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In the l2 sequence space, let $e_n$ have value one in coordinate $n$ and zero elsewhere. This orthonormal sequence is bounded: $\Vert e_n{\Vert }_2=1$. For distinct indices $n,m$, its terms obey $\Vert e_n-e_m{\Vert }_2=\sqrt{2}$. No subsequence can therefore be a Cauchy sequence, so none converges in the norm. Thus even a complete metric space can have a bounded sequence with no convergent subsequence. Finite dimension is essential to the usual Euclidean space extension, and in a general metric space the relevant replacement is compactness or total boundedness together with completeness.

The closed unit ball of a Hilbert space is compact in norm exactly when the space is finite-dimensional. Finite-dimensional compactness follows from the Heine-Borel theorem. In infinite dimension, an orthonormal sequence has pairwise distance $\sqrt{2}$ and no convergent subsequence. Restricting a compact operator that equals a nonzero multiple of the identity on a subspace therefore forces that subspace to be finite-dimensional.

A complete orthonormal sequence $(e_n)$ has $⟨e_n,e_m⟩={\delta }_{nm}$ and dense linear span; equivalently, the only vector orthogonal to every $e_n$ is zero. The Parseval identity says $\Vert x{\Vert }^2={\sum }_n|⟨x,e_n⟩{|}^2$, with expansion $x={\sum }_n⟨x,e_n⟩e_n$ in the Hilbert norm.

Put $G(t,z)=(1-2zt+z^2{)}^{-1/2}$ and $D=1-2zt+z^2$. Direct differentiation givesSubstitution yields $(1-t^2)G_{tt}-2t{G}_t+z^2{G}_{zz}+2z{G}_z=0$. Extracting the coefficient of $z^n$ in the generating function givesthe Legendre differential equation.

A C-star algebra is a complex Banach algebra with a conjugate-linear involution satisfying $(ab{)}^{\ast }=b^{\ast }{a}^{\ast }$, $(a^{\ast }{)}^{\ast }=a$ and $\Vert a^{\ast }a\Vert =\Vert a{\Vert }^2$. It is unital when it has an identity for multiplication. For $B(H)$ use composition and the Hilbert adjoint. Completeness in the operator norm is given; composition is submultiplicative and the adjoint has the required algebraic properties. For a unit vector $\xi$,Taking the supremum and using the opposite bound $\Vert T^{\ast }T\Vert \le \Vert T^{\ast }\Vert \Vert T\Vert =\Vert T{\Vert }^2$ proves the C-star identity. Its identity is $I_H$, proving $B(H)$ is a unital C-star algebra.

Compact operators form a norm-closed linear space and are stable under composition with bounded operators. They are also adjoint stable. To see this directly, choose a finite $\epsilon$-net of the compact image of the unit ball under $T$, and let $P$ project orthogonally onto the span of its centers. Then $\Vert (I-P)T\Vert \le \epsilon$. The finite-rank operator $T^{\ast }P$ approximates $T^{\ast }$ with the same error, so $T^{\ast }$ is compact by norm closure. Thus $\mathcal{K}(H)$ is a C-star algebra, with the inherited norm and operations.

If it has a multiplicative identity $E$, apply $ER=R$ to every rank-one operator with arbitrary prescribed range vector $\xi$. Evaluating that equation on a vector sent to $\xi$ gives $E\xi =\xi$. Hence $E=I_H$. This identity is compact exactly when $H$ is finite dimensional: finite dimension gives compactness of the unit ball, while in infinite dimension an infinite orthonormal sequence has pairwise distance $\sqrt{2}$ and no convergent subsequence. Therefore $\mathcal{K}(H)$ is unital exactly in finite dimension, with the zero-algebra convention understood if $H=0$.

The orthogonal complement $x^{\perp }$ is infinite-dimensional: otherwise $X=\mathrm{span}\{x\}\oplus x^{\perp }$ would be finite-dimensional. Inductively choose an orthonormal sequence $(e_n)$ in $x^{\perp }$, each new vector being orthogonal to the preceding finite linear span. The Bessel inequality gives ${\sum }_n|⟨e_n,z⟩{|}^2\le \Vert z{\Vert }^2$ for every $z\in X$, so $⟨e_n,z⟩\to 0$. Thus $(e_n)$ is a weakly null orthonormal sequence.

SetBy orthogonality, $\Vert x_n{\Vert }^2=\Vert x{\Vert }^2+(1-\Vert x{\Vert }^2)=1$, while $⟨x_n-x,z⟩\to 0$ for every $z$. Hence $x_n\rightharpoonup x$. When $\Vert x\Vert =1$, this formula simply gives the constant sequence $x_n=x$. In particular, the weak closure of the unit sphere contains the whole closed unit ball.

Assume instead that $T$ sends weakly compact sets to norm-compact sets. If $x_n\rightharpoonup 0$, then $K=\{0,x_1,x_2,\dots \}$ is weakly compact: an open cover has a member containing zero, which contains all but finitely many terms of the convergent sequence. Hence $T(K)$ is norm compact. A bounded linear operator is weak-to-weak continuous, because each $g\in F^{\ast }$ pulls back to $g\circ T\in E^{\ast }$. Thus $T{x}_n\rightharpoonup 0$. If $\Vert T{x}_n\Vert$ did not tend to zero, a subsequence with norms bounded below by some $\epsilon >0$ would have a norm-convergent further subsequence with limit $y$. That further subsequence also converges weakly to both $y$ and zero; uniqueness of limits in the Hausdorff weak topology forces $y=0$, contradicting the lower bound. Therefore $\boxed{\Vert T{x}_n\Vert \to 0}$, proving the equivalence and the characterization of completely continuous operators.

Now consider the inclusion of the space of continuous functions on a compact space $C(\mathbb{T})$ into L2 space for normalized Lebesgue measure. It is a bounded linear operator, since $\Vert f{\Vert }_2\le \Vert f{\Vert }_{\infty }$. For a weakly null sequence $(f_n)$ in $C(\mathbb{T})$, point evaluations are continuous linear functionals, so $f_n(t)\to 0$ for every $t$. The Uniform boundedness principle, applied to the canonical embeddings $f_n\in C(\mathbb{T}{)}^{\ast \ast }$, gives a uniform bound $\Vert f_n{\Vert }_{\infty }\le M$: each dual functional is bounded on the weakly convergent sequence. The dominated convergence theorem now yieldsThus $j_2$ satisfies both equivalent conditions.

It is nevertheless not a compact operator. Over complex scalars the functions $f_n(t)=e^{int}$ lie in the unit ball of $C(\mathbb{T})$ and form an orthonormal sequence in $L^2(\mu )$, soThere is no norm-convergent subsequence. Over real scalars use $\cos (nt)$, whose distinct images have distance $1$. Complete continuity controls weakly compact sets, rather than all bounded sets.

Use the printed Fourier transform convention $\hat{\varphi }(t)=\int \varphi (x)e^{-ixt}dx$, interpreted by the Plancherel theorem when $\varphi$ is only square integrable. By the scaling and function translation rules,Taking the Fourier transform of the scaling refinement equation gives $f(t)=m(t/2)f(t/2)$, or equivalentlyConversely this identity, the same scaling rule and injectivity of the Fourier transform recover the scaling refinement equation.

For infinite filters, these statements are equalities in $L^2$, with Fourier series understood in $L^2[0,2\pi ]$. Under the accompanying orthonormality condition, $\{\sqrt{2}\varphi (2\cdot -n)\}$ is an orthonormal sequence, so the refinement series converges in $L^2$ precisely for $a\in {\ell }^2$. Moreover the condition in part (b)(2) makes multiplication by $f$ an isometry from periodic $L^2$ symbols into $L^2(\mathbb{R})$:Applying this to finite filter truncations justifies passage to the limit in both directions. Thus the two paired assertions have a rigorous equivalent interpretation without requiring absolute integrability of $\varphi$.

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.

Use the corrected essential spectrum of a bounded self-adjoint operator and put $A=L-\lambda I$. Essential spectral points are real. If $\ker A$ is infinite-dimensional, choose an orthonormal sequence in the kernel. It converges weakly to zero by the Bessel inequality, and its residuals vanish.

If $\ker A$ is finite-dimensional, membership in the essential spectrum means the range is not closed. Choose unit $v_n\in (\ker A{)}^{\perp }$ with $A{v}_n\to 0$, using the closed-range bound on the kernel complement. A bounded Hilbert space sequence has a weakly convergent subsequence. Its weak limit $v$ satisfies $Av=0$ because bounded operators preserve weak convergence, and $v\in (\ker A{)}^{\perp }$; hence $v=0$. This subsequence is a singular Weyl sequence.

Conversely a singular Weyl sequence first places $\lambda$ in the spectrum of a bounded operator. If it were not essential, the sequential properness for a self-adjoint operator equivalence for $A$ would yield a norm-convergent subsequence. Its weak limit is zero, whereas norm convergence of unit vectors gives a unit norm limit, a contradiction. ThereforeFor nonreal $\lambda$, the resolvent lower bound excludes such a sequence, so the equivalence covers all $\lambda \in \mathbb{C}$.

Throughout this question the Banach spaces and Hilbert spaces are complex. Separability is an additional standing hypothesis, not part of the general definition of a Hilbert space. Here and below unit ball means the closed unit ball. An open unit ball in positive finite dimension is not compact. In finite dimension, equivalence of norms identifies the closed unit ball with a closed bounded subset of Euclidean space, so the Heine-Borel theorem gives compactness in the norm topology.

In an infinite-dimensional Hilbert space, the Gram-Schmidt process constructs an infinite orthonormal sequence $(e_n)$. Put $f_n=e_n/\sqrt{2}$. Then $\Vert f_n\Vert =1/\sqrt{2}\le 1$ and, for $m\ne n$, $\Vert f_n-f_m{\Vert }^2=(1+1)/2=1$. No subsequence is a Cauchy sequence, contradicting sequential compactness. Thus the closed unit ball is norm compact exactly in finite dimension.

For the subsequent essential spectrum arguments, the discrete-spectrum definition must use $\mathrm{ran}(L-\lambda I)$ closed, rather than the printed unshifted range. With that correction, $0\notin {\Sigma }_{\mathrm{e}}(L)$ means exactly that $N=\ker L$ is finite-dimensional and $\mathrm{ran}L$ is closed; this includes the case $0$ is in the resolvent set.

A closed-range bound on the kernel complement supplies the useful equivalenceFor the forward implication, $L:N^{\perp }\to \mathrm{ran}L$ is a bounded bijection between Banach spaces, so the bounded inverse theorem applies. For the reverse implication, any Cauchy sequence of image points has a Cauchy sequence of preimages in $N^{\perp }$, and completeness gives a preimage of its limit.

Now decompose a bounded sequence as $f_n=p_n+v_n$ with $p_n\in N$, $v_n\in N^{\perp }$. If $L{f}_n$ converges, the lower bound makes $(v_n)$ a Cauchy sequence. The finite-dimensional vector space $N$ makes the bounded $p_n$ have a convergent subsequence. Their sum has a norm-convergent subsequence.

Conversely, if every bounded sequence with convergent images has a norm-convergent subsequence, the kernel cannot be infinite-dimensional: an orthonormal sequence in it would have zero images and no convergent subsequence. If the range were not closed, the lower-bound equivalence would provide unit vectors $v_n\in N^{\perp }$ with $L{v}_n\to 0$. Any norm limit would lie in both $N$ and $N^{\perp }$, hence be zero, contradicting its unit norm. This proves the required sequential properness for a self-adjoint operator equivalence.

The spectral-shift repair is essential for later parts. On ${\ell }^2$, take $L=\mathrm{diag}(2,1,1/2,1/3,\dots )$. Its range is not closed, but $2$ is an isolated eigenvalue with one-dimensional eigenspace and closed shifted range. The printed definition would incorrectly place $2$ in the essential spectrum, although no singular Weyl sequence exists there: on the complement of that eigenspace, $\Vert (L-2I)v\Vert \ge \Vert v\Vert$.

Normalize the mutually orthogonal eigenvectors so that $(x_j)$ is an orthonormal sequence. If ${\lambda }_j$ did not tend to zero, some subsequence would satisfy $|{\lambda }_j|\ge \epsilon >0$. For distinct indices in this subsequence, orthogonality givesThus $(T{x}_j)$ would have no Cauchy subsequence, contradicting compactness of $T$. Therefore the orthogonal eigenvectors of a compact operator satisfy

No. Take the Hilbert space $H={\ell }^2$ and its standard orthonormal sequence $(e_n)$. Letthe unit sphere. This set is norm closed but is not convex. By the Bessel inequality, for every $y=(y_j)\in {\ell }^2$,so $e_n\rightharpoonup 0$; equivalently, this is a weakly null orthonormal sequence. Every $e_n$ lies in $C$, but the weak limit $0$ does not. Hence convexity is essential.

A singular Weyl sequence for a bounded self-adjoint operator $L$ at $\lambda$ is a spectral Weyl sequence that also converges weakly to zero. Such a sequence exists exactly when $\lambda$ is in the essential spectrum of a bounded self-adjoint operator. An infinite-dimensional shifted kernel gives a weakly null orthonormal sequence. If that kernel is finite-dimensional but the shifted range is not closed, choose approximate null vectors in the kernel complement and extract a weakly convergent subsequence; its limit belongs to both kernel and complement, so is zero.

Every orthonormal sequence in a Hilbert space converges weakly to zero, by Bessel inequality, but no subsequence converges strongly because distinct terms remain distance $\sqrt{2}$ apart.