Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 7 / 3

If $H$ is a finite-dimensional Hilbert space, a choice of orthonormal basis identifies its closed unit ball with a closed bounded ball in a finite-dimensional Euclidean space. The Heine-Borel theorem makes it compact in the norm topology.

Conversely, if $H$ is infinite-dimensional, choose a unit vector $e_1$ and inductively choose $e_{n+1}$ orthogonal to the span of $e_1,\dots ,e_n$. The finite span is a proper closed subspace, so its orthogonal complement contains a nonzero vector that can be normalized. The resulting orthonormal sequence lies in the closed unit ball and satisfiesIt has no norm-convergent subsequence. Sequential compactness is equivalent to compactness in a metric space, hence the closed ball is not compact. This proves the norm-compact unit ball criterion: a Hilbert space has a norm-compact closed unit ball exactly when it is finite-dimensional.

Work over a complex Hilbert space; for a real space one first complexifies to discuss a complex spectrum. A bounded linear operator $L$ is self-adjoint when $L=L^{\ast }$, where the adjoint operator is defined byEquivalently, $⟨Lh,k⟩=⟨h,Lk⟩$ for every pair of vectors.

The spectrum of a bounded operator isIts complement is the resolvent set. On a Banach space, a bounded bijective operator has a bounded inverse by the bounded inverse theorem, so failure of invertibility is equivalent to failure of bijectivity. An eigenvalue is one possible spectral point, but an injective operator that is not onto can also contribute to the spectrum.

If $|z|>\Vert L\Vert$, the Neumann series givesThe series converges in operator norm because $\Vert L/z\Vert <1$. HenceThe same argument applied to small perturbations of an invertible operator shows that the resolvent set is open and the spectrum is closed.

Suppose $L=L^{\ast }$ and $\mathrm{Im}z\ne 0$. The quadratic form $⟨Lh,h⟩$ is real, so the Cauchy-Schwarz inequality impliesThus $L-zI$ is bounded below by $|\mathrm{Im}z|$: it is injective and has closed range. Its adjoint operator $L-\bar{z}I$ has the same lower bound and trivial kernel. Sincethe range is dense as well as closed, and therefore is all of $H$. The inverse has norm at most $|\mathrm{Im}z{|}^{-1}$. For a bounded self-adjoint operator, $\Sigma (L)\subset \mathbb{R}$. The closed-range and adjoint steps are essential: merely excluding nonreal eigenvalues would not exclude all nonreal spectral points.

Suppose the compact operator $K$ acts on $H$, and $h_n\rightharpoonup h$. The Uniform boundedness principle makes $(h_n)$ bounded, so its images form a relatively compact set. Bounded linearity gives $K{h}_n\rightharpoonup Kh$: for every $g$, $⟨K{h}_n,g⟩=⟨h_n,K^{\ast }g⟩\to ⟨Kh,g⟩$. Every norm-convergent subsequence of $(K{h}_n)$ must therefore converge to $Kh$. If the whole sequence did not converge to $Kh$ in norm, a subsequence staying a fixed positive distance from $Kh$ would have a norm-convergent further subsequence, a contradiction. This proves that compact operators send weak convergence to norm convergence.

Conversely, assume the stated complete continuity property. Its strong limit necessarily equals $Kh$, because bounded linearity still supplies the weak limit. Every bounded sequence in a Hilbert space has a weakly convergent subsequence by weak sequential compactness in a Hilbert space. Applying the assumption to that subsequence gives a norm-convergent subsequence of the images. Thus the image of the unit ball is relatively sequentially compact and hence relatively compact. Complete continuity and compactness are equivalent on a Hilbert space.

We first prove the needed Fredholm alternative for a compact operator without assuming self-adjointness. If $C$ is compact, the finite-rank approximation theorem for compact operators on a Hilbert space follows directly by covering the compact closure of $C$'s unit-ball image by finitely many balls of radius $\epsilon$, spanning their centers by a finite-dimensional space $E$, and using the orthogonal projection $P_E$. Then $\Vert C-P_{E}C\Vert \le \epsilon$.

Choose a finite-rank $F$ with $\Vert C-F\Vert <1$, put $E_0=C-F$, and factorThe first factor is invertible by the Neumann series, and $G$ has finite rank. With $V=\mathrm{ran}G$ and $H=V\oplus V^{\perp }$, the second factor has the triangular formIt is injective precisely when the finite-dimensional block $I_V-G{|}_V$ is injective, and onto precisely when that block is onto. These properties are equivalent in finite dimension. This proves that $I-C$ is injective if and only if it is onto; in that case its inverse is bounded.

For $\lambda \ne 0$, apply this result to $C=K/\lambda$. If $\lambda$ lies in the spectrum of a bounded operator $K$, then $K-\lambda I$ is not invertible, hence is not injective. Therefore every nonzero spectral point of $K$ is an eigenvalue. On its eigenspace $E_{\lambda }$, the operator $K$ equals $\lambda I$. Compactness of $K$ would make the closed unit ball of $E_{\lambda }$ compact after dividing by $\lambda$; part (a) forces $\dim E_{\lambda }<\infty$.

It remains to rule out infinitely many distinct eigenvalues away from zero. Suppose $|{\lambda }_n|\ge \epsilon >0$ are distinct, with respective eigenvectors $v_n$. Distinct eigenvalues give linearly independent eigenvectors. Let $V_n=\mathrm{span}(v_1,\dots ,v_n)$ and choose a unit vector $h_n\in V_n\cap V_{n-1}^{\perp }$. Since $V_n$ is invariant and $(K-{\lambda }_{n}I)V_n\subset V_{n-1}$,For $m<n$, $K{h}_m\in V_{n-1}$, and orthogonality gives $\Vert K{h}_n-K{h}_m\Vert \ge |{\lambda }_n|\ge \epsilon$. This contradicts compactness. Thus for each $\epsilon >0$ there are only finitely many distinct spectral points with modulus at least $\epsilon$.

Taking $\epsilon =1/j$ shows that the nonzero spectrum is countable; if it is infinite, its distinct elements can be enumerated as a sequence tending to zero. Since the spectrum is closed, zero then also belongs to it. In an infinite-dimensional space zero belongs to the spectrum even if there are only finitely many nonzero spectral points: an invertible compact $K$ would make $I=K^{-1}K$ compact, contradicting part (a). The only possible accumulation point is zero, and every nonzero eigenspace is finite-dimensional. Zero need not be an eigenvalue. The argument uses invariant spans, not an orthogonal eigenbasis, which is unavailable for a general non-self-adjoint compact operator.

We prove compact perturbation invariance of Fredholm operators using a two-sided inverse modulo compact operators. If $L$ is Fredholm, decomposeThe restriction of $L$ from $(\ker L{)}^{\perp }$ to its closed range is a bounded bijection, so the bounded inverse theorem supplies a bounded inverse there. Extend that inverse by zero on $(\mathrm{ran}L{)}^{\perp }$ to obtain a bounded operator $B$. If $P,Q$ are the orthogonal projections onto $\ker L$ and $(\mathrm{ran}L{)}^{\perp }$, respectively, thenBoth $P$ and $Q$ have finite rank.

Set $T=L+K$. Products of a compact operator and a bounded operator are compact: on one side compactness preserves compact images, and on the other the bounded operator maps the unit ball into a bounded ball. Hencewhere $C,D$ are compact.

Here is why these identities force $T$ to be Fredholm. On $\ker T$, $C=-I$, so part (a) and compactness make $\ker T$ finite-dimensional. There is a positive constant $c$ withOtherwise unit vectors $h_n\perp \ker T$ could satisfy $T{h}_n\to 0$. From $h_n=-C{h}_n+BT{h}_n$ and compactness, a subsequence would converge strongly to a unit vector $h\perp \ker T$ with $Th=0$, which is impossible. The lower bound proves closed range: for a convergent sequence $T{h}_n$, first discard the kernel components, then the remaining $h_n$ are Cauchy and their limit maps to the desired range limit.

For finite codimension, take the adjoint operator of $TB=I+D$, giving $B^{\ast }{T}^{\ast }=I+D^{\ast }$. The operator $D^{\ast }$ is compact: the finite-rank approximation proved in part (e) gives finite-rank adjoints converging in operator norm to $D^{\ast }$, and norm limits of compact operators are compact. On $\ker T^{\ast }$, $D^{\ast }=-I$, so this kernel is finite-dimensional. Finally,Because the range is closed, its cokernel is isomorphic to this finite-dimensional orthogonal complement. This verifies all three Fredholm operator conditions for $L+K$.

For the converse, start with $L+K$ and perturb by the compact operator $-K$. Therefore $L$ is Fredholm if and only if $L+K$ is Fredholm, with no self-adjointness assumption.

Use the stated Fredholm definition of the essential spectrum, which applies to general bounded operators:For each $z$, part (f) applied to $L-zI$ says that $L-zI$ is Fredholm exactly when $L+K-zI$ is Fredholm. Taking the complement of these equivalent Fredholm conditions givesThis is the general Weyl theorem for bounded compact perturbations. Individual isolated eigenvalues can change; the Fredholm obstruction, and hence this essential spectrum, cannot.