \(\langle Tx_n, z \rangle = \langle x_n, T^*z \rangle \to \langle x, T^*z \rangle = \langle Tx, z \rangle\) for every \(z \in \mathcal{H}\). The extracted subsequence therefore converges to \(Tx\) weakly and to \(y\) strongly, and testing the difference against \(z = y - Tx\) yields \(y = Tx\), a contradiction. Hence \(Tx_n \to Tx\) strongly. Splitting \(\langle Tx_n, x_n \rangle - \langle Tx, x \rangle = \langle Tx_n - Tx, x_n \rangle + \langle Tx, x_n - x \rangle\) and using boundedness of \(\{x_n\}\) then gives \(\langle Tx_n, x_n \rangle \to \langle Tx, x \rangle\) for every bounded sequence with \(x_n \rightharpoonup x\). A maximizing sequence for \(|\langle Tx, x \rangle|\) on \(\overline{B}_\mathcal{H}\) has a weakly convergent subsequence, and its weak limit \(x_*\) attains the supremum. Since the sup over \(\overline{B}\) equals the sup over the unit sphere (by positive homogeneity of \(|\langle Tx, x\rangle|\) in \(\|x\|^2\)), we may assume \(\|x_*\| = 1\). We set \(e_1 = x_*\) and \(\lambda_1 = \langle Te_1, e_1 \rangle\).

Variational argument for \(Te_1 = \lambda_1 e_1\).
For any \(v \perp e_1\) and \(t \in \mathbb{R}\), consider \[ \phi(t) = \langle T(e_1 + tv), e_1 + tv \rangle / \|e_1 + tv\|^2. \] Since \(t = 0\) maximises \(|\phi|\) and \(T\) is self-adjoint, differentiating at \(t = 0\) gives \[ \phi'(0) = 2\operatorname{Re} \langle Te_1 - \lambda_1 e_1, v \rangle = 0. \] Replacing \(v\) with \(iv\) over \(\mathbb{C}\) gives the imaginary part, so \(\langle Te_1 - \lambda_1 e_1, v \rangle = 0\) for all \(v \perp e_1\). Thus \[ Te_1 - \lambda_1 e_1 \in \{e_1\}^{\perp\perp} = \operatorname{span}\{e_1\} \] (since \(\operatorname{span}\{e_1\}\) is one-dimensional, hence closed, and span-invariance together with the double orthogonal complement identifies \(\{e_1\}^{\perp\perp}\) with it). Combined with \(\langle Te_1 - \lambda_1 e_1, e_1\rangle = 0\) by construction, this gives \(Te_1 = \lambda_1 e_1\).

Step 2: Restrict and repeat.
The subspace \(\{e_1\}^\perp\) is invariant under \(T\), because \(v \perp e_1\) gives \(\langle Tv, e_1 \rangle = \langle v, Te_1 \rangle = \lambda_1 \langle v, e_1 \rangle = 0\) by self-adjointness and the eigenvector relation \(Te_1 = \lambda_1 e_1\). The restriction \(T|_{\{e_1\}^\perp}\) is still compact and self-adjoint, so we repeat Step 1 to find \(e_2\) with \(|\lambda_2| = \|T|_{\{e_1\}^\perp}\| \leq |\lambda_1|\). If \(\|T|_{\{e_1, \ldots, e_n\}^\perp}\| = 0\) at some stage, the process stops and \(N = n\). Otherwise every \(\lambda_n\) is nonzero and the construction continues.

Step 3: Convergence to zero.
The sequence \(\{|\lambda_n|\}\) is nonincreasing and bounded below by 0, so it converges. If it converged to some \(\varepsilon \gt 0\), the orthonormal sequence \(\{e_n\}\) would satisfy \(\|Te_m - Te_n\|^2 = \lambda_m^2 + \lambda_n^2 \geq 2\varepsilon^2\), contradicting the compactness of \(T\) (which requires \(\{Te_n\}\) to have a convergent subsequence). Hence \(\lambda_n \to 0\).

Closing the remaining claims.
Steps 1-3 yield (b) directly. Statement (d) follows because the construction enumerates the eigenvalues in nonincreasing order of \(|\lambda_n|\), so any accumulation point must equal \(\lim |\lambda_n| = 0\). For (c), if some nonzero eigenvalue \(\mu\) had infinite multiplicity, an orthonormal basis \(\{u_k\}\) of its eigenspace would give \(\|Tu_k - Tu_l\|^2 = 2\mu^2\) for \(k \neq l\), again contradicting compactness.

For (e), the constructed \(\{e_n\}\) span \(\overline{\operatorname{Range}(T)}\). Any vector \(v\) orthogonal to all \(e_n\) satisfies \[ \|Tv\| \leq \|T|_{\{e_1,\ldots,e_n\}^\perp}\| \cdot \|v\| = |\lambda_{n+1}| \|v\| \to 0, \] forcing \(Tv = 0\), hence \(v \in \ker(T)\). Since \(\mathcal{H}\) is separable, the closed subspace \(\ker(T)\) is also separable and admits a countable orthonormal basis. Concatenating \(\{e_n\}\) with such a basis of \(\ker(T)\) gives a complete orthonormal basis of \(\mathcal{H} = \overline{\operatorname{Range}(T)} \oplus \ker(T)\). The expansion in (a) then follows. Writing \(x = x_R + x_K\) with \(x_R \in \overline{\operatorname{Range}(T)}\) and \(x_K \in \ker(T)\), Parseval gives \(x_R = \sum_n \langle x, e_n\rangle e_n\). Thus, \[ Tx = Tx_R = \sum_n \lambda_n \langle x, e_n\rangle e_n \] by linearity and continuity.

Proof sketch:

By Schauder's theorem, \(T^*\) is compact. Hence \(T^*T\) is compact, self-adjoint (\(\langle T^*Tx, y\rangle = \langle Tx, Ty\rangle = \langle x, T^*Ty\rangle\) by two uses of the defining relation), and positive (\(\langle T^*Tx, x\rangle = \|Tx\|^2 \geq 0\)). No separability of \(\mathcal{H}\) is needed here, even though the Spectral Theorem assumes it. The image of a ball under a compact operator has compact closure and is therefore totally bounded, so it contains a countable dense subset. Since \(\operatorname{Range}(T^*T)\) is a countable union of such images, the closed subspace \(\mathcal{H}_0 = \overline{\operatorname{Range}(T^*T)}\) is a separable Hilbert space. It is invariant under \(T^*T\), and the restriction \(T^*T|_{\mathcal{H}_0}\) is compact, self-adjoint, and injective: if \(x \in \mathcal{H}_0\) satisfies \(T^*Tx = 0\), then \(x \in \ker(T^*T) = \operatorname{Range}(T^*T)^\perp\) by the range-kernel exchange under the adjoint, so continuity of the inner product makes \(x\) orthogonal to all of \(\mathcal{H}_0\), and in particular to itself, so \(x = 0\). The Spectral Theorem applied to \(T^*T|_{\mathcal{H}_0}\) yields nonzero eigenvalues \(\{\sigma_n^2\}_{n=1}^N\) (\(\sigma_n^2 \gt 0\), nonincreasing, \(\sigma_n^2 \to 0\) when \(N = \infty\)) with corresponding orthonormal eigenvectors \(\{e_n\}\) forming an orthonormal basis of \(\mathcal{H}_0\). The basis property is conclusion (e) of that theorem, since the kernel of the restriction is trivial and contributes an empty orthonormal set. Define \(\sigma_n = \sqrt{\sigma_n^2} \gt 0\) and \(f_n = Te_n / \sigma_n\). Direct computation gives \[ \begin{align*} \langle f_n, f_m\rangle &= \langle Te_n, Te_m\rangle/(\sigma_n \sigma_m) \\\\ &= \langle T^*Te_n, e_m\rangle/(\sigma_n\sigma_m) \\\\ &= \sigma_n^2\delta_{nm}/(\sigma_n\sigma_m) \\\\ &= \delta_{nm}, \end{align*} \] so \(\{f_n\}\) is orthonormal.

For the expansion, note that \(\ker(T^*T) = \ker(T)\), since \(T^*Tx = 0\) implies \(\|Tx\|^2 = \langle T^*Tx, x\rangle = 0\), hence \(Tx = 0\), and conversely. Taking orthogonal complements in the range-kernel exchange and applying the double orthogonal complement identifies \(\mathcal{H}_0\) with \(\ker(T^*T)^\perp\). Decompose any \(x \in \mathcal{H}\) as \(x = x_R + x_K\) with \(x_R \in \mathcal{H}_0 = \ker(T^*T)^\perp = \ker(T)^\perp\) and \(x_K \in \ker(T)\). By Parseval on the orthonormal basis \(\{e_n\}\) of \(\mathcal{H}_0\), \(x_R = \sum_n \langle x, e_n\rangle e_n\). Since \(Tx_K = 0\) and \(T\) is continuous, \[ Tx = Tx_R = \sum_n \langle x, e_n\rangle Te_n = \sum_n \sigma_n \langle x, e_n\rangle f_n. \]

Finally, the \(\{f_n\}\) are themselves eigenvectors of \(TT^*\) with the same eigenvalues \(\sigma_n^2\): \(TT^* f_n = TT^*(Te_n/\sigma_n) = T(T^*T)e_n/\sigma_n = T(\sigma_n^2 e_n)/\sigma_n = \sigma_n^2 f_n\). The map \(e_n \mapsto f_n\) preserves orthonormality, so no multiplicity is lost in this direction. Exchanging the roles of \(T\) and \(T^*\) runs the same construction from \(TT^*\) back to \(T^*T\), so the nonzero eigenvalues of the two operators coincide with matching multiplicities. That is the sense in which \(T^*T\) and \(TT^*\) "see" the same singular values from the two sides of \(T\).

Proof sketch:

Apply the Spectral Theorem to the compact self-adjoint covariance operator \(\mathcal{C}\) to obtain orthonormal eigenfunctions \(\{e_n\}\) with eigenvalues \(\lambda_n \geq 0\) (positivity from \(\langle \mathcal{C}f, f\rangle = \mathbb{E}[|\langle X, f\rangle|^2] \geq 0\), real-valuedness of \(\lambda_n\) from self-adjointness). Define \(Z_n = \langle X, e_n\rangle\). Centering of \(X\) gives \(\mathbb{E}[Z_n] = 0\). By Fubini's theorem and the eigenvalue equation \(\mathcal{C}e_m = \lambda_m e_m\), \[ \begin{align*} \mathbb{E}[Z_n\, \overline{Z_m}] &= \int_a^b \!\!\int_a^b C(s,t)\, \overline{e_n(s)}\, e_m(t)\, ds\, dt \\\\ &= \langle \mathcal{C}e_m, e_n\rangle \\\\ &= \lambda_m \langle e_m, e_n\rangle \\\\ &= \lambda_m\, \delta_{nm}, \end{align*} \] giving uncorrelatedness and \(\operatorname{Var}(Z_n) = \mathbb{E}[|Z_n|^2] = \lambda_n\).

For the expansion, note first that \(X\) lies almost surely in \(\overline{\operatorname{Range}(\mathcal{C})}\), whose orthogonal complement is \(\ker\mathcal{C}\) by self-adjointness. Indeed, if \(\mathcal{C}f = 0\) then \(\mathbb{E}[|\langle X, f\rangle|^2] = \langle \mathcal{C}f, f\rangle = 0\), so \(\langle X, f\rangle = 0\) almost surely. Applying this to each member of a countable dense subset of \(\ker\mathcal{C}\), available because \(L^2([a,b])\) is separable, leaves a single exceptional set of probability zero, off which \(\langle X, \cdot\rangle\) vanishes on that dense subset and hence, being continuous, on all of \(\ker\mathcal{C}\). Parseval on the orthonormal basis \(\{e_n\}\) of \(\overline{\operatorname{Range}(\mathcal{C})}\) therefore gives, pathwise, \[ \Bigl\|X - \sum_{n=1}^{N} Z_n e_n\Bigr\|_{L^2([a,b])}^2 = \sum_{n \gt N} |Z_n|^2 , \] and taking expectations turns the right-hand side into \(\sum_{n \gt N} \lambda_n\) (monotone convergence licenses the exchange of expectation and sum). This is the tail of the convergent series \(\sum_n \lambda_n = \mathbb{E}\|X\|_{L^2([a,b])}^2\). Continuity of \(C\) enters this argument only through the finiteness of \(\int_a^b C(t,t)\, dt\), which makes the trace series \(\sum_n \lambda_n\) converge. Its full strength upgrades the conclusion to convergence in mean square at each fixed \(t\), uniformly in \(t\), through the pointwise kernel expansion taken up in the bridge to kernel methods below.