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Let $(u_n)$ be a Cauchy sequence in the graph normThen $(u_n)$ and $(A{u}_n)$ are Cauchy sequences in the Banach space $H$, so for some $u,v\in H$,Because $A$ is a closed linear operator, $u\in D(A)$ and $Au=v$. It follows thatThus every Cauchy sequence converges in $Y=(D(A),\Vert \cdot {\Vert }_Y)$, and

The familyinherits the identity, semigroup property, and strong continuity from $U$, whileIts difference quotient satisfiesfor $u\in D(A)$. Conversely, existence of this limit implies existence of the generator limit for $U$, so $D(\hat{A})=D(A)$ andThis is the exponentially shifted semigroup construction.

The Hille-Yosida theorem in the uniformly bounded case says that a linear operator $B$ on a Banach space generates a C0-semigroup with $\Vert e^{tB}\Vert \le M$ if and only if:

Let $X=C([0,T];L^2(\mathbb{R}))$ and define the map suggested by the variation-of-constants formula:The strong continuity of the evolution family and the continuity of $f$ imply that $\Phi$ maps $X$ into itself. Because $U(t,s)$ is unitary and $\Vert f(t,u){\Vert }_2\le C$,

Equip $X$ with the exponentially weighted supremum normThis equivalent norm makes $X$ a Banach space. Using that $f$ is a globally Lipschitz function and that the unitary operator $U(t,s)$ preserves the norm,Choose $\alpha >L$. Then $\Phi$ is a contraction mapping, so the Banach fixed-point theorem gives a unique fixed point $u\in X$. This fixed point is exactly the required mild solution of an abstract Cauchy problem:The weighted-norm argument works on the whole prescribed finite interval, so no subdivision of $[0,T]$ is needed.

The assertion is understood for every $\eta \in \mathbb{R}$ and for a topology in which sequentially closed sets are closed, in particular the norm topology of a Banach space. Suppose first that $f$ is sequentially lower semicontinuous and $x_n\in {\Omega }_{\eta }=\{x:f(x)\le \eta \}$ converges to $x$. Thenso $x\in {\Omega }_{\eta }$ and the sublevel set is closed. Conversely, if all sublevel sets are closed but lower semicontinuity fails, there are $x_n\to x$ and a real $\eta <f(x)$ with a subsequence satisfying $f(x_{n_k})\le \eta$. Closedness of ${\Omega }_{\eta }$ would put $x$ in that set, a contradiction. This proves the closed-sublevel-set characterization of lower semicontinuity.

Let $(T_m)$ be a Cauchy sequence in the operator norm. For each $v\in V$,so $(T_{m}v)$ is a Cauchy sequence. Completeness of the Banach space $V$ permits the definition
$Tv={\lim }_m{T}_{m}v$. Pointwise passage to the limit proves that $T$ is a linear. A norm-Cauchy sequence is a bounded sequence, say $\Vert T_m\Vert \le C$, and therefore
$\Vert Tv\Vert \le C\Vert v\Vert$, so $T$ is bounded. Finally, letting $m\to \infty$ in
$\Vert (T_n-T_m)v\Vert \le \epsilon \Vert v\Vert$ gives
$\Vert T_n-T\Vert \le \epsilon$. Thus $T_n\to T$ in operator norm and
$\mathcal{B}(V,V)$ is Banach.

A C0-semigroup on a Banach space $X$ is a family $U(t)\in \mathcal{B}(X)$ such thatfor every $x\in X$. Its infinitesimal generator of a semigroup iswith generator domain

For $M\ge 1$ and $\omega \in \mathbb{R}$, write $A\in \mathcal{G}(M,\omega )$ when $A$ generates a $C_0$-semigroup satisfying $\Vert U(t)\Vert \le M{e}^{\omega t}$. The Hille-Yosida theorem states that this holds exactly when $A$ is closed and densely defined,and, for every real $\lambda >\omega$ and every integer $n\ge 1$,The estimates for every resolvent power, rather than only $n=1$, are essential when $M>1$.

The open mapping theorem states that every surjective bounded linear map $S:X\to Y$ between Banach spaces maps open sets to open sets. Equivalently, there is $c>0$ such that

The closed graph theorem states that a linear map $T:X\to Y$ between Banach spaces is continuous if and only if its graphis closed in $X\times Y$.

Assume the open mapping theorem and suppose $\Gamma (T)$ is closed. As a closed subspace of the Banach space $X\times Y$, the graph is itself Banach. The coordinate projectionis a bounded linear bijection. By the open mapping theorem, $P^{-1}$ is bounded. The other coordinate projection $Q:\Gamma (T)\to Y$ is bounded, andThus $T$ is bounded and hence continuous. This is the closed graph theorem from the bounded inverse theorem argument.

PutThe assumed energy estimate for the linear wave equation and the first nonlinear estimate give, for $w\in X_{b,\tau }$,Choose $b\ge 2C_{0}D$ and then choose $\tau >0$ so small that $C_0\beta {\tau }^{1/2}{b}^2\le b/2$. Then $A$ maps the closed ball $X_{b,\tau }$ into itself.

For $w,\tilde{w}\in X_{b,\tau }$, the difference $Aw-A\tilde{w}$ solves the linear equation with zero Cauchy data and forcing $F(w)-F(\tilde{w})$. The second nonlinear estimate therefore givesShrinking $\tau$ once more makes the coefficient strictly smaller than one. Since $X_{\tau }$ is a Banach space and its closed ball is complete, $A$ is a contraction on $X_{b,\tau }$.

If $X$ is a reflexive Banach space, its closed unit ball is weakly compact. A bounded linear map is weak-to-weak continuous, so $T(B_X)$ is weakly compact in $Y$. Compact subsets of a Hausdorff space are closed, hence $T(B_X)$ is weakly closed and therefore norm closed.

Now suppose the two norms on $V$ have the same continuous dual as a set. Each $X_j^{\ast }$ is a Banach space in its dual norm. The identityhas closed graph: if $f_n\to f$ in the first dual norm and $f_n\to g$ in the second, evaluating at each $x\in V$ gives $f(x)=g(x)$. The closed graph theorem makes $I$ bounded, and the same argument for $I^{-1}$ makes the dual norms equivalent. Thus there are $c,C>0$ withThe dual formulafrom the Hahn--Banach theorem transfers these inequalities to the original norms. Hence $\Vert \cdot {\Vert }_1$ and $\Vert \cdot {\Vert }_2$ are equivalent.

The Leray-Schauder fixed point theorem says that a continuous compact map $T$ on a Banach space has a fixed point if the homotopy setis bounded. If $u=tT(u)$ with $t>0$, multiplying the equation for $T(u)$ by $t$ shows that $u$ solvesThe case $t=0$ gives $u=0$. Therefore it is enough to prove one uniform $C^{1,\beta }$ estimate for all solutions of this family and all $t\in [0,1]$. The theorem then yields a fixed point $u=T(u)$, which solves the original quasilinear problem.

Initially the construction gives $u\in C^{2,\alpha \beta }$. This makes $u$ and $Du$ Lipschitz, so composing the original $C^{0,\alpha }$ coefficients with the jet of $u$ produces $C^{0,\alpha }$ coefficients. A second application of the global Schauder estimate gives $u\in C^{2,\alpha }(\overline{\Omega })$. The fixed-point theorem is an existence result and supplies no uniqueness.

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.

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.

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 $X$.

A Schauder basis of a Banach space $X$ is a sequence $(e_n)$ such that every $x\in X$ has a unique norm-convergent expansion $x={\sum }_{n=1}^{\infty }{a}_n{e}_n$.

A bounded operator $T:X\to Y$ between Banach spaces is compact if and only if its dual map of bounded linear operator $T^{\ast }:Y^{\ast }\to X^{\ast }$ is compact.

A Banach space has the Schur property when every weakly convergent sequence converges with respect to the norm topology.

Equip $D(A)$ with its graph norm. The restrictions $U(t){|}_{D(A)}$ form a C0-semigroup on this Banach space; its generator is $A$ restricted to $D(A^2)$.

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.