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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.

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 $3$.

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 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.

For an arbitrary set $\Gamma$ and scalar field $\mathbb{F}$, this is the space of bounded functions $u:\Gamma \to \mathbb{F}$ with the supremum norm $\Vert u\Vert ={\sup }_{\gamma }|u(\gamma )|$. Uniform limits of Cauchy sequences prove it is a Banach space. For $\Gamma =\mathbb{N}$ it is the l-infinity sequence space; for the empty index set it is the zero space, with norm zero.

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.

The continuous functions from the p-adic integers to the P-adic numbers form a Banach space over ${\mathbb{Q}}_p$ with supremum norm $\Vert f{\Vert }_{\infty }={\sup }_{x\in {\mathbb{Z}}_p}|f(x){|}_p$. Compactness of the domain bounds the norm, and uniform convergence in the complete codomain proves completeness. Translation and the forward difference operator $\Delta f(x)=f(x+1)-f(x)$ are bounded operators, with $\Vert \Delta f{\Vert }_{\infty }\le \Vert f{\Vert }_{\infty }$.

A convex block of a sequence in a Banach space is a sequence of convex combinations supported on successively disjoint finite intervals, with $p_n<q_n<p_{n+1}$, $a_i\ge 0$, and ${\sum }_{i=p_n}^{q_n}{a}_i=1$. Zero coefficients can fill gaps and ensure strict endpoint inequalities. By the Mazur theorem, every weakly null sequence has convex blocks tending to zero in norm.

Let $T:X\to X$ be a bounded linear operator on a Banach space, and let $B$ be the open unit ball. If $B\subseteq \overline{T(B)}$, then $B\subseteq T(B)$. Scale the density statement and correct a target by a geometrically decreasing sequence of approximate preimages. Completeness makes their sum converge inside $B$, and continuity of $T$ makes its image equal the target.

For a bounded operator on a complex Banach space, partition a disconnected spectrum into two nonempty compact pieces. The holomorphic function equal to one near the first piece and zero near the second defines a Riesz projection. The holomorphic spectral mapping theorem makes it neither zero nor identity. Its closed range is a proper nonzero invariant vector subspace. This need not be an orthogonal projection, and the argument does not require a normal operator.

The duality mapping takes a vector to its supporting dual functionals for half its squared norm. Equivalently $p\in \mathcal{J}(x)$ when $\Vert p{\Vert }_{\ast }=\Vert x\Vert$ and $⟨x,p⟩=\Vert x{\Vert }^2$. This follows from the dual-norm inequality and equality in the conjugate relation for squared norms. In a Hilbert space, the Riesz representation theorem identifies it with the identity; in a general Banach space it can be set-valued.

Let $D$ be a dense linear subspace of a normed space $X$, let $Y$ be a Banach space, and let $T:D\to Y$ be linear with $\Vert Tx\Vert \le C\Vert x\Vert$. Then $T$ has a unique bounded extension $\tilde{T}:X\to Y$, defined by $\tilde{T}x={\lim }_{n}T{x}_n$ for any sequence $x_n\in D$ converging to $x$, and $\Vert \tilde{T}\Vert \le C$.

A proper lower-semicontinuous convex functional on a real Banach space equals its biconjugate, with the canonical embedding in the bidual understood. One inequality follows from the Fenchel–Young inequality; separation of a point below the closed convex epigraph gives the opposite inequality. This is the closed-convex analogue of recovering a function from all its affine supporting functions.

A norm-closed finite-codimensional vector subspace $Y\subseteq X^{\ast }$ that is weak-star dense is a norming subspace. Its annihilator $F=Y^{\perp }\subseteq X^{\ast \ast }$ is finite-dimensional and disjoint from $JX$. The positive distance $d$ of the unit sphere from $F$ gives a triple-dual linear functional vanishing on $F$ and taking a value at least $d$ at a unit vector. Apply the finite-dimensional interpolation form of Goldstine's theorem to the Banach space $X^{\ast }$, then normalize and take a supremum to obtain norming constant $d$.

A function $u\in L^1(\Omega )$ has bounded variation when $\mathrm{TV}(u)<\infty$. The space $BV(\Omega )$ is a Banach space with norm $\Vert u{\Vert }_{L^1}+\mathrm{TV}(u)$.