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Let $K$ be a nonempty compact convex set in a Banach space and $T:K\to K$ a continuous map. For every $\epsilon >0$, there is a finite-dimensional linear subspace $L_{\epsilon }$ and a continuous map $T_{\epsilon }:K\to K\cap L_{\epsilon }$ that approximates $T$ uniformly to within $\epsilon$ in norm.

Choose $y_1,\dots ,y_N\in T(K)$ whose open radius-$\epsilon$ balls cover the compact image $T(K)$. Set $w_i(x)=\max \{0,\epsilon -\Vert Tx-y_i\Vert \}$ andThe denominator is positive. This convex combination belongs to $K\cap \mathrm{span}\{y_1,\dots ,y_N\}$ and depends continuously on $x$. Every contributing $y_i$ has distance less than $\epsilon$ from $Tx$, so the triangle inequality gives $\Vert T_{\epsilon }(x)-Tx\Vert <\epsilon$.

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 $x={\sum }_{n=1}^{\infty }{a}_n{e}_n$.

Every continuous map from a nonempty compact convex set $K$ in a Banach space to itself has a fixed point. A useful equivalent form takes a nonempty closed bounded convex set $C$ and a continuous map $T:C\to C$ whose image is relatively compact. To obtain this version from the compact-domain version, restrict $T$ to $\overline{\mathrm{conv}}T(C)\subset C$ and use closed convex hull of a compact set.

For a proof on $K$, the Schauder approximation lemma gives continuous maps $T_{\epsilon }:K\to K\cap L_{\epsilon }$ with finite-dimensional $L_{\epsilon }$ and $\Vert T-T_{\epsilon }{\Vert }_{\infty }\le \epsilon$. The Brouwer fixed-point theorem on the nonempty compact convex set $K\cap L_{\epsilon }$ gives $x_{\epsilon }=T_{\epsilon }{x}_{\epsilon }$, so $\Vert T{x}_{\epsilon }-x_{\epsilon }\Vert \le \epsilon$. A convergent subsequence as $\epsilon \downarrow 0$ and the continuity of $T$ give an exact fixed point.

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 algebra is semisimple when its Jacobson radical is zero. This does not assert that every quasinilpotent element is zero. For example, the algebra of bounded operators on a Banach space of dimension greater than one is semisimple, but contains nonzero square-zero rank-one operators. This notion also does not require the finite-dimensional direct-sum decomposition associated with a semisimple algebra.

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 an everywhere-defined linear map $T:E\to F$ between Banach spaces, its separating space measures the possible limiting defect in continuity at zero. It is a closed linear subspace: linear combinations combine witnessing sequences, and diagonal choices with $\Vert x_j\Vert <1/j$ and $\Vert T{x}_j-y_j\Vert <1/j$ handle a convergent sequence $y_j$ of defects. The closed graph theorem gives $T$ continuous exactly when $S(T)=\{0\}$, since a limit $(x_n,T{x}_n)\to (x,y)$ has $y-Tx\in S(T)$.

A subset $K$ of a normed space is sequentially weakly closed when every weakly convergent sequence in $K$ has its limit in $K$. Every weakly closed set has this property. For a convex set in a Banach space, it is equivalent to closure in the norm topology: Mazur lemma produces norm convergent convex combinations from any weakly convergent sequence, and conversely norm convergence implies weak convergence. For a Hilbert space, the weak Banach–Saks theorem in a Hilbert space supplies these convex combinations as subsequence averages. No equivalence with weak closedness is asserted here for arbitrary nonconvex sets.

Let $K:U\to V$ be a bounded linear operator from a real Banach space to a real Hilbert space, let $J$ be a proper lower-semicontinuous convex function, and let $\alpha >0$ and $\Vert f^{\delta }-K{u}^{†}\Vert \le \delta$. For an existing minimizer of $\Vert Ku-f^{\delta }{\Vert }^2/2+\alpha J(u)$ and a source condition in variational regularization $p^{†}=K^{\ast }w\in \partial J(u^{†})$, let $e=f^{\delta }-K{u}^{†}$ and $q=K(u_{\alpha }-u^{†})$. Its optimality subgradient is $p_{\alpha }=K^{\ast }(f^{\delta }-K{u}_{\alpha })/\alpha$. Then $\alpha D_J^{\mathrm{sym}}=⟨e-\alpha w,q⟩-\Vert q{\Vert }^2\le \Vert e-\alpha w{\Vert }^2/4$. The displayed estimate follows from $\Vert e-\alpha w{\Vert }^2\le 2{\delta }^2+2{\alpha }^2\Vert w{\Vert }^2$.

For a compact Hausdorff space $K$, $C(K)$ consists of continuous scalar-valued functions with the supremum norm. It is a Banach space and, with complex scalars, pointwise multiplication and conjugation make it a commutative unital C-star algebra.

Why does completeness hold? A Cauchy sequence in the supremum norm is Cauchy at every point, hence has a pointwise limit. The uniform Cauchy estimate passes to that limit, giving uniform convergence; the uniform limit theorem makes the limit continuous. Thus the norm and the chosen regularity of the functions cooperate to keep every limit inside $C(K)$.

Changing to the L2 norm on $C([0,1])$ loses this property. For integers $n\ge 2$, the continuous functions $h_n(x)=\min (1,\max (0,n(x-1/2)))$ approximate the step function $h(x)=0$ for $x\le 1/2$ and $h(x)=1$ for $x>1/2$. Direct integration givesThe triangle inequality makes $(h_n)$ a Cauchy sequence, but no continuous function agrees with $h$ almost everywhere: continuity would force value zero on the whole left open interval and one on the whole right open interval. Hence this normed space is incomplete. The missing limit exists in L2 space, its completion of a normed space, showing exactly what the larger space adds.

For a bounded linear operator on a complex Banach space, either a spectral scalar has approximate unit eigenvectors or it is an eigenvalue of the Banach-space adjoint. Indeed outside the approximate point spectrum, $S=T-\lambda I$ is bounded below, hence injective with closed range. If the adjoint has no zero vector in its kernel other than zero, the Hahn-Banach theorem makes the range dense: any proper closed range would have a nonzero continuous functional annihilating it. Thus the range is all of the space and the lower bound gives a bounded inverse. Conversely invertibility rules out both alternatives. With complex-linear duals there is no conjugation of $\lambda$ in the displayed formula; Hilbert-adjoint conventions must be distinguished.

Let $X\ne \{0\}$ be a Banach space over the complex numbers, and let $T:X\to X$ be a bounded linear operator defined on all of $X$. Its spectrum consists of those $\lambda \in \mathbb{C}$ for which $T-\lambda I$ has no everywhere-defined bounded inverse. Its complement is the resolvent set of an operator $\rho (T)$. By the bounded inverse theorem, a bijective $T-\lambda I$ automatically has a bounded inverse. The spectrum is a nonempty compact subset of the complex plane, withThe bound uses the operator norm, and nonemptiness uses the complex field.^[ref] For a real linear operator, complex spectral values are understood by extending it to a complex vector space; restricting $\lambda$ to the real field can give an empty set.

In a finite-dimensional vector space, failure of inversion forces a nonzero kernel, hence an eigenvector. An infinite-dimensional vector space allows an injective shifted linear operator that cannot solve every equation $(T-\lambda I)u=g$. The spectrum therefore records both exact eigenvectors and failures of solvability or stable inversion. Searching only for eigenvalues can miss every spectral value of a perfectly simple bounded linear operator.

Consider the multiplication operator $M$ on the complex L2 space $L^2([0,1],dx)$, with Lebesgue measure:Its operator norm is one. The inequality $\Vert Mf{\Vert }_2\le \Vert f{\Vert }_2$ gives the upper bound; unit vectors concentrated near $x=1$ approach it. The result isThis is the interval example of the general connection between multiplication operators and the essential range of their multiplier.^[ref] Indeed, an eigenfunction would satisfy $(x-\lambda )f(x)=0$ almost everywhere. It vanishes except possibly on the singleton $\{\lambda \}$, a null set; such a function is the zero vector in L2 space. Concentrating at one point cannot produce a nonzero L2 function.

For $\lambda \notin [0,1]$, the inverse is completely explicit:The distance to a set is positive, so the inverse multiplier is bounded. Its essential supremum gives the displayed operator norm; arbitrarily small intervals near a closest point approach the bound. Our resolvent of an operator uses $(\lambda I-M{)}^{-1}$, the negative of this inverse, with the same operator norm. For $\lambda =-1/4$, $5/4$, and $1/2+i/10$, the inverse norms are respectively $4$, $4$, and $10$. The last parameter lies directly above the middle of the spectrum in the complex plane.

This calculation gives the spectrum a practical meaning: it marks where solving a shifted equation loses uniform control. At $\lambda =1/2+i\eta$, the largest amplification of an input error is $1/|\eta |$. Changing $\eta$ from $0.1$ to $0.01$ increases that factor from ten to one hundred. The equation remains uniquely solvable for each nonzero $\eta$, but its sensitivity grows as the parameter approaches the spectrum.

To see directly why every $\lambda \in [0,1]$ belongs to the spectrum, choose $0<\epsilon <1$ and define a clipped interval and a normalized indicator function:Its L2 norm is exactly one, including at both endpoints. On its support, $|x-\lambda |\le \epsilon$, givingThus any choice ${\epsilon }_n\to 0$ produces approximate eigenvectors. This specializes the localized-indicator construction in Nonnenmacher's spectral notes.^[ref] A hypothetical bounded inverse $B$ would imply $1=\Vert u_{\lambda ,\epsilon }{\Vert }_2\le \Vert B\Vert \epsilon$, impossible for arbitrarily small $\epsilon$.

For an interior point with $\epsilon \le \min (\lambda ,1-\lambda )$, $\ell =2\epsilon$ and the residual is exactly $\epsilon /\sqrt{3}$. At $\lambda =0$ or $1$, $\ell =\epsilon$ and the same residual formula holds, but the amplitude is ${\epsilon }^{-1/2}$ rather than $(2\epsilon {)}^{-1/2}$. At $\lambda =1/2$, taking $\epsilon =0.2,0.1,0.05$ gives amplitudes approximately $1.5811,2.2361,3.1623$ and residuals $0.11547,0.057735,0.028868$. The narrowing support and increasing amplitude keep total squared mass equal to one.

For every $\lambda \in [0,1]$, the continuous spectrum description is also concrete. Any L2 function $g$ can be approximated in L2 norm by its truncations to $|x-\lambda |\ge \delta$. Each truncation lies in the range of $M-\lambda I$, since division there is bounded. The range is therefore dense. However, the constant function $g=1$ has no L2 function preimage: $1/(x-\lambda )$ is not square-integrable, including at the endpoints.

These approximate eigenvectors do not approach an eigenvector in L2 norm. Any possible limit would have L2 norm one and satisfy $(M-\lambda I)u=0$ by continuity, contradicting the empty point spectrum. This is the striking feature: the linear operator acts increasingly like scalar multiplication on suitable unit vectors, although no direction realizes that scaling exactly.

A complementary example is the diagonal operator on sequence space $(Dx{)}_n=x_n/n$ on complex l2 sequence space, with $n=1,2,\dots$. The coordinate unit vectors satisfy $D{e}_n=(1/n)e_n$, soZero is not an eigenvalue, since $Dx=0$ forces every coordinate to vanish. Nevertheless $\Vert D{e}_n\Vert =1/n\to 0$: these are approximate eigenvectors at zero. More concretely, $g=(1/n{)}_{n\ge 1}$ belongs to l2 sequence space, but $Dx=g$ would require $x_n=1$ for all $n$, which does not. Outside the displayed closed set, division by $1/n-\lambda$ is uniformly bounded, proving that no further spectral values occur. Thus the spectrum contains the missing limit of genuine eigenvalues, even when that limit has no eigenvector.

Let $P$ be a bounded linear operator on a complex Banach space with $P^2=P$, $P\ne 0$ and $P\ne I$. Its image of a linear map and kernel contain nonzero vectors, giving eigenvalues one and zero. For $\lambda \notin \{0,1\}$,Direct multiplication proves the formula, so the spectrum is exactly $\{0,1\}$ and its spectral radius is one. Since $P^j=P$ and $\Vert P\Vert \ge 1$, the spectral radius formula is explicit: ${\inf }_{j\ge 1}\Vert P^j{\Vert }^{1/j}={\lim }_{j\to \infty }\Vert P{\Vert }^{1/j}=1$.

A normalized basic sequence generates a spreading model when the norms of all finite scalar combinations converge as the first index tends to infinity, uniformly over increasing choices of the indices. The limiting norm on finitely supported scalar sequences is unchanged by increasing relabellings of coordinates. Completing this norm gives a Banach space with its canonical basic sequence. A diagonal application of the finite-colour infinite Ramsey theorem produces such a model from a subsequence of every normalized basic sequence.

A strictly monotone operator $A:X\to X^{\ast }$ on a real Banach space satisfies strict operator monotonicity for every pair of distinct inputs. It is injective: $Au=Av$ would make the pairing zero and hence force $u=v$. In particular $Au=f$ has at most one solution for each $f\in X^{\ast }$.

A function into a Banach space is strongly measurable if, outside a null set, it is a pointwise norm limit of measurable simple functions. A measurable function with values in a separable Banach space is strongly measurable: approximate values by a countable collection of balls of shrinking radii and then truncate the resulting countably valued approximants to simple functions. A strongly measurable function is Bochner integral integrable exactly when its norm is integrable. Finite second moments under a probability measure therefore imply existence of its Bochner integral expected value by the Cauchy-Schwarz inequality.

A map from a subset of a real Banach space to its continuous dual space is strongly monotone if the displayed inequality holds for every pair in its domain, with one constant $m>0$. The brackets denote the dual pairing. This implies strict monotonicity for distinct arguments. Fixing $v=u_0$ and applying the dual norm inequality gives $⟨A(u),u-u_0⟩\ge m\Vert u-u_0{\Vert }^2-\Vert A(u_0)\Vert \Vert u-u_0\Vert$, so it is a nonlinear coercive operator with respect to $u_0$.

Let $T:X\to Y$ be a bounded linear operator with $X$ a Banach space. Suppose every $\Vert y\Vert <1$ has an approximation $Tx$ with $\Vert x\Vert <1$ and error less than $\delta \in (0,1)$. Repeatedly approximate the scaled residual to construct $x_n$ with $\Vert x_n\Vert <{\delta }^n$ and residual less than ${\delta }^{n+1}$. Completeness sums the series, and continuity gives $T\sum x_n=y$. Since $\Vert \sum x_n\Vert <(1-\delta {)}^{-1}$, scaling proves $B_Y(0,1-\delta )\subseteq T(B_X(0,1))$. Hence $T$ is a quotient operator, even without assuming completeness of $Y$.