Banach space

A Banach space is a normed vector space that is complete in its norm metric.

For a closed vector subspace $Y$ of a Banach space $X$, the quotient vector space $X/Y$ is complete with its quotient norm. Closedness makes the quotient seminorm a norm. Completeness follows by choosing representatives of a rapidly Cauchy subsequence whose successive differences have summable norms in $X$.

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 $3$.

The quotient norm measures the smallest norm among representatives of a coset. Its infimum need not be attained. Nevertheless, for every $\epsilon >0$, an element $z\in X/Y$ has a representative $v\in X$ with $\Vert v\Vert <\Vert z\Vert +\epsilon$. The quotient map is contractive.

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.

If a bounded sequence $(y_n)$ in $X$ converges weak-star to $\varphi \in X^{\ast \ast }$ under the canonical embedding into the bidual, then any convex blocks $(u_n)$ have the same scalar limit on every $f\in X^{\ast }$. Thus $y_n-u_n$ is a weakly null sequence, even when $\varphi$ does not belong to the embedded space $X$.

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.

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.

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 Banach space is separable when it has a countable dense subset for its norm topology. It then has a countable base of norm balls and a countable norming family. Its continuous dual space need not be norm separable.

For $1\le p<\infty$, the space ${\ell }^p$ consists of scalar sequences $x=(x_n)$ satisfyingIt is a Banach space, and ${\ell }^q\subset {\ell }^p$ when $1\le q<p$.

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.

The space ${\ell }^2$ consists of scalar sequences $x=(x_n)$ satisfying ${\sum }_n|x_n{|}^2<\infty$. With inner product $⟨x,y⟩={\sum }_n{x}_n\overline{y_n}$, it is a Hilbert space.

The l2 sequence space is a uniformly convex Banach space. If $\Vert x{\Vert }_2=\Vert y{\Vert }_2=1$, the parallelogram law givesThus $\Vert x-y{\Vert }_2\ge \epsilon$ permits the modulus $\delta (\epsilon )=1-\sqrt{1-{\epsilon }^2/4}$.

A finitely supported sequence has only finitely many nonzero coordinates. Their vector space is commonly denoted $c_{00}$.

For $1\le p<\infty$, coordinate truncations converge in ${\ell }^p$, so $c_{00}$ is dense. It is not dense in ${\ell }^{\infty }$: every finitely supported sequence is at distance at least one from the constant-one sequence.

The space ${\ell }^{\infty }$ consists of bounded scalar sequences with norm $\Vert x{\Vert }_{\infty }={\sup }_n|x_n|$.

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.

The left shift is a bounded linear operator of operator norm one on the l-infinity sequence space. For $y=x-Sx$, telescoping gives $N^{-1}{\sum }_{n=1}^N{y}_n=(x_1-x_{N+1})/N\to 0$. This identity is useful in constructing a Banach limit.

If $Y$ is a proper closed subspace of a normed space $X$ and $0<\alpha <1$, there is a unit vector $x\in X$ whose distance from $Y$ exceeds $\alpha$.

A normed vector space has compact closed unit ball if and only if it is finite-dimensional. In an infinite-dimensional space, repeated application of Riesz lemma produces unit vectors separated pairwise by a fixed positive distance, contradicting sequential compactness.

A Banach space $X$ is uniformly convex when, for every $\epsilon >0$, some $\delta >0$ satisfies

For $2\le p<\infty$, the Clarkson inequality isIt gives the uniform convexity of $L^p$.

If a family of bounded linear operators from a Banach space to a normed space is pointwise bounded, then their operator norms are uniformly bounded.

Let $(x_n)$ lie in a normed space $X$ and suppose ${\sum }_n|f(x_n)|<\infty$ for every $f\in X^{\ast }$. Then some $C$ satisfiesApply the uniform boundedness principle on the Banach space $X^{\ast }$ to the finite signed sums $f\mapsto {\sum }_n{a}_{n}f(x_n)$ with $|a_n|\le 1$.