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Work in the complete Banach space $C([0,1])$ with the uniform norm. Each $E_m$ is closed. Indeed if $f_j\to f$ uniformly and $x_j$ witnesses membership in $E_m$, pass to a subsequence with $x_j\to x$. For each $y\in [0,1]$, uniform convergence and continuity giveThus $x$ witnesses membership for $f$.

Each $E_m$ has empty interior. Approximate an arbitrary $f$ uniformly by a piecewise-linear function $q$ on a uniform mesh, and let $D$ bound the absolute slopes of $q$. On a sufficiently fine uniform refinement of mesh $\delta$, add a triangular oscillation of fixed small amplitude $b$, alternating between $b$ and $-b$ at successive vertices. The resulting function $g$ is still as close to $f$ as desired, and every segment has absolute slope at least $2b/\delta -D$. For any $x$, choose the farther endpoint of a segment containing $x$. Its distance $h$ is between $\delta /2$ and $\delta$, sowhen $\delta$ is small. Therefore $g\notin E_m$.

By the Baire category theorem, the complement of ${\bigcup }_m{E}_m$ is dense and nonempty. If the local limsup in the question were finite at some $x$, its nearby differences would satisfy a finite bound. The remaining differences, with $|h|\ge \delta >0$, are bounded by $2\Vert f{\Vert }_{\infty }{\delta }^{-\alpha }$. Combining the bounds would put $f$ in some $E_m$, a contradiction. Hence the required limsup is infinite at every point, including one-sided approaches at the endpoints.

Non-characteristic. The first-order principal symbol is ${\nu }_t-i{\nu }_x$, which equals one on $(0,1)$.

Set $z=x+it$. The equation is equivalent to ${\partial }_{\bar{z}}\varphi =({\varphi }_x+i{\varphi }_t)/2=0$. Because $\varphi$ is continuously differentiable, the Cauchy-Riemann equations imply that it is a holomorphic function of $z$. Its convergent complex Taylor series restricts to a real power series along $t=0$, so $g$ is a real analytic function.

For smooth-data instability of the Cauchy-Riemann Cauchy problem, chooseThese are entire functions and satisfy the equation. Their initial derivatives obey ${\sup }_x|{\partial }_x^j{g}_n|\le n^j{e}^{-\sqrt{n}}$, so every finite sum tends to zero. ButThus analytic solutions can exist while continuous dependence on initial data fails for this smooth-data topology.

The Cauchy estimate follows directly from the differentiated Cauchy integral formula:The circle has length $2\pi r$, giving the bound for every integer $j\ge 0$.

Put $w=\varphi -\psi$ and integrate on the straight segment $z=\tau t/|t|$. For fixed $|x|<sR$, a disc centred at $x$ with radius approaching $R(\sigma (\tau )-s)$ stays inside $|\xi |\le \sigma (\tau )R$. Applying the first-derivative Cauchy estimate there givesIntegrating its modulus and taking the supremum over $|x|<sR$ is exactly the asserted integral estimate. The source cancels from $T\varphi -T\psi$.

To obtain a contraction mapping, use the weighted holomorphic norm on a shrinking time domain. Its finite-norm space consists of holomorphic functions on ${\mathcal{C}}_{\alpha }$ with zero initial value; the apparent quotient at $t=0$ is interpreted by a limit. Completeness follows because convergence in this norm implies locally uniform convergence of holomorphic functions, including near $t=0$, and the limiting pointwise bounds give convergence in the norm.

Here is the explicit contraction estimate on a shrinking holomorphic domain. Write $N=\Vert w{\Vert }_{\alpha }$, $r=|t|$, $A=\alpha (1-s)$, and chooseThen $\sigma -s=(A-\tau )/(2\alpha )$ and $\alpha (1-\sigma )-\tau =(A-\tau )/2$. The whole auxiliary polydisc lies inside ${\mathcal{C}}_{\alpha }$, and its supremum of $|w|$ is at most $2N\tau /(A-\tau )$. ConsequentlyThus $\Vert T\varphi -T\psi {\Vert }_{\alpha }\le (4\alpha /R)\Vert \varphi -\psi {\Vert }_{\alpha }$. Choose $0<\alpha <\min (\eta ,R/4)$. The source is bounded on the closed polydisc, say by $M_f$, and $\Vert T0{\Vert }_{\alpha }\le \alpha M_f$. Therefore $T$ maps the Banach space into itself and is a strict contraction mapping. The Banach fixed-point theorem gives a holomorphic fixed point with $\varphi (x,0)=0$. Differentiating the integral identity yields ${\varphi }_t-i{\varphi }_x=f$, establishing this case of the Cauchy-Kovalevskaya theorem.

Finally, near each real initial point, a real analytic function $g$ has a holomorphic extension $G$. Apply the same construction to $\varphi =G(x)+v$ with $v(x,0)=0$ and source $i{G}^{\prime }(x)$, restricting the discs if necessary. This gives a local real analytic function of $(x,t)$ with the prescribed initial value. Equivalently the solution is $\boxed{\varphi (x,t)=G(x+it)}$ wherever the extension is defined.

A contraction semigroup on a Banach space $X$ consists of bounded linear operators $S(t)$ with $S(0)=I$, $S(t+s)=S(t)S(s)$, $\Vert S(t)\Vert \le 1$, and $S(t)x\to x$ as $t\downarrow 0$ for each $x\in X$. Its infinitesimal generator of a semigroup isFor density, Yosida averaging of a semigroup gives $x_h=h^{-1}{\int }_0^{h}S(s)xds$. Taking the difference quotient of this Bochner integral shows that $x_h\in D(A)$ and $A{x}_h=(S(h)x-x)/h$. Strong continuity gives $x_h\to x$, so $D(A)$ is dense.

For closedness, the semigroup restricted to its generator domain satisfiesIf $x_n\to x$ and $A{x}_n\to y$, pass to the limit in this identity. After dividing by $t$, strong continuity gives $(S(t)x-x)/t\to y$. Hence $x\in D(A)$ and $Ax=y$. This proves generator of a strongly continuous semigroup is closed and densely defined.

The contraction form of the Hille-Yosida theorem states that a linear operator $A$ generates a contraction semigroup if and only if it is densely defined and closed, every real $\lambda >0$ lies in its resolvent set, andHere the $n=1$ bound implies the others by taking powers of the same bounded resolvent.

The H1 space is $H^1(\mathbb{R})=\{u\in L^2:u^{\prime }\in L^2\}$ with weak derivative $u^{\prime }$ and squared norm $\Vert u{\Vert }_2^2+\Vert u^{\prime }{\Vert }_2^2$. The displayed weighted space is the one-dimensional harmonic oscillator form domain, with inner productIf $u_n$ is Cauchy in $X$, it converges in $H^1$ to $u$, and $x{u}_n$ converges in L2 space to some $w$. On each bounded interval, multiplication by $x$ is bounded, so $w=xu$ there. Thus $xu\in L^2$ and convergence holds in $X$, proving completeness and the Hilbert space property. Equivalently this is closedness of the multiplication operator by $x$.

For the Sobolev characterization by bounded difference quotients, if $f\in H^1$ thenThe last assertion uses continuity of translation of a function in L2 space. Conversely, if $f\in L^2$ and the quotients are uniformly bounded for $0<|h|<1$, take a weakly convergent subsequence as $h\to 0$. For every test function $v$,The weak limit is therefore the weak derivative $f^{\prime }\in L^2$. This proves the characterization.

For a convex subset of a Banach space, weak closure equals norm closure. The Mazur theorem states that for every convex set $C\subseteq X$,Since every bounded linear functional is norm-continuous, the weak topology is weaker than the norm topology, giving ${\overline{C}}^{\Vert \cdot \Vert }\subseteq {\overline{C}}^w$.

For the reverse inclusion, use this form of the Hahn-Banach separation theorem: if $D$ is a nonempty norm-closed convex set in a real normed vector space and $x\notin D$, there is a continuous real-linear functional $F$ such thatApply it to $D={\overline{C}}^{\Vert \cdot \Vert }$. In a complex Banach space, use the underlying real space; every continuous real-linear $F$ is the real part of the complex bounded linear functional $f(v)=F(v)-iF(iv)$. The separating strict inequality gives a neighbourhood open for the weak topology of $x$ disjoint from $D$, so $x\notin {\overline{C}}^w$. The empty set case is immediate. This proves the Mazur theorem.

A weakly null sequence has disjoint convex blocks converging to zero in norm. If $x_n\stackrel{w}{\to }0$, then for every starting index $m$,The Mazur theorem places $0$ in the norm closure of that tail convex hull. Thus a finite convex combination of vectors from any tail can have norm less than any prescribed $\epsilon >0$.

Choose the convex blocks recursively. After selecting the previous terminal index $q_{n-1}$, let $p_n>q_{n-1}$, and choose a finite convex combination from $\{x_i:i\ge p_n\}$ with norm less than $2^{-n}$. Choose $q_n$ beyond its largest used index, padding the intervening coefficients and the final coefficient with zero. This ensures the printed strict condition $p_n<q_n$ as well as $q_n<p_{n+1}$. Setting unused coefficients to zero yieldsTherefore $u_n\to 0$ in norm. The recursive tail selection is what makes these convex blocks, rather than merely unrelated convex combinations.

Next suppose $X^{\ast }$ is separable and $(x_n)$ is bounded. Use the canonical embedding into the bidual $J_X:X\to X^{\ast \ast }$, where $J_{X}x(f)=f(x)$. If $\Vert x_n\Vert \le M$, then $J_X{x}_n\in M{B}_{X^{\ast \ast }}$. The Banach-Alaoglu theorem makes this ball compact in $\sigma (X^{\ast \ast },X^{\ast })$, and weak-star metrizability of the dual ball makes it metrizable because its predual $X^{\ast }$ is separable. A compact space that is a metric space is sequentially compact, so some subsequence $(y_n)$ satisfiesThe limit need not belong to $J_{X}X$; it is precisely the use of the bidual space that supplies compactness without assuming reflexivity.

For a convex block $u_n={\sum }_{i=p_n}^{q_n}{a}_i{y}_i$, every $f\in X^{\ast }$ satisfiesBoth $f(y_n)$ and $f(u_n)$ tend to $\varphi (f)$, soThis is the useful convex-block cancellation of a weak-star limit; it does not require the limit to be a vector in $X$.

The quotient sequence has approximate lifts in $3B_X$ that are weakly null after passing to a subsequence. Let $q:X\to Z=X/Y$ be the quotient map, with the quotient norm. Since $\Vert z_n\Vert \le 1$, choose $v_n\in X$ such thatThis uses the infimum defining the quotient norm and does not assume that it is attained. By the preceding compactness argument, pass to a subsequence $y_n=v_{k_n}$ with a common weak-star limit in $X^{\ast \ast }$. Its quotient images $q(y_n)=z_{k_n}$ form a weakly null sequence.

Apply the convex block construction to $(q(y_n))$ in the quotient Banach space. Using the same coefficients on $(y_n)$ gives convex blocksSet $x_n=y_n-u_n$. The convex-block cancellation of a weak-star limit proves $x_n\stackrel{w}{\to }0$, whileThus the approximate weakly null lifting through a quotient is achieved with the required bound:

For a proper convex function $E$ on a real Hilbert space, its subdifferential at $y$ isIn a Banach space, interpret the pairing with an element of the dual space. For a chosen subgradient $p\in \partial E(y)$, the Bregman distance isIf also $q\in \partial E(x)$, its symmetric Bregman distance isThe subgradient inequality proves nonnegativity. These quantities depend on the chosen subgradients and generally are not metrics: they need not distinguish different points or obey the triangle inequality.

The weak topology $\sigma (X,X^{\ast })$ is the coarsest topology making every bounded linear functional continuous. A neighbourhood basis at $x$ consists of sets $\{y:|f_j(y-x)|<ϵ,1\le j\le m\}$ for finite families in $X^{\ast }$.

Mazur theorem says that for any convex subset $C$ of a real or complex normed vector space,The norm closure is convex. If $x$ is outside it, the Hahn-Banach separation theorem gives a bounded linear functional $f$ and a real $a$ with $\mathrm{Re}f(x)>a\ge {\sup }_{y\in C}\mathrm{Re}f(y)$. In the real case omit the real part. Thus $x$ has a weak neighbourhood missing $C$, proving that the weak closure lies in the norm closure. The other inclusion follows because the weak topology is weaker than the norm topology. In the complex case real separation is converted to a complex functional by $f(z)=h(z)-ih(iz)$.

The sequential formulation, Mazur lemma, follows as well. If $x_n\rightharpoonup x$, then $x$ lies in the weak closure of each tail and hence in the norm closure of its convex hull. Select a finite convex combination of the $n$th tail at norm distance less than $1/n$ from $x$.

A weakly bounded set $D$ satisfies ${\sup }_{x\in D}|f(x)|<\infty$ for every $f\in X^{\ast }$. Regard $J(D)$ as a pointwise bounded family of functionals on $X^{\ast }$. The completeness of the dual space holds even if $X$ is incomplete: a norm-Cauchy sequence of functionals has a pointwise bounded linear limit and then converges uniformly on the unit ball. The Uniform boundedness principle therefore gives ${\sup }_{x\in D}\Vert Jx\Vert <\infty$, and $\Vert Jx\Vert =\Vert x\Vert$ proves norm boundedness.

One can see the precise Baire argument here. The closed sets $F_n=\{f\in X^{\ast }:{\sup }_{x\in D}|f(x)|\le n\}$ cover the Banach space $X^{\ast }$. The Baire category theorem makes some $F_n$ contain a ball $f_0+r{B}_{X^{\ast }}$, after shrinking the radius. For $\Vert g\Vert \le r$, both $f_0$ and $f_0+g$ lie in $F_n$, so ${\sup }_D|g(x)|\le 2n$. Scaling and the dual norm formula yield ${\sup }_D\Vert x\Vert \le 2n/r$. A weakly compact set is weakly bounded because each functional has compact, hence bounded, image. Consequently it is norm bounded.

For a Banach space, let $J:X\to X^{\ast \ast }$ be its canonical isometry. We use two explicitly stated weak-star facts: Banach-Alaoglu theorem makes a dual unit ball weak-star compact, and Goldstine theorem makes $J(B_X)$ weak-star dense in $B_{X^{\ast \ast }}$. The weak topology on $X$ is carried by $J$ to $\sigma (X^{\ast \ast },X^{\ast })$ on its image, since their coordinates are the same evaluations $f(x)$.

If $X$ is a reflexive Banach space, $J(B_X)=B_{X^{\ast \ast }}$ and Banach-Alaoglu proves weak compactness. Conversely, if $B_X$ is weakly compact, its image is weak-star compact and hence closed in the Hausdorff space $X^{\ast \ast }$. Goldstine density then forces $J(B_X)=B_{X^{\ast \ast }}$, which implies $JX=X^{\ast \ast }$. This proves the weak compactness characterization of reflexivity:Finally let $K\subset X$ be weakly compact and let $(f_n)\subset B_{X^{\ast }}$ separate points. If $K$ is empty there is nothing to prove; otherwise defineThe summands are bounded by $2^{-n}$ and separation makes $d(x,y)=0$ only for $x=y$. The coordinate maps are weakly continuous, so the series, being uniformly convergent, makes the identity from weak $K$ to metric $K$ continuous. A continuous bijection from a compact space to a Hausdorff space is a homeomorphism. Thus $d$ induces precisely the weak topology on $K$, the countable separating family metrizes a weakly compact set result. No norm-density of the separating family in $X^{\ast }$ is claimed or required.

For a real Banach space $X$ and a proper convex function $E:X\to \mathbb{R}\cup \{+\infty \}$, the convex conjugate is the extended-valued function on the continuous dual spaceHere proper means that $E$ never takes $-\infty$ and is finite somewhere. The pairing is the duality pairing, or the inner product after identification in a Hilbert space. The conjugate is convex and lower semicontinuous because it is a supremum of continuous affine functions. Its definition gives the Fenchel–Young inequality $E(x)+E^{\ast }(p)\ge ⟨x,p⟩$; the Fenchel-Moreau theorem recovers $E$ as $E^{\ast \ast }$ under the stated closed-convex hypotheses.

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.

On a real Hilbert space, for a proper lower-semicontinuous convex function $E$, the proximal operator isThe quadratic makes this objective strongly convex and coercive; the affine lower bound available for a proper closed convex function and weak lower semicontinuity give existence. The minimizer is unique. The subdifferential optimality condition isFor a positive parameter $t$, ${\mathrm{prox}}_{tE}$ uses $tE(x)$ in the same formula. On a general Banach space, the analogous norm-squared minimization is Banach-space proximal minimization; it can be set-valued and the identity-valued Hilbert derivative must be replaced by the duality mapping.

First use the usual Hilbert space convention for the proximal operator, identifying the dual element $z$ with its representing vector. For $c\ne 0$, completing the square givesSet $v=cx-y$ and multiply the objective by $c^2$. Its variable part becomes $\Vert v-[c(q-z)-y]{\Vert }^2/2+\alpha c^{2}J(v)$. Thus the proximal operator under affine rescaling isNegative $c$ is allowed. If $c=0$ and $J(-y)<\infty$, the penalty is constant and the answer is $q-z$. If $c=0$ and $J(-y)=\infty$, $E$ is identically infinite, so there is no proper proximal problem.

The PDF actually assumes an arbitrary Banach space, so the Hilbert input shift cannot be asserted literally under its hypotheses. Define ${\mathrm{Prox}}_F(s)=\arg {\min }_v\{F(v)+\Vert v-s{\Vert }^2/2\}$ as a possibly empty or set-valued minimizer set. Absolute homogeneity of the norm still gives the valid change of variablesThis retains the linear term, rather than shifting by a dual vector. Equivalently its optimality condition uses the duality mapping $\mathcal{J}=\partial (\Vert \cdot {\Vert }^2/2)$:For a concrete failure of the Hilbert shortcut, take $({\mathbb{R}}^2,\Vert \cdot {\Vert }_1)$, $J=0$, $q=y=0$, $c=1$ and $z=(1,1)$. Every $x_1,x_2\le 0$ with $x_1+x_2=-1$ minimizes $\Vert x{\Vert }_1^2/2+x_1+x_2$ with value $-1/2$. The naive shifted answer $(-1,-1)$ has value zero and is not a minimizer. Hence the simple shifted formula requires Hilbert geometry; general Banach proximal minimizers need not be unique. For $c=0$ with finite constant penalty the Banach condition is $-z\in \mathcal{J}(x-q)$, not generally $x=q-z$.

For a convex functional $J:U\to \mathbb{R}\cup \{+\infty \}$ on a real Banach space, its subdifferential at $v$ is empty unless $J(v)$ is finite; otherwiseGiven $p_v\in \partial J(v)$, the Bregman distance isIt can be infinite if $J(u)$ is infinite. If $J$ is differentiable at $v$, the choice is its derivative; otherwise the base subgradient must be specified. Given also $p_u\in \partial J(u)$, add both directional distances to obtain the symmetric Bregman distanceThe symmetric version here is the sum, not half the sum. These nonnegative quantities are not generally metrics and can vanish at distinct points.

For real $u\in L^1(\Omega )$, its total variation seminorm on a domain isThe vector fields have compact support inside $\Omega$, so no boundary term is charged. Taking both $\xi$ and $-\xi$ makes this equivalent to a supremum of absolute pairings. When finite, the distributional derivative $Du$ is a finite vector-valued Radon measure, and the displayed supremum is its total variation norm of a measure. For a smooth function it equals ${\int }_{\Omega }|\nabla u{|}_{2}dx$.

The function of bounded variation on a domain belongs to the Banach spaceThe L1 norm is needed because variation alone vanishes on constant functions and is only a seminorm. No smoothness of $u$ is part of this definition.

Take $Y=X^{\ast \ast }$, which is a Banach space by the completeness of the dual just proved. The canonical embedding into the bidual isFor fixed $x$ this is a bounded linear functional on $X^{\ast }$, since $|f(x)|\le \Vert f\Vert \Vert x\Vert$, and $\Phi$ is linear in $x$. This gives $\Vert \Phi (x)\Vert \le \Vert x\Vert$. For $x\ne 0$, the norming functional from the Hahn-Banach theorem has $\Vert f\Vert =1$ and $f(x)=\Vert x\Vert$, giving the reverse inequality. At zero equality is immediate. ThusIn particular $\Phi$ is injective and isometric. If a dense isometric embedding is wanted, rather than merely the stated embedding, take the norm closure of $\Phi (X)$ in $X^{\ast \ast }$; a closed subspace of a Banach space is complete and supplies a completion of a normed space.

One form of the Baire category theorem is: a nonempty complete metric space cannot be a countable union of closed nowhere dense subsets. Equivalently, a countable intersection of open dense subsets is dense. A Banach space is a complete metric space for its norm metric, so this form applies directly to the next part.

Suppose $(x_k{)}_{k\ge 1}$ were a countable Hamel basis. Put $E_n=\mathrm{span}(x_1,\dots ,x_n)$. Each $E_n$ is a finite-dimensional, hence closed, linear subspace of $X$. It is proper because $X$ is infinite dimensional, and therefore a nowhere dense set. A Hamel basis means that every vector is a finite linear combination, so $X={\bigcup }_n{E}_n$. This contradicts the Baire category theorem for the Banach space $X$. Hence an infinite-dimensional Banach space has no countable Hamel basis. The same proof excludes any countable algebraic spanning set.

If the bounded injective Fourier transform $L^1\to C_0$ were surjective, the open mapping theorem for Banach spaces would make its inverse bounded, yielding $\Vert h{\Vert }_1\le C\Vert \hat{h}{\Vert }_{\infty }$. But the preceding $g_n$ have uniform norm $2$, whereas $\Vert h_n{\Vert }_1$ diverges.

For a quantitative lower bound, on $0<x\le 1/4$ we have $\sin (2\pi x)\ge 4x$. On each interval $I_j=[(j+1/12)/n,(j+5/12)/n]$, $|\sin (2\pi nx)|\ge 1/2$. For $1\le j\le \lfloor n/8\rfloor$ these intervals lie below $1/4$ for large $n$, soThis contradicts boundedness of the inverse. Therefore the Fourier transform has proper dense image in $C_0(\mathbb{R})$.

The weak-star topology $\sigma (X^{\ast },X)$ is generated by the seminorms $p_x(g)=|g(x)|$, $x\in X$. Thus every evaluation is continuous, so $f_n\stackrel{w^{\ast }}{\to }f$ implies $f_n(x)\to f(x)$ for each $x$. Conversely, a basic neighborhood of $f$ imposes finitely many conditions $|(g-f)(x_j)|<\epsilon$. Pointwise convergence eventually satisfies all of them simultaneously. HenceThe Uniform boundedness principle, applied on the Banach space $X$, gives $M={\sup }_n\Vert f_n\Vert <\infty$. If $x_n\to x$ in the norm topology, thenTherefore $\boxed{f_n(x_n)\to f(x)}$. Completeness of $X$ is used in this uniform boundedness step.

If $C\subseteq X^{\ast }$ is norm compact, the identity map from $(C,\Vert \cdot \Vert )$ to $(C,w^{\ast })$ is a continuous bijection. The target is Hausdorff, since distinct functionals differ at some $x\in X$. The compact-to-Hausdorff continuous bijection theorem gives

The Banach-Alaoglu theorem states that the closed dual unit ball $B_{X^{\ast }}$ is compact in the weak-star topology; in fact $X$ need only be a normed vector space. To prove it, work over $\mathbb{F}=\mathbb{R}$ or $\mathbb{C}$ and formEach $D_x$ is compact, so $Q$ is compact by the Tychonoff theorem. Embed $B_{X^{\ast }}$ by $g\mapsto (g(x){)}_{x\in X}$. Its image consists exactly of those families $(z_x)$ satisfyingThese are closed conditions in the product topology. Conversely, any family satisfying them defines a linear functional bounded by $|z_x|\le \Vert x\Vert$, hence an element of $B_{X^{\ast }}$. The image is therefore closed and compact. The induced product topology is precisely the weak-star topology, proving the theorem. This is a compactness proof for arbitrary nets, with no separability assumption.

For the dual map of bounded linear operator $T^{\ast }:Y^{\ast }\to X^{\ast }$, each $x\in X$ satisfiesEvaluation at the fixed vector $Tx$ is weak-star continuous, so $\boxed{T^{\ast }\text{ is weak-star-to-weak-star continuous}}$.

Suppose $T$ is a compact operator. By the allowed Schauder theorem for compact operators, $T^{\ast }$ is compact. If $S\subseteq Y^{\ast }$ is bounded, the norm closure $C=\overline{T^{\ast }(S)}$ is norm compact. We already proved that its norm and weak-star topologies coincide. The weak-star continuous map $T^{\ast }{|}_S$ into $C$ is consequently norm continuous. Conversely, if $T^{\ast }$ is weak-star-to-norm continuous on bounded subsets, its restriction to $B_{Y^{\ast }}$ maps a weak-star compact set, by Banach-Alaoglu theorem, to a norm compact set. Thus $T^{\ast }$ is a compact operator, and the Schauder theorem for compact operators gives