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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.

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.

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.