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A Banach limit is a positive linear functional on the real l-infinity sequence space that extends the ordinary limit on the convergent sequence space, has operator norm one, and is invariant under the left shift on bounded sequences. The Hahn-Banach theorem constructs one by separating the constant sequence $1$ from the subspace $(I-S){\ell }^{\infty }$, whose distance from $1$ is one. The functional is generally not unique.

A map from an open subset of $\mathbb{C}$ to a complex Banach space is holomorphic when the displayed limit exists in the Banach-space norm at every point. A norm-convergent power series is holomorphic inside its radius of convergence: on every smaller disk the series of derivatives converges uniformly by the geometric coefficient bound, allowing differentiation term by term. Applying a continuous linear functional produces an ordinary scalar holomorphic function. Scalar Cauchy identities therefore imply the corresponding vector-integral identities when bounded linear functionals separate points, as they do by the Hahn-Banach theorem.

The bidual of a normed vector space $X$ is the continuous dual $X^{\ast \ast }=(X^{\ast }{)}^{\ast }$. Evaluation defines the canonical embedding into the bidual $J_X(x)(f)=f(x)$. The norm inequality gives $\Vert J_{X}x\Vert \le \Vert x\Vert$, and a norming functional from the Hahn-Banach theorem gives equality. Thus $J_X$ is isometric, and its image is dense in its closure, a completion of a normed space.

The displayed sublinear functional on the real l-infinity sequence space agrees with the ordinary limit on convergent sequences. A dominated extension by the Hahn-Banach theorem therefore defines a generalized limit bounded between the lower and upper limits of the Cesàro means. For the left shift on bounded sequences $S$, the averages of $Sx-x$ equal $(x_{N+1}-x_1)/N$, which tend to zero. Domination applied to both signs gives $L(Sx-x)=0$, hence shift invariance. Positivity, agreement on constant sequences and the operator norm one follow from the bounds, so $L$ is a Banach limit.

Suppose a linear map $S:c_0\to Y$ has compact image of the closed unit ball. It is bounded, and each $g\in Y^{\ast }$ attains its maximum on this compact image. Thus $S^{\ast }g$ attains its norm, so norm attainment on the vanishing sequence space implies $S^{\ast }(Y^{\ast })\subseteq c_{00}$. The Baire category theorem applied to the inverse images of the first $n$ coordinate spans makes this adjoint finite rank: these are closed linear subspaces covering the complete continuous dual space $Y^{\ast }$, so one has interior and is all of $Y^{\ast }$. Finally the Hahn-Banach theorem identifies $\ker S$ as the common kernel of $S^{\ast }(Y^{\ast })$, proving finite rank of $S$. Ordinary compact operators require only relative compactness of the ball image, so they need not have finite rank.

The completion of a normed vector space $X$ is a Banach space containing an isometric dense copy of $X$. Construct it from Cauchy sequences modulo sequences whose difference tends to zero; define the norm by $\Vert [(x_n)]\Vert ={\lim }_n\Vert x_n\Vert$. Equivalently, the canonical evaluation map into the bidual space is an isometry by the Hahn-Banach theorem, and the closure of its image supplies a completion.

A Banach space is precisely a normed vector space in which every Cauchy sequence converges in norm to an element of the same space. Completion of a normed space fills missing limits without changing distances between existing elements. For example, $C([0,1])$ with the L2 norm completes to L2 space on the interval with Lebesgue measure. The bounded integral functional extends uniquely to that completion of a normed space: define $F(f)={\lim }_{k}F(f_k)$ for any approximating sequence of continuous functions. Its bound makes this independent of the approximation. The unbounded point evaluation has no continuous extension agreeing with its values on all continuous functions.^[ref]

A bounded complex-linear linear functional on a complex linear subspace of a normed space extends to the whole space with the same operator norm. To deduce this from the real Hahn-Banach theorem, extend the real part $u$ under the bound $\Vert \ell \Vert \Vert x\Vert$ on the underlying real vector space, then put $L(x)=u(x)-iu(ix)$. This is complex-linear and agrees with $\ell$ on its original domain. Given $x$, choose a complex number $a$ of modulus one with $aL(x)=|L(x)|$. Then $|L(x)|=\mathrm{Re}L(ax)=u(ax)\le \Vert \ell \Vert \Vert x\Vert$. Restriction gives the reverse operator norm inequality.

In a Hausdorff locally convex space, a separating family of continuous seminorms has $p(x)>0$ for any chosen nonzero $x$ and some $p$. Define a linear functional on its one-dimensional span attaining $p(x)$ and extend it under that seminorm by the real Hahn-Banach theorem. The extension is continuous because it is seminorm-bounded, and distinguishes $x$ from zero. The Hausdorff condition is essential.

A real convex function locally bounded on an open convex set has a continuous supporting linear functional at every interior point. Apply the dominated Hahn-Banach theorem to its sublinear directional derivative $p_x$ to obtain $l\le p_x$. Increasing secant slopes imply $p_x(z)\le f(x+z)-f(x)$ whenever the endpoint is in the domain. A local Lipschitz continuity bound gives both $l(y)\le L\Vert y\Vert$ and $-l(y)=l(-y)\le L\Vert y\Vert$, proving continuity.

A bounded linear operator $T:X\to {\ell }_{\infty }(\Gamma )$ is exactly a uniformly bounded family of bounded linear functionals $f_{\gamma }\in X^{\ast }$ through $(Tx)(\gamma )=f_{\gamma }(x)$. Evaluation shows $\Vert f_{\gamma }\Vert \le \Vert T\Vert$; conversely the family bound gives $\Vert Tx\Vert \le (\sup \Vert f_{\gamma }\Vert )\Vert x\Vert$, proving equality. Taking all functionals in the dual unit ball gives an isometric embedding by the Hahn-Banach theorem.

A real separable Banach space has a sequence $({\phi }_n)$ in its dual unit ball with $\Vert x\Vert ={\sup }_n{\phi }_n(x)$ for every $x$. Choose a dense sequence $(u_n)$ in the unit sphere and apply the Hahn-Banach theorem to obtain ${\phi }_n(u_n)=1$. In the complex case use ${\sup }_n\mathrm{Re}{\phi }_n(x)$.

This linear functional extracts one coefficient of a B-spline expansion using local polynomial data. With ${\psi }_i(x)={\prod }_{\ell =1}^{k-1}(x-t_{i+\ell })/(k-1)!$ and a point $\xi$ inside a knot cell in the support, it is ${\sum }_{r=0}^{k-1}(-1{)}^r{s}^{(r)}(\xi ){\psi }_i^{(k-1-r)}(\xi )$. The local polynomial identity for the Marsden dual functional gives the desired coefficient. Its bounded restriction to the local spline space can be extended to continuous data on $[t_i,t_{i+k}]$ by the Hahn-Banach theorem, with the same bound. Such extensions supply spline quasi-interpolation coefficients without requiring derivatives of the data.

On ${\mathbb{R}}^n$, integration embeds $L^1$ isometrically into $(L^{\infty }{)}^{\ast }$, but this is not surjective. For example, the functional taking the limit at infinity on $\mathbb{C}1\oplus C_0({\mathbb{R}}^n)$ extends to $L^{\infty }$ by the Hahn-Banach theorem; it vanishes on compactly supported functions and is one on the constant function, so no $L^1$ density represents it.

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.

A generalized limit on real bounded sequences is a positive linear functional of operator norm one that extends the ordinary limit of a sequence. Such a functional can be chosen by the Hahn-Banach theorem: extend the limit on the convergent sequence space while dominating it by the sublinear functional $p(x)={\mathrm{limsup}}_n{x}_n$. Applying the same domination to $-x$ gives the displayed lower bound. A Banach limit additionally has invariance under the left shift on bounded sequences.

For a nonempty radially open convex set $C$ disjoint from a vector subspace $U$, there is a nonzero linear functional $L$ whose kernel of a linear map contains $U$ and avoids $C$. Choose $a\in C$ and the Minkowski functional $p$ of $C-a$. On $U+\mathbb{R}a$, set $g(u+ta)=-t$. Disjointness implies $p(u-a)\ge 1$, which gives $g\le p$ for negative $t$; nonnegativity of $p$ gives it for nonnegative $t$. The algebraic Hahn-Banach theorem extends $g$ to $L\le p$. For $x\in C$, $L(x)+1=L(x-a)<1$, so $L(x)<0$. Also $L(a)=-1$ and $L{|}_U=0$.

The closed linear span of these vectors is all of $c_0$. Otherwise the Hahn-Banach theorem gives a nonzero continuous functional annihilating them, represented by $a\in {\ell }_1$. At the first index $m$ where $a_m\ne 0$, divide ${\sum }_j{a}_j{t}_n^j=0$ by $t_n^m$. The remaining tail has magnitude at most $t_n{\sum }_{j>m}|a_j|\to 0$, forcing $a_m=0$, a contradiction. This is an elementary coefficient-isolation argument for density by annihilators.

Suppose the distinct complex numbers ${\beta }_k$ satisfy $|{\beta }_k|\le \delta <1$. The geometric sequences in the displayed formula belong to the absolutely summable sequence space and have dense linear span. Otherwise the Hahn-Banach theorem gives a nonzero annihilating functional, represented by a bounded sequence $b_j$. Its power series $F(z)={\sum }_{j\ge 1}{b}_j{z}^{j-1}$ is holomorphic in the unit disk and vanishes at every ${\beta }_k$. Compactness supplies an accumulation point inside that disk. The identity theorem forces $F=0$ and hence all its coefficients zero, a contradiction. This is a useful form of density by annihilators that converts a Banach-space problem into uniqueness of an analytic function.

For a holomorphic scalar function on a neighborhood of a closed annulus, $M(r)$ is the maximum modulus on its radius-$r$ circle. The logarithm of its modulus is subharmonic; compare it with the harmonic function affine in $\log |z|$ which takes the logarithms of the two boundary maxima. The maximum principle for subharmonic functions proves the displayed bound. Zero boundary maxima are handled by adding a positive bound and taking its limit. The same bound holds for the norm of a Banach space-valued holomorphic function by applying scalar functionals and the Hahn-Banach theorem.

For a normed vector space $X$, take $K=B_{X^{\ast }}$ with the weak-star topology. Banach-Alaoglu theorem makes it compact Hausdorff, and each displayed evaluation function is continuous. Its supremum norm is $\Vert x\Vert$ by the Hahn-Banach theorem, giving a linear isometric embedding into $C(K)$.