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A Schauder basis of a real Banach space $X$ is a sequence $(e_n)$ for which every $x\in X$ has a unique norm-convergent expansion $x={\sum }_n{a}_n{e}_n$. Its basis projection isand its basis constant is $K_b={\sup }_n\Vert P_n\Vert <\infty$.

For the sequence space $Y$ in the question, defineConvergence in the definition of $Y$ makes $J$ well defined, and uniqueness of basis coefficients makes it a linear bijection. Moreover,whereasThus $J$ is an isomorphism and, in particular, the displayed supremum really is a complete norm on $Y$.

The coordinate functional of a Schauder basis isIt is bounded because $a_n{e}_n=(P_n-P_{n-1})x$. On finite linear combinations of the $e_n^{\ast }$, the nth partial-sum operator is the restriction of $P_n^{\ast }$, sinceThe operators have norms at most $K_b$, so the standard basis criterion shows that the dual sequence of a Schauder basis $(e_n^{\ast })$ is a basic sequence in $X^{\ast }$. For every $x^{\ast }\in X^{\ast }$ and $x\in X$,which is precisely $P_n^{\ast }{x}^{\ast }\to x^{\ast }$ in the weak-star topology.

Suppose now that $X$ is a reflexive Banach space. If $\Vert x^{\ast }-P_n^{\ast }{x}^{\ast }\Vert$ did not tend to zero, approximation by finite basis blocks would give a bounded block sequence $(u_j)$ and an $\epsilon >0$ such that $|x^{\ast }(u_j)|\ge \epsilon$. Reflexivity gives a weakly convergent subsequence. Every fixed coordinate functional is eventually zero on a block sequence, so its weak limit has every basis coordinate zero and is therefore zero. This contradicts $|x^{\ast }(u_j)|\ge \epsilon$. Hence $P_n^{\ast }{x}^{\ast }\to x^{\ast }$ in norm for every $x^{\ast }$, so the basis is shrinking and $(e_n^{\ast })$ is a basis of $X^{\ast }$.

The converse fails. The standard basis of $c_0$ is shrinking because its dual sequence is the standard basis of ${\ell }^1=(c_0{)}^{\ast }$, but $c_0$ is not reflexive.

Finally assume $(e_n^{\ast })$ is a basis of $X^{\ast }$. Map $x^{\ast \ast }\in X^{\ast \ast }$ toFor $s_n={\sum }_{k=1}^n{a}_k{e}_k$,so $(a_n)\in Z$. Conversely, if $(a_n)\in Z$ and $M={\sup }_n\Vert s_n\Vert$, defineThe limit exists: for $m>n$,because $(e_n^{\ast })$ is a basis. Also $|F(x^{\ast })|\le M\Vert x^{\ast }\Vert$, so $F\in X^{\ast \ast }$ and $F(e_n^{\ast })=a_n$. These two constructions are inverse and satisfyThus $X^{\ast \ast }$ and $Z$ are isomorphic.

Let $(X_n)$ be Cauchy in the normChoose a subsequence $(X_{n_k})$ for whichTonelli theorem impliesalmost surely. The subsequence therefore converges uniformly on $[0,\infty )$, outside one null event, to a continuous process $X$. For each $t$, $X_t$ is the almost-sure limit of ${\mathcal{F}}_t$-measurable variables; completeness of the filtration makes the chosen version adapted. The same summable bound and monotone convergence theorem show that $\mathbb{E}{\sup }_t|X_t|<\infty$ and that $X_{n_k}\to X$ in norm.

Since the original sequence is Cauchy, the usual triangle argument upgrades convergence of the subsequence to $X_n\to X$ in norm. Thus the space of indistinguishability classes is a Banach space.

One form of Schauder's theorem says that if $M$ is a nonempty closed bounded convex subset of a Banach space and $T:M\to M$ is continuous with relatively compact image, then $T$ has a fixed point.

For a Schauder solution of a small Hammerstein integral equation, take $X=C([0,1])$ with the supremum norm and $M=\{u:\Vert u{\Vert }_{\infty }\le R\}$. This set is nonempty, closed, bounded and convex. PutBoth maxima are finite by continuity on compact sets. Define $(T_{\mu }u)(x)=\mu {\int }_0^{1}G(x,y)f(y,u(y))dy$. Continuity and uniform continuity of the kernel give $T_{\mu }u\in X$, and for $0<\mu \le {\mu }_0$,so it is a self-map. This choice of ${\mu }_0$ also handles the case when either maximum is zero without division by zero.

If $u_n\to u$ uniformly in $M$, uniform continuity of $f$ on $[0,1]\times [-R,R]$ makes $f(\cdot ,u_n(\cdot ))\to f(\cdot ,u(\cdot ))$ uniformly. The integral estimate then proves continuity of $T_{\mu }$ in the supremum norm. Moreoverwhich tends to zero with $|x-x^{\prime }|$ uniformly in $u\in M$. Thus the image is uniformly bounded and equicontinuous. Arzelà-Ascoli states that such a family on a compact interval has compact closure in the supremum norm. Hence $T_{\mu }(M)$ is relatively compact, and Schauder gives a fixed point for each $\mu$ in the specified interval. Therefore a continuous solution with supremum norm at most $R$ exists. The argument imposes no Lipschitz hypothesis on $f$ and makes no uniqueness assertion.

This is the fixed point theorem for summably Lipschitz iterates. Put $x_n=A^n{x}_0$ and $d=\Vert x_1-x_0\Vert$. For $n\ge 1$,Consequently, for $m>n\ge 1$,Summability makes the tails tend to zero, so $(x_n)$ is a Cauchy sequence. Since $M$ is closed in a Banach space, $x_n\to x\in M$. The $n=1$ hypothesis makes $A$ a Lipschitz map, hence continuous; passing to the limit in $x_{n+1}=A{x}_n$ gives $x=Ax$.

If $y$ is another fixed point, then $\Vert x-y\Vert \le {\alpha }_n\Vert x-y\Vert$ for every $n$. Since ${\alpha }_n\to 0$, eventually ${\alpha }_n<1$ and thus $x=y$. Finally let $m\to \infty$ in the tail estimate:The map itself need not be a contraction mapping; its successive iterates provide the summable error bound.

The space $C([a,b])$ with the supremum norm is a Banach space. Continuity of the kernel makes $K$ a bounded linear operator. Induction on $n$ givesso $A^{n}y-A^{n}z=K^n(y-z)$. The ordered integration region gives the factorial bound for a Volterra iterate:For completeness, if this holds at $n$, then integrating $M{M}^n(s-a{)}^n/n!$ from $a$ to $t$ gives the bound at $n+1$; the initial case is the defining integral. Thusand ${\sum }_{n\ge 1}{\alpha }_n=e^{M(b-a)}-1<\infty$. Part (b) now gives uniform convergence of every iteration sequence to the unique fixed point. Since $y=Ay$ is precisely the Volterra integral equation, this is its unique continuous solution. No restriction $M(b-a)<1$ is needed: the factorial controls all iterates on the whole finite interval.

Here $T$ is a nonlinear compact operator: it is continuous and sends bounded subsets of $M$ to relatively compact subsets of $X$. Equivalently, the images of any bounded input sequence have a norm-convergent subsequence.

The Schauder fixed-point theorem states that if $M$ is a nonempty closed bounded convex set in a Banach space and $T:M\to M$ is continuous and compact, then $\boxed{T\text{ has a fixed point in }M}$. Linearity, a contraction estimate, and uniqueness are not assumptions or conclusions of this theorem.

A summability kernel on the circle is a sequence of continuous periodic functions withThe frequently used stronger condition of uniform convergence to zero off each neighborhood of zero implies the last condition. For $f$ in a homogeneous Banach space,Given $\epsilon >0$, translation continuity bounds the difference inside a sufficiently small neighborhood by $\epsilon$; its integral is at most $C\epsilon$. Outside, use $2\Vert f{\Vert }_B$ and the vanishing tail condition. Hence $f\ast k_n\to f$ in $B$.

With the indexing needed in part (c), the Fejér kernel isIn particular $K_0=1$. The squared-modulus expression proves $K_n\ge 0$, and its constant Fourier coefficient proves $\int K_{n}dm=\Vert K_n{\Vert }_1=1$. The geometric series yieldsThus these are summability kernels. For $T_f:g\mapsto f\ast g$, part (a) gives $\Vert T_f\Vert \le \Vert f{\Vert }_B$. Since $\Vert K_n{\Vert }_1=1$ and $T_f{K}_n\to f$ in $B$, the reverse bound follows:

Let $Z=(Z_n{)}_{n\ge 0}$ be a Markov chain with transition kernel $K$, and let $N$ be an independent Poisson process of constant rate $r>0$. A Pseudo-Poisson process is $X_t=Z_{N_t}$. On the Banach space $B_b(E)$ of bounded Borel measurable functions, with the supremum norm, writeIts infinitesimal generator, transition semigroup, and resolvent operator are respectivelyand, for $\lambda >0$,The transition semigroup averages $K^n$ against the Poisson distribution of the number of updates. Since $K$ is a Markov operator, $\Vert K\Vert \le 1$; both series converge in operator norm. Updates with $Z_{n+1}=Z_n$ are allowed and do not count as changes of the observed state. The degenerate rate $r=0$ gives $L=0$, $P_t=I$, and $R_{\lambda }=I/\lambda$.

Let $E$ be the underlying Banach space. By the Laplace-transform formula for a semigroup resolvent,These are Bochner integrals, defined for every $f\in E$ and $\lambda >0$. The contraction semigroup bound and semigroup property giveInterchanging and regrouping these Bochner integrals is justified by the integrable bound $e^{-\lambda s-\mu t}\Vert f\Vert$, whose integral is $\Vert f\Vert /(\lambda \mu )$. For $\lambda \ne \mu$, set $v=s+t$ and evaluate the inner scalar integral:Since $f$ was arbitrary, the resolvent identity follows:For $\lambda =\mu$, both sides vanish, so that case holds as well.

Let $U\subset \mathbb{R}\times {\mathbb{R}}^d$ be an open set, $d<\infty$, and $f:U\to {\mathbb{R}}^d$ a continuous function. Through every $(t_0,y_0)\in U$ there is a local solution of a differential equation: a continuously differentiable function $y:I\to {\mathbb{R}}^d$ on an open interval containing $t_0$, satisfying the displayed initial value problem pointwise. No Lipschitz condition is assumed.^[ref]

Continuity of a proposed rate law therefore guarantees at least one trajectory for a short time. For a physical model, this is a first consistency check: can the stated rule and initial condition actually be followed? Prediction also needs uniqueness. The delayed solutions of a square-root differential equation below show how the same rule and initial condition can leave a departure time undetermined. The Picard-Lindelöf theorem supplies a sufficient additional hypothesis for uniqueness; failure of that hypothesis alone does not prove nonuniqueness.

For the initial value problem $y^{\prime }=2\sqrt{|y|}$, $y(0)=0$, a forward solution of a differential equation can stay zero until any $c\ge 0$, then follow $y=(t-c{)}^2$. The delayed solutions of a square-root differential equation verify this freedom exactly.

Local existence gives neither a global solution of a differential equation nor a bound on all future values. The continuous function $f(t,y)=y^2$ already permits finite-time blowup; the finite-dimensional continuation criterion explains the additional issue of extending a trajectory within its domain.

Proof outline: choose a compact set $K=[t_0-a,t_0+a]\times {\overline{B}}_r(y_0)\subset U$. By the extreme value theorem, $M={\max }_K\Vert f\Vert <\infty$. Take $0<h\le a$ with $Mh\le r$. Forward Euler method polygons remain in the closed ball and satisfy $\Vert y_n(t)-y_n(s)\Vert \le M|t-s|$. The Arzelà-Ascoli theorem yields a convergent subsequence with uniform convergence. Uniform continuity of $f$ on $K$ makes their integral equation defect tend uniformly to zero, so the limit satisfiesThe fundamental theorem of calculus gives a continuously differentiable function. Repeat backward and join at $t_0$, where both derivatives equal $f(t_0,y_0)$.^[ref]

The compact set here uses a finite-dimensional vector space; bounded closed balls in an arbitrary Banach space need not be compact. When $M=0$, the constant solution of a differential equation $y=y_0$ works on the entire time interval in $K$. In general the proof selects a subsequence, without asserting that all polygons converge to one unique trajectory.

For a Banach space $E$ satisfying $E\cong ({⨁}_{n\ge 1}E{)}_p$, a complemented subspace $W$ is absorbed by $E$: write $E\cong W\oplus V$, expand the countable sum, and absorb one additional copy of $W$ into the countably many $W$ summands. Hence $E\oplus W\cong E$. If $Z$ is complemented in $E$ and contains a complemented copy of $E$, write $Z\cong E\oplus W$ with $W$ complemented in $E$; the preceding argument gives $Z\cong E$. The decompositions are Banach space isomorphisms, not merely algebraic decompositions.

Suppose $f$ is continuous and Lipschitz in its state variable on a closed cylinder $|t-a|\le h$, $\Vert y-b\Vert \le k$, with bounds $\Vert f\Vert \le M$ and Lipschitz constant $L_f$. On a smaller interval of radius $\eta$, the displayed Picard operator maps continuous curves in that closed ball into itself when $\eta M\le k$, and it is a contraction mapping in the supremum norm when $\eta L_f<1$. This function space is complete because it is a closed subset of a Banach space of continuous vector-valued functions. The contraction mapping theorem gives a unique fixed point, and the fundamental theorem of calculus makes it a solution of the ordinary differential equation. Choosing $\eta M<k$ also prevents any solution with the same initial data from first exiting the cylinder during that interval.

For a bounded linear operator $T:X\to X$ on a complex Banach space, the point spectrum is precisely its set of eigenvalues:Here $I$ is the identity operator. Thus membership requires an actual nonzero eigenvector, whereas the spectrum of a bounded operator only requires failure of a bounded inverse. Every eigenvalue belongs to the spectrum, but the inclusion can be strict. The multiplication operator discussed below has an entire interval in its spectrum and empty point spectrum.^[ref]

If $E$ is a nonzero closed linear subspace of a normed ambient space and $F$ is finite-dimensional with $E\cap F=\{0\}$, the unit sphere of $E$ has positive distance from $F$. Otherwise a sequence of nearly coincident points yields a bounded sequence in $F$, hence a convergent subsequence; closedness of $E$ puts its norm-one limit in both vector subspaces. In applications to a bidual space, completeness of the Banach space makes its canonical image closed.

A primitive ideal of a unital associative algebra is the annihilator of a module for a nonzero simple module. Every maximal proper two-sided ideal is primitive: its quotient is a simple ring, and any simple module over that quotient has zero annihilator there. In a unital Banach algebra, a simple module can be realized as $A/L$ for a maximal left ideal; this ideal is closed, making the representation a bounded algebraically irreducible action on a Banach space.

A primitive ideal of a unital associative algebra is the annihilator of a module for a nonzero simple module. Every maximal proper two-sided ideal is primitive: its quotient is a simple ring, and any simple module over that quotient has zero annihilator there. In a unital Banach algebra, a simple module can be realized as $A/L$ for a maximal left ideal; this ideal is closed, making the representation a bounded algebraically irreducible action on a Banach space.

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 $X$.

For a bounded linear operator into a Banach space, compactness is equivalent to the displayed condition, where ${\pi }_Y$ is the quotient map by a finite-dimensional subspace of the target. A finite net for the image of the unit ball gives such a subspace by taking its span. Conversely, if the image lies within $\epsilon$ of a finite-dimensional subspace, choose approximants there; they remain bounded because the original image is bounded. A finite net for these approximants gives a slightly larger finite net for the image. This proves total boundedness and hence compactness. This condition does not require choosing the approximants by a linear map, and therefore does not assume an approximation property for the Banach space.

A quotient operator is a surjective bounded linear operator that is an open map. It is equivalent to requiring a ball about zero in the image of the open unit ball. Translation and scaling then prove being an open map at every point. For a Banach space domain, a bounded linear operator is a quotient operator exactly when its Banach-space adjoint is an isomorphic embedding: a lower bound on the adjoint gives approximate ball-surjectivity by Hahn-Banach separation, and successive approximation for quotient operators gives exact preimages.

A locally convex topological vector space is reflexive when its canonical evaluation map to its continuous bidual is an isomorphism with the specified strong dual topologies. For a Banach space, this reduces to surjectivity of the canonical map and gives a reflexive Banach space.

Every bounded sequence in a Banach space has either a weakly Cauchy sequence as a subsequence or a subsequence equivalent to the unit vector basis of ${\ell }^1$. The real-scalar proof represents vectors as continuous functions on a compact dual ball and uses infinite Ramsey theory to select independent lower and upper level sets when pointwise convergence cannot be obtained.