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In an infinite-dimensional Hilbert space, every bounded sequence has a weakly convergent subsequence, but it need not have a norm-convergent one. For example an orthonormal sequence $e_n$ has pairwise distance $\sqrt{2}$ and therefore no strongly convergent subsequence, although $e_n\rightharpoonup 0$ by Bessel inequality.

More generally, weak sequential compactness of bounded sequences in a reflexive Banach space ensures a weakly convergent subsequence with its limit in that space. Conversely, this property for all bounded sequences characterizes reflexivity, using the Eberlein-Smulian theorem and weak compactness of the closed unit ball. Thus reflexivity replaces finite-dimensional norm compactness by weak sequential compactness.

In a nonreflexive Banach space there is no such general weak-subsequence guarantee. The coordinate unit vectors in ${\ell }^1$ give an example: by the Schur property, weak convergence would imply norm convergence, impossible for their pairwise distance two. Banach-Alaoglu theorem supplies weak-star compactness of a dual unit ball, which gives convergent subnets, not automatically convergent subsequences. For a separable predual, the weak-star compact dual ball is metrizable and subsequences do suffice. For an arbitrary Banach space one can also regard a bounded sequence in its bidual and extract a weak-star convergent subnet there; the limit need not belong to the original space. These qualifications distinguish weak from weak-star compactness.

Use the standard inverse-problem setting of a bounded linear $K:{\ell }^2\to {\ell }^2$ and exact data having a finite-penalty solution. Let $u^{†}$ denote the unique $J$-minimizing exact solution for $J(u)=\Vert u{\Vert }_1+\beta \Vert u{\Vert }_2^2$, $\beta >0$. This is the limiting target of elastic net regularization in Hilbert sequence space; it need not be the Moore–Penrose minimum-norm solution unless additional hypotheses identify them, for example injectivity of $K$.

For each $\alpha >0$, the reconstruction objective is weakly lower semicontinuous, coercive by $\alpha \beta \Vert u{\Vert }_2^2$, and strictly convex. The direct method on the reflexive l2 sequence space gives existence, and strict convexity gives uniqueness. For two data sets, monotonicity of the one-norm subgradients and the quadratic penalty give, with $d=u_1-u_2$ and $g=f_1-f_2$,Completing the square yields $\Vert d\Vert \le \Vert g\Vert /(2\sqrt{2\alpha \beta })$, so each reconstruction map is continuous.

Minimality against $u^{†}$ givesChoose $\alpha \to 0$ and ${\delta }^2/\alpha \to 0$, for example $\alpha =\delta$. This bounds the Hilbert norms, drives the data residual to zero and gives $\mathrm{limsup}J(u_{\alpha }^{\delta })\le J(u^{†})$. Every weakly convergent subsequence has an exact solution as its limit, by boundedness of $K$; weak lower semicontinuity and uniqueness of the penalty minimizer identify that limit with $u^{†}$. Separately applying weak lower semicontinuity to the one-norm givesWeak convergence and convergence of Hilbert norms imply strong convergence by the Radon-Riesz theorem. This proves norm convergence of elastic net regularization:The argument applies to every sequence of noise levels and admissible data, hence gives the usual uniform noise-ball convergence for a fixed admissible exact datum.

For nonlinearity of elastic net reconstruction, zero is the minimizer exactly when $\Vert K^{\ast }{f}^{\delta }{\Vert }_{\infty }\le \alpha$. This includes a neighborhood of zero. If $K\ne 0$, choose $e_j$ with $K{e}_j\ne 0$ and $f^{\delta }=tK{e}_j$ for $t\Vert K{e}_j{\Vert }^2>\alpha$; zero then fails its optimality condition. A linear map vanishing on a neighborhood would vanish everywhere, so the elastic-net reconstruction is nonlinear whenever $K\ne 0$. For $K=I$ the explicit formula is $[R_{\alpha }f{]}_j=\mathrm{sign}(f_j)(|f_j|-\alpha {)}_{+}/(1+2\alpha \beta )$, exhibiting soft thresholding.

A concrete distinction from Moore–Penrose convergence is $K(u)=u_1+2u_2$ (embedded in the first output coordinate of ${\ell }^2$), exact datum one, and $\beta =1$. Its minimum-norm solution is $(1/5,2/5,0,\dots )$, with penalty $4/5$, while the unique penalty-minimizing exact solution is $(0,1/2,0,\dots )$, with penalty $3/4$. For $0<\alpha <2$ the exact-data elastic-net reconstruction is $(0,(2-\alpha )/(4+2\alpha ),0,\dots )$, converging to the latter. Thus it is not a Moore–Penrose-consistent regularization for every noninjective operator under the earlier definition.

The PDF supplies neither a nonzero assumption on $K$ nor a target/domain qualification here. Its unconditional nonlinearity claim is false for $K=0$, whose reconstruction is the zero linear map. The convergence proof above states the finite-penalty exact-solution hypotheses explicitly; it establishes convergence to the penalty-minimizing solution, not an automatically identical minimum-norm least-squares solution.

Use the standard bounded linear operator setting for $K:U\to V$, so the adjoint operator $K^{\ast }:V\to U^{\ast }$ in the updates is defined on every datum. Put $q=K^{\ast }K{u}_{\lambda }$. The generalized singular vector relation is $\lambda q\in \partial J(u_{\lambda })$ with $\Vert K{u}_{\lambda }\Vert =1$ and $\lambda =J(u_{\lambda })\ge 0$. For an absolutely one-homogeneous functional, a subgradient $p$ at $u$ is characterized by $⟨p,u⟩=J(u)$ and $⟨p,v⟩\le J(v)$ for all $v$. Consequently $\lambda q$ is also a subgradient at every positive multiple of $u_{\lambda }$, and $bq\in \partial J(0)$ for every $0\le b\le \lambda$.

Use the initialization $u^0=p^0=0$ and start the first update at $k=0$; the PDF lists $k=1,2,\dots$ despite specifying only index-zero initial data. Suppose $p^k=b_{k}q$ with $0\le b_k\le \lambda$. DefineIf it is positive, $u^{k+1}=c_{k+1}{u}_{\lambda }$ and $p^{k+1}=\lambda q$ satisfy the optimality equation and the dual update. If it is zero, take $u^{k+1}=0$ and $p^{k+1}=(b_k+\gamma /\alpha )q\in \partial J(0)$. Thus this compatible branch hasInduction proves the finite termination of Bregman iteration on a singular vector formulasOnce the dual coefficient reaches $\lambda$, the next reconstruction is $\gamma u_{\lambda }$. Therefore a guaranteed stopping index for this branch isThe formulas also cover $\lambda =0$, where exact reconstruction occurs at the first update. The subgradient conditions are sufficient for global minimization because each objective is convex.

There is a genuine uniqueness issue in the printed claim about arbitrary iterates. Strict convexity of data fidelity in $Ku$ ensures that any two minimizers at a given step have the same image under $K$. Induction therefore makes every minimizer branch have the same dual variables and attain exact data at the stated finite index. It does not make their states equal to the specified $\gamma u_{\lambda }$ if $K$ has a null space.

For a counterexample take $U={\mathbb{R}}^2$, $V=\mathbb{R}$, $K(x,y)=x$, $J(x,y)=|x|$, $u_{\lambda }=(1,0)$, and $\lambda =1$. Every step leaves $y$ completely unconstrained. Choosing the minimizing state's second coordinate to be $k$ gives valid updates and subgradients but never reaches $(\gamma ,0)$ and does not converge as a state sequence. Thus finite exact-data recovery holds for all minimizer branches; exact recovery of the stated vector needs uniqueness or the aligned branch. If $K$ is injective, equal reconstructed data identify the state, and the literal vector conclusion follows.

Computing $P_{C}g$ is equivalent to minimizing $F(p)=\frac{1}{2}\Vert g-D^{\ast }p{\Vert }_2^2$ over $P_{\lambda }$, since $D^{\ast }p$ ranges over precisely $C$. Differentiating with the adjoint operator gives $\nabla F(p)=D(D^{\ast }p-g)$, whose Lipschitz constant is $L=\Vert D{D}^{\ast }\Vert =\Vert D{\Vert }^2$. Starting with any $p^0\in P_{\lambda }$, for example zero, the projected-gradient dual total variation algorithm isThe block formula is the Euclidean projection onto a convex set $P_{\lambda }$, since it clips each two-dimensional block to a Euclidean ball of radius $\lambda$. Both components must be clipped together for the isotropic block mixed norm.

For the unscaled forward differences, $\Vert D_x^{+}u{\Vert }_2^2\le 4\Vert u{\Vert }_2^2$ and likewise in the other direction, so $L\le 8$. More exactly, for $N>1$ the one-dimensional path difference has eigenvalues $4{\sin }^2(k\pi /(2N))$, $0\le k<N$, and the square-grid operator sums two such spectra. Thus $L=8{\cos }^2(\pi /(2N))$. Finite-dimensional projected gradient descent for a convex objective with an $L$-Lipschitz gradient converges from every feasible starting point to a dual minimizer for fixed $0<\tau <2/L$. The theorem assumes a nonempty closed convex set, a convex differentiable objective with globally Lipschitz gradient, and a nonempty minimizer set. All hold here: $P_{\lambda }$ is compact and convex, and $F$ is a continuous convex quadratic. Finite-dimensional convergence is of the full sequence, rather than only its objective values.

Consequently any fixed step with $0<\tau <1/4$ is safe on this grid; the more conservative $0<\tau \le 1/8$ is also safe. The dual minimizer need not be unique because $D^{\ast }$ has a null space, but $D^{\ast }{p}^k\to P_{C}g$ and $u^k\to u$ uniquely. For $N=1$, the gradient penalty vanishes and $u=g$, so no positive-$L$ step restriction is needed.

The continuous dual space $X^{\ast }$ consists of all bounded real linear functionals on $X$, with operator norm $\Vert f\Vert ={\sup }_{\Vert x\Vert \le 1}|f(x)|$. Let $(f_n)$ be Cauchy in this norm. For each $x$, $|f_n(x)-f_m(x)|\le \Vert f_n-f_m\Vert \Vert x\Vert$, so $f_n(x)$ converges in $\mathbb{R}$; define its limit to be $f(x)$.

Pointwise limits of the linear identities show that $f$ is linear. The sequence of norms is bounded, say by $M$, so $|f(x)|\le M\Vert x\Vert$ and $f\in X^{\ast }$. Given $\epsilon >0$, choose $N$ so that $\Vert f_n-f_m\Vert \le \epsilon$ for all $m,n\ge N$. Holding $n$ fixed and taking the pointwise limit in $m$ gives $|f_n(x)-f(x)|\le \epsilon \Vert x\Vert$ for all $x$. Thus $\Vert f_n-f\Vert \le \epsilon$, provingCompleteness of $X$ is not needed; completeness of the scalar field supplies the pointwise limit.

For $1<p<\infty$, apply Fatou lemma to the nonnegative sequence $|f_n{|}^p$, which converges pointwise to $|f{|}^p$:For $p=\infty$, each inequality $\Vert f_n{\Vert }_{\infty }\le M$ fails pointwise only on a null set. Outside the union of these countably many null sets, $|f_n(x)|\le M$ for every $n$, so the pointwise limit also satisfies $|f(x)|\le M$. Therefore in both cases

Use singular homology with integer coefficients. The Künneth theorem for triangulable spaces gives a split short exact sequence with tensor terms ${⨁}_{i+j=k}{H}_i(X)\otimes H_j(Y)$ and torsion terms ${⨁}_{i+j=k-1}{\mathrm{Tor}}_1(H_i(X),H_j(Y))$. When one factor's homology groups are free abelian, the torsion terms vanish and the tensor map is an isomorphism.

The circle has $H_0(S^1)=H_1(S^1)=\mathbb{Z}$ and all other groups zero. Inductively, the homology groups of a torus are free, andPascal's identity givesGeometrically the generators correspond to choosing $k$ of the $n$ circle factors and taking their product, consistent with the cellular model having one $k$-cell for each such subset and cancelling opposite boundary faces.

A compact space is one for which every open cover has a finite subcover. A metric space is sequentially compact when every sequence in it has a convergent subsequence whose limit belongs to the space.

Let $(x_n)$ be any sequence in a complete, totally bounded metric space $X$. Cover $X$ by finitely many balls of radius $2^{-1}$; one contains infinitely many terms. From those terms choose an infinite subsequence lying in one ball of radius $2^{-2}$, and continue. The diagonal argument produces a subsequence $(x_{n_j})$ whose tail from its $j$th term onward lies in a ball of radius $2^{-j}$. Hencewhenever $k,\ell \ge j$, so the subsequence is Cauchy. Completeness makes it converge in $X$.

Every sequence in $X$ therefore has a convergent subsequence, so $X$ is sequentially compact. Sequential compactness is equivalent to compactness for metric spaces, and thus $X$ is compact. This proves that a complete totally bounded metric space is compact.

Suppose $a_n\to L$. Given $\epsilon >0$, choose $N$ such that $|a_k-L|<\epsilon /2$ for $k>N$. ThenThe first term tends to zero because its numerator is fixed, and the second is at most $\epsilon /2$. Thus the Cesaro mean tends to $L$.

The converse is false. For $a_n=(-1{)}^n$, the Cesaro means tend to $0$, while the original sequence alternates between $-1$ and $1$ and does not converge.

Using multi-index notation, the Schwartz space isA sequence ${\phi }_m$ converges to $\phi$ in this Fréchet space when $p_{\alpha ,\beta }({\phi }_m-\phi )\to 0$ for every $\alpha ,\beta$. The space ${\mathcal{S}}^{\prime }({\mathbb{R}}^n)$ of tempered distributions is the continuous dual of $\mathcal{S}({\mathbb{R}}^n)$, and $u_m\to u$ there means weak convergence of distributions, namely $⟨u_m,\phi ⟩\to ⟨u,\phi ⟩$ for every $\phi \in \mathcal{S}$.

Continuity of a linear functional immediately implies that ${\phi }_m\to 0$ entails $⟨u,{\phi }_m⟩\to 0$. Conversely, enumerate the Schwartz seminorms as $p_1,p_2,\dots$. If $u$ were not continuous, then for each $m$ one could choose ${\phi }_m$ such thatEvery fixed seminorm tends to zero along this sequence, so ${\phi }_m\to 0$ in $\mathcal{S}$, contradicting the assumed sequential property. This is the sequential continuity criterion for a linear map on a metrizable topological vector space.

The intermediate value theorem says that a continuous $f:[a,b]\to \mathbb{R}$ takes every value between $f(a)$ and $f(b)$. For $f(a)<y<f(b)$, the closed sets where $f\le y$ and $f\ge y$ cannot separate the connected interval; equivalently, taking the supremum of $\{x:f(x)\le y\}$ and using continuity gives a point with value $y$.

Set $\varphi (a)=0$ and $\varphi (x)=\sin (1/(x-a))$ for $x>a$. Every interval $[a,b]$ contains a zero $c<b$; continuity on $[c,b]$ makes $\varphi$ take every value between $0=\varphi (a)$ and $\varphi (b)$, yet $\varphi$ is discontinuous at $a$.

A monotone function can be discontinuous only by a jump. If it had a jump at $c$, any number strictly between the left and right limits would lie between $f(a)$ and $f(b)$ but would not be attained, contrary to the hypothesis. Thus it is continuous.

For the last assertion, pass to subsequences with $x_n,y_n$ both within $1/n$ of $a$ and $g(x_n)$ near $l$, $g(y_n)$ near $L$. For $\lambda \in (l,L)$, the intermediate value theorem on the interval joining $x_n$ and $y_n$ gives $z_n$ with $g(z_n)=\lambda$; then $z_n\to a$. The endpoint cases use the original sequences.

The arithmetic--geometric mean inequality gives $x_n\ge 1$ from $n=2$ onward. For $x>1$,so the tail decreases to a limit $L\ge 1$. Passing to the recurrence gives $2L=L+L^{-1}$, hence $L=1$.

For the subadditive sequence, $0\le x_n/n\le x_1$, so it is bounded. The Fekete lemma can be proved directly here as follows. Let $\alpha ={\inf }_k{x}_k/k$. Fix $k$ and write $n=qk+r$, $0\le r<k$. ThenThus $\mathrm{limsup}{x}_n/n\le \alpha$, while the definition gives $\mathrm{liminf}{x}_n/n\ge \alpha$. Hence $x_n/n\to \alpha$.

All binary sequences are uncountable, while part (ii) accounts for the sequences which fail to have infinitely many symbols of both kinds and is countable. Removing that countable subset leaves an uncountable set.

Sequences with finitely many ones correspond to finite subsets of $\mathbb{N}$, a countable union of the countable sets of $k$-element subsets. Complementation gives the same result for finitely many zeros. Their union is therefore countable.

The set is uncountable by Cantor's diagonal argument: from any proposed list, form a sequence whose $n$th bit differs from the $n$th bit of the $n$th listed sequence.