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A finite regular complex measure of total variation one can be approximated on a finite-dimensional vector subspace of continuous functions by a finite sum of phased point evaluations with ${\sum }_i|t_i|=1$. The Krein-Milman theorem and the extreme points of the dual unit ball of C(K) give weak-star approximation by convex combinations of phased point masses. A finite norm net of the vector subspace unit ball turns finitely many scalar approximations into one uniform estimate. Repeated nodes may be kept separate to retain the exact coefficient-magnitude sum.

For a bounded operator $K$ between Hilbert spaces, compactness is equivalent to $f_n\rightharpoonup f$ implying $K{f}_n\to Kf$ in norm. The Uniform boundedness principle bounds a weakly convergent sequence. Relative compactness of its images and the unique possible weak limit then give norm convergence. Conversely, every bounded domain sequence has a weakly convergent subsequence, so the image of the unit ball is relatively compact.

A dual norm measures how strongly a bounded linear functional can respond to a unit-sized input. If $p$ is a bounded linear functional on a normed vector space $X$, write $⟨p,x⟩=p(x)$ and defineIt is the operator norm of $p$ when its scalar output is measured by absolute value. On ${\mathbb{R}}^n$, represent $p(x)=a^{T}x$ using a coefficient vector $a$. The unit ball is symmetric, so maximizing $|a^{T}x|$ gives the same value as maximizing $a^{T}x$: negate an input whenever its output is negative.^[ref]

For an original two-coordinate example, fix $a=(2,-1{)}^T$, so $p(x)=2x_1-x_2$. The input norm changes the allowable unit ball and hence the dual norm:Under the L1 norm, $|2x_1-x_2|\le 2(|x_1|+|x_2|)\le 2$, and $(1,0{)}^T$ attains the bound. Under the Euclidean norm, the Cauchy-Schwarz inequality gives $|a^{T}x|\le \Vert a{\Vert }_2\Vert x{\Vert }_2\le \sqrt{5}$, with equality at the displayed normalized vector. Under the maximum norm, each coordinate can reach unit absolute value, so $|2x_1-x_2|\le 2+1=3$; $(1,-1{)}^T$ attains it. Thus the dual norms of these three vector norms are respectively the maximum norm, the Euclidean norm, and the L1 norm. For $1\le p\le \infty$, Lp norms on a finite coordinate space pair with their Holder conjugate exponents.^[ref]

This geometry explains the usefulness of a dual norm: it translates a bound on inputs into the sharp bound on a linear output. The support function of the unit ball gives its value in the coefficient direction $a$. Changing the input budget changes the set of possible inputs; the three different answers do not conflict. In finite dimensions a closed unit ball is compact, so a continuous linear functional achieves its maximum absolute value. In an infinite-dimensional normed vector space, the supremum remains the definition and need not be achieved.

Suppose a perturbation $\delta x$ satisfies $\Vert \delta x\Vert \le \epsilon$, with $\epsilon \ge 0$. Absolute homogeneity of a norm and the dual norm definition giveFor nonzero $\delta x$, divide it by its norm to obtain an input in the unit ball; for zero perturbation the inequality is immediate. This is an exact error bound for a linear functional, for every perturbation in the stated input set.

For $p(x)=2x_1-x_2$ and a budget $\epsilon =0.1$, the sharp output-error bounds are $0.2$ for $\Vert \delta x{\Vert }_1\le 0.1$, $\sqrt{5}/10\simeq 0.224$ for $\Vert \delta x{\Vert }_2\le 0.1$, and $0.3$ for $\Vert \delta x{\Vert }_{\infty }\le 0.1$. Scale the maximizing vectors from the preceding example by $0.1$ to attain the positive endpoints; their negatives attain the negative endpoints. The L1 norm budget constrains a total absolute error; the maximum norm budget permits each coordinate to have that error separately. These budgets define nested sets and carry different information.

The coordinates must use the stated scale and units. Rescaling one coordinate changes both its coefficient in $a$ and the meaning of its input budget. For a nonlinear differentiable function with scalar values, its Fréchet derivative at the base point is a linear functional governing the first-order response. An exact finite-perturbation error bound additionally needs control of the Taylor remainder or of the Fréchet derivative along the path.

On the real space of sequences converging to zero $c_0$, use the supremum norm and define the linear functionalThe geometric series gives $|p(x)|\le \Vert x{\Vert }_{\infty }$, so $p$ is a bounded linear functional and $\Vert p{\Vert }_{\ast }\le 1$. The sequence with its first $N$ entries equal to one and all later entries zero belongs to $c_0$, has unit supremum norm, and gives $p(x)=1-2^{-N}$. Hence $\Vert p{\Vert }_{\ast }=1$. But if $\Vert x{\Vert }_{\infty }\le 1$ and $x_n\to 0$, not all coordinates can equal one or all equal minus one. Since every weight $2^{-n}$ is positive, $-1<p(x)<1$. The supremum is approached but never attained on the unit ball.

The Euclidean unit ball is the closed unit ball of the Euclidean norm, centered at zero. Its radial function is identically $1$, making it a natural reference body for John ellipsoids and volume ratios.

If $Z$ is a Banach space and $X$ is a norm-closed reflexive linear subspace of $Z^{\ast }$, evaluation gives the displayed surjection. Restriction $R:Z^{\ast \ast }\to X^{\ast }$ is weak-star continuous and maps the closed unit ball onto the target ball by the Hahn-Banach theorem. The Goldstine theorem then makes $T(B_Z)$ weak-star dense in $B_{X^{\ast }}$, where $T=R{J}_Z$. Reflexivity of $X$ makes this target weak-star topology equal to its weak topology. Convexity and the Hahn-Banach separation theorem convert weak density of this ball image to norm density.

To pass from density to surjectivity, approximate a desired $f$ by $T{z}_1$ with $\Vert z_1\Vert \le \Vert f\Vert$ and residual at most $\Vert f\Vert /2$. Repeat on each residual. The chosen vectors have summable norms, with sum at most $2\Vert f\Vert$, and the residuals tend to zero geometrically. Completeness of $Z$ produces $z={\sum }_n{z}_n$ with $Tz=f$. Using any smaller residual ratio proves the infimum lifting norm equals $\Vert f\Vert$, although that infimum need not be attained. This argument proves onto-ness, rather than merely density of the evaluation functionals.

For a finite-dimensional vector subspace $E\subseteq X^{\ast }$ and $x^{\ast \ast }\in B_{X^{\ast \ast }}$, exact interpolation on $E$ is possible by an $x\in X$ of norm less than $1+\epsilon$. The restriction map $X\to E^{\ast }$ is onto and open. Separation of the image of the radius-$1+\epsilon$ open ball would contradict $|x^{\ast \ast }(e)|\le \Vert e\Vert$. Scaling the interpolant slightly back into the closed unit ball proves the Goldstine theorem.

A finite-rank operator is a linear map whose image is finite-dimensional. A bounded finite-rank operator between normed spaces is compact: the image of its unit ball is bounded inside a finite-dimensional subspace, where bounded sets have compact closure.

For example, $Tf=\left({\int }_0^{1}f(t)dt\right)1$ on $L^2[0,1]$ has one-dimensional image, consisting of constant functions. Boundedness follows from Cauchy-Schwarz. Finite rank alone does not supply continuity for an arbitrary everywhere-defined linear map on an infinite-dimensional normed space; the compactness assertion requires the boundedness hypothesis.

The John ellipsoid of a convex body is its inscribed ellipsoid of largest volume. For the unit ball $K$ of a $d$-dimensional normed vector space, change coordinates so that this ellipsoid is the Euclidean unit ball. The resulting coordinates are called John position. In these coordinates the John contact decomposition implies $B_2^d\subseteq K\subseteq \sqrt{d}{B}_2^d$, or $\Vert x{\Vert }_E\le |x|\le \sqrt{d}\Vert x{\Vert }_E$.

For a unital star-subalgebra $A\subseteq B(H)$, its unit ball is dense in the unit ball of $A^{\prime \prime }$ in the strong operator topology. The same holds for self-adjoint unit balls and for positive unit balls, and the full unit balls may be approximated strongly together with their adjoints. Passing first to the norm closure allows continuous functional calculus; resolvent identities prove strong continuity of continuous functions vanishing at infinity, even for unbounded approximating nets of self-adjoint operators. A cutoff function supplies the norm bound, and a two-by-two self-adjoint matrix removes the self-adjoint restriction.

For measurable functions $f,g$ with values in the real numbers or complex numbers on the same measure space $(X,\mathcal{A},\mu )$, and $1\le p\le \infty$, membership in the same Lp space givesThus adding two functions cannot make their combined Lp norm exceed the sum of their separate Lp norms. The measure need not be finite or a probability measure. The hypotheses require finite Lp norms; $p=\infty$ uses the essential supremum of the absolute value, not an ordinary pointwise supremum. For complex numbers, the absolute value is the complex modulus. As usual, Lp spaces identify functions equal almost everywhere. ^[ref]

The Minkowski inequality makes the Lp norm a workable notion of size: it supplies the triangle inequality needed to turn $d_p(f,g)=\Vert f-g{\Vert }_p$ into a metric on Lp space. Applied to $(f-h)+(h-g)$, it says that travelling through an intermediate function $h$ cannot shorten the direct distance measured by this metric. Applied to two error functions, it gives a bound on the total error even when the errors reinforce one another; Minkowski inequality for error aggregation gives explicit examples.

The geometry is already visible in the Euclidean plane. Under counting measure on two coordinates, the closed unit balls for the L1 norm, Euclidean norm, and L-infinity norm are respectively $|x|+|y|\le 1$, $x^2+y^2\le 1$, and $\max (|x|,|y|)\le 1$. Every one is a convex set: if $u,v$ have Lp norm at most one, the Minkowski inequality and absolute homogeneity of a norm give $\Vert tu+(1-t)v{\Vert }_p\le t+(1-t)=1$ for $0\le t\le 1$. The Euclidean norm has no flat boundary segments, whereas the other two norms do. This is the distinction captured by a strictly convex normed space and developed in equality in Minkowski inequality.

A short Hölder's inequality proof outline handles $1<p<\infty$. First $|f+g{|}^p\le 2^{p-1}(|f{|}^p+|g{|}^p)$, so $f+g$ has finite Lp norm before any division. With conjugate exponents $p$ and $q=p/(p-1)$, set $h=|f+g{|}^{p-1}$; then $\Vert h{\Vert }_q=\Vert f+g{\Vert }_p^{p-1}$. The scalar triangle inequality followed by Hölder's inequality givesDivide if $\Vert f+g{\Vert }_p>0$; otherwise the desired bound is immediate. For $p=1$, integrate $|f+g|\le |f|+|g|$. For $p=\infty$, the almost everywhere bounds $|f|\le \Vert f{\Vert }_{\infty }$ and $|g|\le \Vert g{\Vert }_{\infty }$ give the result by the essential supremum. The same scalar triangle inequality works for real numbers and complex numbers.

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 vector subspace $Z\subseteq X^{\ast }$ is norming if some $c>0$ satisfies the displayed inequality. It retains enough continuous linear functionals to control the norm of every vector. Every such vector subspace is dense for the weak-star topology: otherwise a nonzero evaluation linear functional would annihilate its weak-star closure, contradicting the inequality. In the real case its symmetric unit ball makes the absolute-value and signed supremum versions equivalent.

A bounded linear operator is compact when the closure of the image of its unit ball is compact. Equivalently every bounded sequence has a subsequence whose images converge; on a Banach target it also suffices that the image of its unit ball be totally bounded.

For a finite-rank operator that image is bounded in a finite-dimensional subspace, whose closed bounded sets are compact. Thus every finite-rank operator is compact.

Suppose $T_n$ are compact and $\Vert T_n-T\Vert \to 0$. For any $\epsilon >0$ take $n$ with $\Vert T_n-T\Vert <\epsilon$. A finite $\epsilon$-net of $T_n$ of the unit ball is then a finite $2\epsilon$-net of its $T$ image. That image is totally bounded, so $T$ is compact by the stated criterion. Hence $\mathcal{K}(X)$ is norm closed in $B(X)$.

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 $H=0$.

A subset is totally bounded if for every $\epsilon >0$ it has a finite cover by balls of radius $\epsilon$. If its closure is compact, the cover by all such balls has a finite subcover, proving total boundedness.

Conversely, cover $S$ by finitely many balls of radius $2^{-r}$ at stage $r$. Starting with all indices, at least one ball contains infinitely many of the currently indexed terms. Retain those indices as ${\sigma }_r$. These sets are infinite and nested, and two terms indexed by ${\sigma }_r$ have distance less than $2^{1-r}$. Choose increasing $n_r\in {\sigma }_r$; then $(x_{n_r})$ is Cauchy, because all its terms after stage $r$ lie in that same ball. If $M$ is complete it converges. To cover sequences in the closure as well, approximate their $n$th terms by points of $S$ at distance less than $1/n$ and apply this argument. The corresponding subsequence in the closure has the same limit. Sequential compactness therefore proves total boundedness implies relative compactness in a complete ambient space.

The Arzelà-Ascoli theorem says a subset of $C[0,1]$ is relatively compact exactly when it is uniformly bounded and equicontinuous. Pointwise boundedness together with equicontinuity is an equivalent version here. In the specified derivative-bounded family, the mean-value estimate gives $|f(t)-f(s)|\le |t-s|$, uniformly over the family, while $\Vert f{\Vert }_{\infty }\le 1$. Thus this family is relatively compact.

For $V$, the image of the unit ball has sup norm at most one and derivative $f$ of sup norm at most one; it lies in that family, so $V$ is compact.

For $W$, take $f_n=n{1}_{[0,1/n]}$, each of $L^1$ norm one. Then $W{f}_n(t)=\min (nt,1)$. Every subsequence converges pointwise to zero at $t=0$ and to one at every $t>0$. This limit is discontinuous, so no subsequence converges uniformly to a continuous function. Thus $W$ is not compact. The same integration formula has different compactness properties because the domain unit balls impose different controls.

The weak-star topology on $X^{\ast }$ is the coarsest topology making every evaluation $\varphi \mapsto \varphi (x)$ continuous. Finite intersections of evaluation inequalities form a neighborhood basis. Distinct functionals differ on some $x$, so disjoint disks around their evaluation values give disjoint neighborhoods; thus the topology is Hausdorff. The Banach-Alaoglu theorem states that the closed unit ball of $X^{\ast }$ is compact in this topology.

If $f\in X^{\ast \ast }$ is weak-star continuous, some $x_1,\dots ,x_n$ and $\epsilon >0$ make $|f(\varphi )|<1$ whenever $|\varphi (x_i)|<\epsilon$ for every $i$. If all evaluations vanish, every scalar multiple of $\varphi$ lies in this neighborhood, forcing $f(\varphi )=0$. Thus $f$ vanishes on the kernel of the map $E(\varphi )=(\varphi (x_1),\dots ,\varphi (x_n))$. It factors linearly through the finite-dimensional range of $E$. Extending that factor algebraically to ${\mathbb{K}}^n$ gives scalars $a_i$ withConsequently $f=J_{X}x$ for $x={\sum }_i{a}_i{x}_i$.

For the two induced topologies, let $B$ be the common norm-closed unit ball and write ${\tau }_j=\sigma (Y,X_j)$. Under either topology $B$ is compact by Banach-Alaoglu, and the ambient space is Hausdorff. If the two topologies are comparable, the identity map from $B$ with the finer topology to $B$ with the coarser one is a continuous bijection from a compact space to a Hausdorff space. Such a map is a homeomorphism, since it carries closed compact sets to closed sets. Thus in the comparable case the topologies agree on $B$, giving alternative (a).

The open mapping theorem says that a bounded surjective linear map $T:E\to F$ between Banach spaces is open. Here is a proof using only Baire and completeness. Surjectivity gives $F={\bigcup }_{n\ge 1}T(n{\overline{B}}_E)$, hence also a cover by their closures. Baire makes one such closure contain a ball $B(y_0,r)$. Taking differences gives $B(0,r)\subseteq \overline{T(2n{\overline{B}}_E)}$. Scaling yields some $\delta >0$ withIf $\Vert y\Vert <\delta /2$, recursively use scaled versions of this inclusion to choose $x_k$ with $\Vert x_k\Vert \le 2^{-k}$ and residual $\Vert y-T{\sum }_{j=1}^k{x}_j\Vert <\delta 2^{-k-1}$. The sum converges in $E$ to an element of norm at most one, whose image is $y$. Thus $B(0,\delta /2)\subseteq T({\overline{B}}_E)$, and scaling gives $B(0,\delta /4)\subseteq T(\frac{1}{2}{\overline{B}}_E)\subseteq T(B_E)$. Translating and scaling proves openness on every open set. This proof does not use the closed graph or inverse mapping theorem.

Apply it to the evaluation map $E:X\to {\ell }^{\infty }$, $Ef=(f(z_n){)}_n$. It is bounded with norm at most one and onto by hypothesis. Hence its image of the unit ball contains a ball about zero, yielding a constant $c$ and, for every $\alpha$, an interpolant $g$ withFor completeness, scale a ball of radius $\eta$ inside $E(B_X)$ by $2\Vert \alpha {\Vert }_{\infty }/\eta$ to obtain $c=2/\eta$; zero data use $g=0$.

For each $n$, interpolate the data which are one at index $n$ and zero elsewhere, obtaining $g_n$ of norm at most $c$. The supplied factorization property gives $g_n=k_n{h}_n$ with $\Vert k_n{\Vert }_{\infty }=\Vert g_n{\Vert }_{\infty }\le c$. Evaluation at $z_n$ gives $1=|k_n(z_n)h_n(z_n)|\le c|h_n(z_n)|$, soThe three enumerated properties of $h_n$ are supplied hypotheses for this argument, rather than additional requests to prove them.