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Suppose $J$ is nonnegative and sequentially lower semicontinuous, and its sublevel sets are strongly sequentially compact. For a solvable exact equation $Au=f$, every sequence of residual-method minimizers with $\delta \to 0$ has a strongly convergent subsequence, and every limit is a J-minimizing solution. Indeed, comparison with an exact penalty minimizer bounds all reconstructions in one compact sublevel set; the residual bound gives $\Vert A{u}_{\delta }-f\Vert \le 2\delta$. Continuity of $A$ and lower semicontinuity of $J$ identify each subsequential limit as an exact minimizer. Thus the distance to a set of exact penalty minimizers tends to zero. Uniqueness of the exact minimizer upgrades this to full strong convergence.

Uniqueness cannot be omitted: for $A=0$, $f_{\delta }=f=0$, and $J(u)=\max \{|u|-1,0\}$ on $\mathbb{R}$, the sublevel sets are compact intervals but the minimizers $u_{1/k}=(-1{)}^k$ do not converge.

In a metric space $X$, let the nonempty compact set $K$ lie in the open set $\Omega$. If $F=X\setminus \Omega$ is nonempty, the distance to a set $d(x,F)$ is a continuous function of $x$: the triangle inequality gives $|d(x,F)-d(y,F)|\le d(x,y)$. Since $F$ is a closed set, this distance is strictly positive on $K$. The extreme value theorem attains its minimum on $K$, soIf $F$ is empty, the extended-real convention gives $d(K,F)=+\infty$. This result provides a single positive neighbourhood radius that stays inside $\Omega$ at every point of $K$. Compactness is essential: in the real line $K=\{1/n:n\ge 1\}\subset (0,2)$ has zero distance between two sets from $(-\infty ,0]\cup [2,\infty )$, although each point of $K$ has positive distance to a set from that complement.

For any nonempty complete metric space $X$, the family of its nonempty closed bounded subsets is a complete metric space under the Hausdorff distance. Boundedness of the entire ambient space is unnecessary in this general form.

For a Cauchy sequence $(F_n)$ of these sets, define $A_n=\overline{{\bigcup }_{m\ge n}{F}_m}$ and ${\delta }_n={\sup }_{p,q\ge n}{d}_H(F_p,F_q)$. A Cauchy sequence in any metric space is bounded, so its tail sets are contained in a common finite-radius neighbourhood of one bounded $F_N$; hence every $A_n$ is bounded. The distance to a set is one-Lipschitz, so estimates pass to the closure, giving $d_H(F_n,A_n)\le {\delta }_n$ and $d_H(A_n,A_m)\le {\delta }_n$ for $m\ge n$. Choose $n_j$ with ${\delta }_{n_j}\le 2^{-j}$. The Summably nested Hausdorff intersection theorem gives $A_{n_j}\to G\ne \varnothing$. The triangle inequality now gives $F_n\to G$, proving completeness.

For a nonempty subset $F$ of a metric space, the distance to a set is $d(x,F)={\inf }_{y\in F}d(x,y)$. The triangle inequality gives $d(x,F)\le d(x,z)+d(z,F)$; swapping $x,z$ proves $|d(x,F)-d(z,F)|\le d(x,z)$. Thus it is one-Lipschitz. Its zero set is exactly the closure of $F$: every neighbourhood of $x$ meets $F$ precisely when this infimum vanishes. For a closed set, zero distance therefore implies membership. When the empty set is allowed, use the convention $d(x,\varnothing )=+\infty$; the finite-valued Lipschitz assertion applies to nonempty sets.

For a real function $g$ in the bounded continuous functions of norm $M$ on a closed subset $Y$ of a metric space, separate its closed level sets $A=\{g\le -M/3\}$ and $B=\{g\ge M/3\}$ by $u=(M/3)(d(\cdot ,A)-d(\cdot ,B))/(d(\cdot ,A)+d(\cdot ,B))$. The distance to a set functions make $u$ continuous with the displayed bounds. Empty level sets are handled by the constants $M/3$, $-M/3$ or $0$. Iterating on the residual yields a uniformly convergent continuous extension with no increase in the supremum norm.

For a finite open cover $U_0,\dots ,U_d$ of $S^d$, take continuous nonnegative functions ${\varphi }_i$ summing to one and positive only in $U_i$. Such a partition of unity comes from normalizing the distance to a set functions $d(x,S^d\setminus U_i)$. The Borsuk-Ulam theorem applied to $({\varphi }_1,\dots ,{\varphi }_d)$ gives equal values at an antipodal pair. The sum condition makes ${\varphi }_0$ equal too, and some common positive value places both points in one $U_i$.

If $x\in F$, using $y=x$ in the infimum gives $0\le f(x)\le 0$, so $f(x)=0$.

Conversely, if $f(x)=0$, the definition of the infimum allows a choice $y_j\in F$ with $|x-y_j|<1/j$. The sequence $y_j$ is therefore a convergent sequence with limit $x$. Since $F$ is a closed set, its limit $x$ belongs to $F$. HenceClosedness matters: without it, zero distance to a set characterizes its closure, which can contain additional points.

The distance to a set is nonnegative and finite here: any fixed $y_0\in F$ gives $0\le f(x)\le |x-y_0|$. For every $y\in F$, the triangle inequality givesTaking the infimum over $y\in F$ yields $f(x)\le |x-x^{\prime }|+f(x^{\prime })$. Interchanging $x$ and $x^{\prime }$ provesThus $f$ is a Lipschitz function with Lipschitz constant one, so it is continuous. In the epsilon-delta definition of continuity, choose $\delta =\epsilon$.

Start with the approximate surjectivity with geometric correction argument. For a fixed $y$, put $r_0=y$. Choose $x_j$ successively so thatInduction gives $\Vert r_j\Vert \le k^j\Vert y\Vert$ and $\Vert x_j\Vert \le R{k}^{j-1}\Vert y\Vert$. The series ${\sum }_j{x}_j$ is absolutely convergent in the Banach space $V$, because the sum of its norms is bounded by the convergent geometric series $R\Vert y\Vert /(1-k)$. Write its sum as $x$. The bounded linear operator $T$ commutes with this limit, while $T{\sum }_{j=1}^m{x}_j=y-r_m\to y$. ConsequentlyThe choices need not depend linearly on $y$; no bounded linear right inverse is being asserted.

For the bounded extension, use the restriction bounded linear operator $S:C_b(X;\mathbb{R})\to C_b(Y;\mathbb{R})$ on the Banach spaces of bounded continuous functions, each with its supremum norm. We will construct an approximation to any $g$ with $\Vert g{\Vert }_{\infty }=M>0$. The two setsare disjoint closed sets in $X$, since $Y$ is closed. If both are nonempty, the distance to a set construction defines a continuous functionThe denominator is positive at every point: zero distance to a closed set means membership in that set. Thus $\Vert u{\Vert }_{\infty }\le M/3$, with $u=-M/3$ on $A$ and $u=M/3$ on $B$. On $A$ and $B$, and also on the intervening range $|g|<M/3$, this impliesIf only $A$ is empty take $u=M/3$, if only $B$ is empty take $u=-M/3$, and if both are empty take $u=0$; the same bounds hold. For $M=0$ take $u=0$. This proves the approximation hypothesis with $R=1/3$ and $k=2/3$. Applying the argument above to $S$ yields an element $h$ of the bounded continuous functions with $h{|}_Y=g$ and $\Vert h{\Vert }_{\infty }\le \Vert g{\Vert }_{\infty }$. Restriction gives the reverse inequality, so the extension preserves the exact supremum norm:This proves the bounded metric-space case of the Tietze extension theorem rather than presupposing it.

For an arbitrary real continuous function $f$ on $Y$, let $g=\tanh f$. Its values lie strictly between $-1$ and $1$, although its supremum norm can be $1$. The bounded result supplies an extension $h$ with $|h|\le 1$. Let $D=\{x:|h(x)|=1\}$, a closed set disjoint from $Y$. If $D$ is nonempty and $Y$ is nonempty, putBoth factors are continuous functions; $\eta =1$ on $Y$ and $\eta =0$ on $D$. Off $D$ we have $|h|<1$, and on $D$ we have $q=0$, so $|q(x)|<1$ everywhere. Therefore a real continuous extension isOn $Y$ it equals $f$. If $D$ is empty use $q=h$; if $Y$ is empty simply take $F=0$. This handles unbounded $f$ without requiring a uniform margin between $|g|$ and $1$.

For nonempty $F$, the triangle inequality gives $d(x,z)\le d(x,y)+d(y,z)$ for every $z\in F$. Taking the infimum yields $d(x,F)\le d(x,y)+d(y,F)$. Interchanging $x,y$ provesThus the distance to a set is a Lipschitz function, hence continuous. If $x\notin F$ and $F$ is closed, its open complement contains a ball $B(x,ϵ)$; every point of $F$ is at least $ϵ$ away. Hence $\boxed{d(x,F)>0}$ for $x\notin F$.

For two nonempty disjoint closed sets, let $h(x)=d(x,F_1)-d(x,F_2)$. It is continuous, negative on $F_1$ and positive on $F_2$. The disjoint open sets $U_1=\{h<0\}$ and $U_2=\{h>0\}$ contain them. Empty sets cause no difficulty, using an empty neighborhood for the empty set. This proves every metric space is a normal topological space.

In any normal space, a closed $F$ contained in an open $U$ can be separated from $X\setminus U$: choose disjoint open $W\supset F$ and $V\supset X\setminus U$. Because $X\setminus V$ is closed and contains $W$, $\overline{W}\subset X\setminus V\subset U$. First choose disjoint open $U_1,U_2$ around $F_1,F_2$, then apply this argument separately to each pair $F_i\subset U_i$. It gives ${\overline{W}}_i\subset U_i$, and therefore $\boxed{{\overline{W}}_1\cap {\overline{W}}_2=\varnothing }$. This is closed-neighborhood shrinking in a normal space.

For the two specified hyperbola branches, explicit choices areThey are open and contain the branches with products $-1$ and $1$. Their closures lie respectively in $\{x\le 0,xy\le -1/2\}$ and $\{x\ge 0,xy\ge 1/2\}$. Any common point would have $x=0$, making both product inequalities impossible. Thus their closures are disjoint even though the original branches can approach one another arbitrarily closely at large height.

We prove the required Hahn-Banach theorem for a seminorm directly. First suppose the vector space is real. At an intermediate extension stage, let $g:E\to \mathbb{R}$ be a linear functional with $|g(y)|\le p(y)$, and let $z\notin E$. Extending to $E+\mathbb{R}z$ amounts to choosing $c=f(z)$. The upper domination $f\le p$ requiresThe interval is nonempty: for all $y,w\in E$,Its endpoints are finite since inserting zero gives bounds $-p(z)$ and $p(z)$, while every lower candidate is below every upper candidate. Choose $c$ in the interval and put $f(y+tz)=g(y)+tc$. For $t>0$, the upper bound with $w=y/t$ proves $f(y+tz)\le p(y+tz)$. For $t<0$, writing $s=-t>0$ and applying the lower bound with $y/s$ proves the same inequality. The case $t=0$ is already known. Applying this domination to $-x$ and using $p(-x)=p(x)$ gives $|f(x)|\le p(x)$.

Partially order all dominated linear functional extensions of the original $g$ by extension of their domains and values. They form a nonempty set. The union along any chain is a well-defined dominated linear functional on a vector subspace, so every chain has an upper bound. The Zorn lemma gives a maximal extension. If its domain were not $X$, the preceding one-dimensional construction would enlarge it, a contradiction. Thus the required extension exists on $X$.

For a complex vector space, apply the proved real result to $\mathrm{Re}g$ on the underlying real vector subspace. Let $U:X\to \mathbb{R}$ be the resulting real linear functional, with $|U(x)|\le p(x)$, and setIt is additive and real-linear, and $f(ix)=if(x)$, so it is complex-linear. On $Y$, the identity $\mathrm{Re}g(iy)=-\mathrm{Im}g(y)$ proves $f(y)=g(y)$. Choose $\alpha$ with $|\alpha |=1$ and $\alpha f(x)=|f(x)|$ when $f(x)\ne 0$. ThenThe bound is immediate if $f(x)=0$. This proves the real and complex seminorm forms of Hahn-Banach theorem without invoking any version of that theorem.

For the distance to a set assertion, let $d=d(z,Y)>0$. Define on $Y+\mathbb{K}z$ the linear functional $g(y+\alpha z)=\alpha d$. The decomposition is unique because $z\notin Y$, and for $\alpha \ne 0$,For $\alpha =0$ the bound is immediate. Apply the just-proved Hahn-Banach theorem with $p(x)=\Vert x\Vert$ to get $f\in X^{\ast }$, $f{|}_Y=0$, $f(z)=d$, and $\Vert f\Vert \le 1$. Choose $y_n\in Y$ with $\Vert z-y_n\Vert \to d$. Sinceand $d>0$, passage to the limit gives $\Vert f\Vert \ge 1$. Thus $\boxed{f{|}_Y=0,f(z)=d(z,Y),\Vert f\Vert =1}$. This version of the Hahn-Banach distance formula needs neither a closed $Y$ nor an attained distance to a set.

For the Banach limit construction below, take the real l-infinity sequence space, let $S(x_1,x_2,\dots )=(x_2,x_3,\dots )$ be its left shift on bounded sequences, put $Y=(I-S){\ell }^{\infty }$, and let $1=(1,1,\dots )$. This is a vector subspace and its distance to a set from $1$ determines the construction.

The assumptions prove convergence to the set of J-minimizing solutions, and subsequential convergence to a member of that set. Convergence to a single fixed member requires uniqueness or a suitable selection rule.

Let $u^{†}$ be any J-minimizing solution, whose existence is granted. It is feasible for every noisy-data constraint because $A{u}^{†}=f$ and $\Vert f-f_{\delta }\Vert \le \delta$. Minimality therefore gives $J(u_{\delta })\le J(u^{†})$. All reconstructions lie in this strongly sequentially compact space sublevel set. For any sequence ${\delta }_k\downarrow 0$, pass to a subsequence with $u_{{\delta }_k}\to u_{\ast }$. Boundedness of $A$ and the triangle inequality implyso $A{u}_{\ast }=f$. By sequential lower semicontinuity,Thus $u_{\ast }$ is a J-minimizing solution. The same compactness argument shows the distance to a set of exact minimizers tends to zero: otherwise a subsequence at a fixed positive distance would converge to an exact minimizer, a contradiction. If that minimizer is unique, every convergent subsequence has the same limit, giving the asserted full strong convergence.

For a counterexample to full convergence under the printed hypotheses alone, take $\mathcal{U}=\mathcal{V}=\mathbb{R}$, $A=0$, $f=f_{\delta }=0$, and $J(u)=\max \{|u|-1,0\}$. This is nonnegative and continuous, with compact sublevel sets $[-1-C,1+C]$ for $C\ge 0$. Every point of $[-1,1]$ is both an exact and a residual-method minimizer. Choosing $u_{1/k}=(-1{)}^k$ gives a valid sequence of minimizers that does not converge. The reusable corrected statement is compact-sublevel convergence of the residual method.

For a harmonic function $u$, letThe divergence theorem givesSince $M(r)\to u(x)$ as $r\downarrow 0$, $M(r)=u(x)$. Integrating the spherical averages in the radial variable gives the corresponding ball average, proving the mean value property for harmonic functions. If $u$ attains its maximum at an interior point, the average of the nonnegative function $\max u-u$ on every sufficiently small centred sphere is zero. Continuity makes $u$ constant on those spheres, and connectedness propagates that value through the domain. Thus the weak maximum principle for elliptic operators gives

For the derivative estimate, choose $r>0$ smaller than half the distance from ${\Omega }^{\prime }$ to $\partial \Omega$, and let ${\rho }_r$ be a smooth radial mollifier supported in $B_r(0)$. Writing its convolution in polar coordinates and using the spherical mean value property shows that $u\ast {\rho }_r=u$ on ${\Omega }^{\prime }$. Hence, for every multi-index $\alpha$,so Holder inequality gives

First suppose $F$ is a closed set. If $x\notin F$, the definition in part (ii) gives an open ball $B(x,r)$ disjoint from $F$. Thus every $y\in F$ has $d(x,y)\ge r$, and the distance to a set satisfies $\mathrm{dist}(x,F)\ge r>0$. Its contrapositive is the required implication.

Conversely, suppose that zero distance to a set forces membership. For $x\notin F$ we have $\mathrm{dist}(x,F)>0$. Choose $0<r<\mathrm{dist}(x,F)$. Then $B(x,r)\cap F=\varnothing$, proving $X\setminus F$ is an open set. HenceFor $F=\varnothing$, use the extended-real convention $\inf \varnothing =+\infty$: both assertions hold, with the implication vacuous. Part (b) only uses nonempty sets, so all its distances to a set are ordinary finite real numbers.

Let $(F_n)$ be a Cauchy sequence for the Hausdorff distance, and putThe bounded set hypothesis makes ${\delta }_n$ finite, and the Cauchy sequence property gives ${\delta }_n\to 0$. Each $A_n$ is nonempty and a closed set, with $A_{n+1}\subseteq A_n$.

We need estimates that survive taking the closure. For every nonempty $F$, the distance to a set is Lipschitz continuous with constant one: the triangle inequality gives $\mathrm{dist}(x,F)\le d(x,y)+\mathrm{dist}(y,F)$, and interchanging $x,y$ provesFor every $m\ge n$, points of $F_m$ have distance at most ${\delta }_n$ from $F_n$. The same holds throughout their union and its closure, by this continuity. Since $F_n\subseteq A_n$,More generally, if $m\ge n$, every point in ${\bigcup }_{k\ge n}{F}_k$ is within ${\delta }_n$ of $F_m\subseteq A_m$. Passing to its closure again, and using $A_m\subseteq A_n$, givesChoose strictly increasing indices $n_j$ with ${\delta }_{n_j}\le 2^{-j}$. Then $(A_{n_j})$ satisfies part (ii), so it converges in Hausdorff distance to the nonempty closed set $G={\bigcap }_j{A}_{n_j}={\bigcap }_n{A}_n$, with $\rho (A_{n_j},G)\le 2^{1-j}$.

For $n\ge n_j$, the triangle inequality gives the explicit full-sequence boundIt tends to zero as $j\to \infty$. ThereforeThis proves Completeness of the Hausdorff hyperspace using the proposed tail closures, rather than assuming their diameters shrink.

The intersection $G$ is a closed set because arbitrary intersections of closed sets are closed. Its nonemptiness needs a construction: bounded closed sets need not be compact sets in this setting.

Fix $n$, $x_n\in F_n$, and $\eta >0$. Recursively choose $x_{k+1}\in F_{k+1}$, for $k\ge n$, using the approximation property of an infimum so thatThis is possible because $x_k\in F_k$ and $\mathrm{dist}(x_k,F_{k+1})\le \rho (F_k,F_{k+1})\le 2^{-k}$. The two bounds form convergent geometric series, so the triangle inequality shows $(x_k{)}_{k\ge n}$ is a Cauchy sequence. By completeness it converges to some $x\in X$.

For each fixed $j$, all sufficiently late $x_k$ lie in $F_j$, by nesting. Closedness of $F_j$ gives $x\in F_j$, and hence $x\in G$. In particular, starting with any $x_1\in F_1$ proves $G\ne \varnothing$. Moreover,For each starting point $x_n$, let $\eta \downarrow 0$ in this bound on its distance to a set $G$. This does not assert that a nearest point exists. It proves $\mathrm{dist}(x_n,G)\le 2^{1-n}$. Since $G\subseteq F_n$, the other directed Hausdorff distance is zero, andThis is the Summably nested Hausdorff intersection theorem; the summable movement replaces any need for compactness.

Write $h(A,B)={\sup }_{a\in A}\mathrm{dist}(a,B)$, so that $\rho (A,B)=\max \{h(A,B),h(B,A)\}$ is the Hausdorff distance. Since $X$ is a bounded set, there is a finite $M$ with $d(x,y)\le M$. Nonemptiness of $A,B$ therefore gives $0\le h(A,B)\le M$, so $\rho$ is finite. Its symmetry and nonnegativity are immediate, as is $\rho (A,A)=0$.

If $\rho (A,B)=0$, then every $a\in A$ has zero distance to a set $B$. Since $B$ is a closed set, part (a)(iii) gives $A\subseteq B$. Interchanging the sets gives $B\subseteq A$. Thus $\rho (A,B)=0$ implies $A=B$. Closedness is essential here: without it a set and its closure have Hausdorff distance zero.

For the triangle inequality, fix $a\in A$ and $ϵ>0$. The approximation property of an infimum supplies $b\in B$ with $d(a,b)<\mathrm{dist}(a,B)+ϵ$, and then $c\in C$ with $d(b,c)<\mathrm{dist}(b,C)+ϵ$. Consequently,Let $ϵ\downarrow 0$ and take the supremum over $a$ to obtain $h(A,C)\le h(A,B)+h(B,C)$. Applying this inequality in the reverse direction as well yieldsAll metric axioms follow. Boundedness was used to ensure finiteness, and closedness to distinguish different sets; neither was needed for the triangle inequality.

The distance to a set is finite and nonnegative because $A$ is nonempty. For $x,x^{\prime }\in X$ and every $a\in A$, the triangle inequality givesTaking the infimum over $a$ yields $D_A(x)\le d(x,x^{\prime })+D_A(x^{\prime })$. Interchanging $x$ and $x^{\prime }$ gives the opposite inequality. Together these implyThus the distance to a set is $1$-Lipschitz continuous. No closedness, compactness or attainment of the infimum was used.

The intended hypothesis is that $A$ is not connected. The original PDF has the same grammatical ambiguity as the transcription. If only $X$ were assumed not connected, the assertion would fail for the discrete space $X=\{0,1\}$ with its usual metric and the closed subset $A=\{0\}$: two disjoint sets cannot both meet that singleton.

Under the intended reading, choose a separation of a topological space $A=B\bigsqcup C$. Both $B$ and $C$ are nonempty and closed in $A$, because each is the complement of the other relatively open part. Since $A$ is closed in $X$, both are closed in $X$.

For a nonempty subset $F$ of the metric space $(X,d)$, the distance to a set satisfiesIndeed, the triangle inequality gives $d(x,F)\le d(x,z)+d(z,F)$, and swapping $x,z$ proves the bound. Thus the two distance to a set functions below are continuous. SetThese are disjoint open sets. For $b\in B$, we have $d(b,B)=0$ and $d(b,C)>0$, since $C$ is closed and $b\notin C$. Hence $B\subseteq U$. Similarly $C\subseteq V$. Therefore $A\subseteq U\cup V$ and both $U$ and $V$ meet $A$, as required. The construction is a concrete instance of the normality of every metric space.

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.