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For $1\le p<\infty$, a bounded linear operator $T:E\to F$ between Banach spaces is absolutely p-summing if $({\sum }_j\Vert T{x}_j{\Vert }^p{)}^{1/p}\le C{\sup }_{\varphi \in B_{E^{\ast }}}({\sum }_j|\varphi (x_j){|}^p{)}^{1/p}$ for every finite family. The least $C$ is its p-summing norm ${\pi }_p(T)$. Direct substitution gives the ideal inequality ${\pi }_p(BTA)\le \Vert B\Vert {\pi }_p(T)\Vert A\Vert$ for compatible bounded linear operators.

The Banach space ${\ell }^1$ consists of absolutely summable scalar sequences, with norm $\Vert x{\Vert }_1={\sum }_n|x_n|$. Its continuous dual is ${\ell }^{\infty }$, under the pairing ${\sum }_n{x}_n{y}_n$. Finitely supported sequences are norm dense, which identifies a linear functional from its values on coordinate vectors. The space of sequences converging to zero inside this dual is a norming subspace of infinite codimension.

For example, $x=(2^{-1},2^{-2},\dots )$ belongs to the absolutely summable sequence space and has norm one, by the geometric series formula. Merely tending to zero is insufficient: $(1/n)$ tends to zero but its series of absolute values diverges. Thus this sequence space records total absolute size, rather than just the eventual size of individual coordinates. Its completeness ensures that a Cauchy sequence in this total-error norm has an absolutely summable limit.^[ref]

A normed vector space is a Banach space exactly when every absolutely convergent series converges in the space. Completeness implies the assertion because the triangle inequality makes the partial sums a Cauchy sequence. Conversely, select from any Cauchy sequence a subsequence whose consecutive differences have norm at most $2^{-k}$. The series of these differences converges by hypothesis, so the subsequence converges. The original Cauchy sequence then has the same limit. This provides a useful way to demonstrate incompleteness by a summable-norm series whose candidate sum lies outside the space.

Let $A$ be differentiable on a neighbourhood of a closed radius-$r$ ball in a Banach space, with $\Vert DA\Vert \le \kappa <1$ on that ball. If $\Vert A(p_0)-p_0\Vert \le (1-\kappa )r$, the mean-value estimate gives $A(B)\subseteq B$ and $\Vert A(p)-A(q)\Vert \le \kappa \Vert p-q\Vert$. The contraction mapping theorem supplies exactly one fixed point in $B$, and its distance from $p_0$ is at most $\Vert A(p_0)-p_0\Vert /(1-\kappa )$.

// Target: dynamical-systems.bigb

A bounded linear operator between normed spaces is approximable if it is an operator norm limit of bounded finite-rank operators. If the target is a Banach space, every approximable operator is a compact operator. For a given tolerance, approximate the operator in norm by one finite-rank operator, cover that operator's bounded unit-ball image by finitely many small balls in its finite-dimensional range, and transfer the cover across the operator-norm error. This makes the original image totally bounded; its closure is complete in the Banach target and therefore compact. The converse on arbitrary Banach spaces would require an additional approximation hypothesis and is not part of this assertion.

Let $T:V\to W$ be a bounded linear operator with $V$ a Banach space. If every $y\in W$ has an approximate preimage $x$ with $\Vert x\Vert \le R\Vert y\Vert$ and $\Vert Tx-y\Vert \le k\Vert y\Vert$, where $k<1$, then every $y$ has an exact preimage with the displayed norm bound. Repeatedly correct the residual: the corrections have norms at most $R{k}^j\Vert y\Vert$ and form an absolutely convergent series; its image under $T$ equals $y$. This proves surjectivity but does not provide a bounded linear right inverse.

If a Banach space $X$ has separable dual and $q:X\to X/Y$ is a quotient by a closed subspace, every weakly null sequence in the quotient's closed unit ball has a subsequence with approximate weakly null lifts of norm at most $3$. Choose representatives of norm below $3/2$, pass to a common bidual weak-star limit, and subtract convex blocks whose quotient images tend to zero in norm. Convex-block cancellation of a weak-star limit gives weak nullity and the triangle inequality gives the bound $3$.

For isomorphic finite-dimensional normed vector spaces, the Banach-Mazur distance is the infimum of $\Vert S\Vert \Vert S^{-1}\Vert$ over their linear isomorphisms. If an $N$-dimensional subspace $E$ of a Banach space of cotype 2 satisfies the summing estimate ${\pi }_2(T)\le L\Vert T\Vert$ for operators from ${\ell }_{\infty }^N$, then $d(E,{\ell }_{\infty }^N)\ge \sqrt{N}/L$: compose an isomorphism with the inclusion and use ${\pi }_2(I_E)=\sqrt{N}$.

On $C(X)$ for compact Hausdorff $X$, any complete norm making every point evaluation functional continuous is equivalent to the uniform norm. If $f_n$ converges in the new norm to $f$ and uniformly to $g$, evaluation gives $f(x)=g(x)$ for every $x$, so the identity map between the two Banach spaces has closed graph. The closed graph theorem and open mapping theorem make the identity and its inverse bounded. Completeness of the proposed norm is essential to this argument.

A Banach space has the Banach-Saks property if every bounded sequence contains a subsequence whose arithmetic averages converge in norm. The weak Banach-Saks property requires this only for weakly convergent sequences. The weak Banach–Saks theorem in a Hilbert space establishes that weak version in Hilbert spaces.

A Banach space is a normed vector space over the real numbers or complex numbers that is complete for the metric $d(x,y)=\Vert x-y\Vert$. Explicitly, whenever a sequence $(x_n)$ satisfiesthere is an element $x$ of the same space with $\Vert x_n-x\Vert \to 0$. The hypothesis says that the approximations eventually agree to any requested accuracy; completeness says that they describe an actual object in the space. The Cauchy sequence condition can be checked before anyone knows what that object is.

This makes a Banach space a natural setting for constructing solutions by successive improvements. If corrections $v_k$ have ${\sum }_k\Vert v_k\Vert <\infty$, their partial sums form a Cauchy sequence by the triangle inequality, and completeness supplies their sum. In fact, convergence of every such absolutely summable vector series characterizes a Banach space. The contraction mapping theorem uses precisely this mechanism: each correction is smaller than the preceding one by a fixed factor.^[ref]

Every normed vector space of finite dimension is a Banach space. Infinite-dimensional examples include the absolutely summable sequence space and the l-infinity sequence space; the choice of norm remains part of the specification. A collection of increasingly accurate finite descriptions can require an infinite description for its limit, as the finitely supported sequence example below demonstrates. Completion of a normed space adds these missing limits while preserving all original distances.

No infinite-dimensional Banach space has a countable Hamel basis. If $(e_n)$ were such a basis, the space would be the countable union of the finite-dimensional subspaces $\mathrm{span}(e_1,\dots ,e_n)$. Each is closed and has empty interior, contradicting the Baire category theorem.

If $X$ is a normed vector space and $Y$ is a Banach space, then the bounded linear maps $\mathcal{B}(X,Y)$ form a Banach space under the operator norm. Completeness of the domain is unnecessary.

On a Banach space one can define the set of minimizers of $E(x)+\Vert x-q{\Vert }^2/2$, but existence and uniqueness need additional geometry or coercivity hypotheses. The optimality condition is $0\in \partial E(x)+\mathcal{J}(x-q)$, with the duality mapping $\mathcal{J}=\partial (\Vert \cdot {\Vert }^2/2)$. The Hilbert completion-of-square shift by a dual vector does not generalize unchanged. On $({\mathbb{R}}^2,\Vert \cdot {\Vert }_1)$ with $E(x)=x_1+x_2$ and $q=0$, every $x_1,x_2\le 0$ with $x_1+x_2=-1$ is a minimizer; the vector $(-1,-1)$ is not one.

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.

Let $X$ be a normed space, $Y$ a Banach space, and $A_n:X\to Y$ be bounded linear operators with $C={\sup }_n\Vert A_n\Vert <\infty$. Convergence of $A_{n}d$ for every $d$ in a dense subset $D$ implies convergence of $A_{n}x$ for every $x\in X$ to a bounded linear operator $A$, with $\Vert A\Vert \le {\mathrm{liminf}}_n\Vert A_n\Vert$. Indeed, $\Vert A_{n}x-A_{m}x\Vert \le 2C\Vert x-d\Vert +\Vert A_{n}d-A_{m}d\Vert$ makes $(A_{n}x)$ a Cauchy sequence; completeness gives its limit, and linearity and the bound pass to that limit. If $X$ is a Banach space, pointwise convergence conversely implies the uniform bound by the Uniform boundedness principle. If the limit bounded linear operator is specified in advance, completeness of $Y$ is unnecessary.

A surjective linear isometry $U:C(K)\to C(L)$ between compact Hausdorff function spaces has the form $(Uf)(s)=u(s)f(h(s))$, where $h:L\to K$ is a homeomorphism and $u$ is continuous with modulus one. Thus the Banach space structure of $C(K)$ determines $K$ up to homeomorphism.

A bounded linear operator $T:X\to Y$ between Banach spaces is a weakly compact operator exactly when its second adjoint takes values in the canonical copy of $Y$ inside $Y^{\ast \ast }$. The Goldstine theorem proves necessity; the Banach-Alaoglu theorem, continuity of the second adjoint, and the Mazur theorem prove sufficiency.

The Bochner integral integrates a strongly measurable function taking values in a Banach space. It is obtained as a norm limit of integrals of simple vector-valued functions and satisfies the usual linearity and norm bound.

Bounded continuous functions on a topological space form a Banach space with the supremum norm. A uniform limit is continuous and bounded, proving completeness. This space differs from the bounded uniformly continuous functions: free-transport composition is jointly continuous in time and position, but need not be continuous in the global sup norm as time varies.