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A nonlinear compact operator is a continuous map that sends bounded sets of its domain to relatively compact subsets of the target space; linearity is not part of this general definition.

For the specified diagonal operator on sequence space on l2 sequence space,so it is well-defined, linear and bounded. Let $T_N$ retain only its first $N$ output coordinates. This is a finite-rank operator, andThe upper bound follows by estimating each tail multiplier by $1/(N+1)$; equality follows on the unit coordinate vector $e_{N+1}$. Finite-rank operators between Banach spaces are compact, and a limit of compact linear operators in operator norm is compact, the norm-closedness of compact operators. Therefore $\boxed{T\text{ is compact}}$.

Define a diagonal operator on sequence space byIt is bounded because $\Vert Sx{\Vert }_2^2\le \Vert x{\Vert }_{\infty }^2{\sum }_{n\ge 1}{4}^{-n}=\Vert x{\Vert }_{\infty }^2/3$. Its truncation $S_{m}x=(2^{-1}{x}_1,\dots ,2^{-m}{x}_m,0,\dots )$ is a finite-rank operator, andEquality follows by using finitely supported sequences equal to $1$ on longer and longer finite tails. An operator norm limit of compact operators into the Banach space ${\ell }_2$ is compact; therefore $S$ is a compact operator. The vectors $S{e}_n=2^{-n}{e}_n$ are linearly independent, so its rank is infinite.

There is no conflict with the preceding result about a compact closed-ball image. Indeed, take $x^{(m)}=(1,\dots ,1,0,\dots )$ with $m$ initial ones. Then $S{x}^{(m)}\to (2^{-n}{)}_{n\ge 1}$ in ${\ell }_2$, but that limit has no preimage in $c_0$: its only possible preimage is the constant sequence $1$. Thus $S(B_{c_0})$ is relatively compact but is not closed and hence is not compact.

The exact condition for this diagonal operator on sequence space isFirst, $T$ is bounded precisely when ${\sup }_n|{\lambda }_n|<\infty$, with operator norm equal to that supremum: the upper bound follows by summing $|{\lambda }_n{x}_n{|}^2$, and the coordinate unit vectors $e_n$ give the lower bound. If the supremum is infinite, choose distinct indices with $|{\lambda }_{n_j}|\ge j$ and set $x_{n_j}=1/j$, other coordinates zero. Then $x\in {\ell }^2$ but $Tx\notin {\ell }^2$, so the displayed prescription would not even define an everywhere-defined map into ${\ell }^2$.

If $T$ is compact but ${\lambda }_n\nrightarrow 0$, infinitely many indices satisfy $|{\lambda }_n|\ge ϵ>0$. Along those indices,contradicting compactness. Conversely, if ${\lambda }_n\to 0$, truncate to $T_{N}x=({\lambda }_1{x}_1,\dots ,{\lambda }_N{x}_N,0,\dots )$. This is a finite-rank compact operator, and $\Vert T-T_N\Vert ={\sup }_{n>N}|{\lambda }_n|\to 0$. To see compactness directly, approximate the image of the unit ball by $T_N$ to within $ϵ/2$, and cover its bounded finite-dimensional image by finitely many $ϵ/2$-balls. Hence $T$'s unit-ball image is totally bounded; its closure is compact because ${\ell }^2$ is complete.

No. Reuse the diagonal operator on sequence space $L(x_n)=(x_n/n)$ on ${\ell }^2$. Its kernel is zero, hence finite-dimensional. Let $y^{(N)}$ be the truncation of $(1/n)$ to its first $N$ coordinates. Each $y^{(N)}=L(1,\dots ,1,0,\dots )$ lies in the range, while $y^{(N)}\to (1/n)$ in ${\ell }^2$. The limit is not in the range because its formal preimage is not square summable. Thus finite-dimensional kernel does not imply closed range. Equivalently the unit vectors $e_n\in (\ker L{)}^{\perp }$ violate every positive closed-range bound on the kernel complement.

First let $L$ be a bounded linear operator on $H$ satisfying the symmetric positive-definiteness convention in part (a). For any direction $w\in H$, expansion of the quadratic functional givesThus vanishing of the first variation in every direction is precisely the weak equation $⟨Lu,w⟩=⟨f,w⟩$ for every $w$. In this whole-space bounded-operator setting it is equivalent to $Lu=f$, the Euler-Lagrange equation.

If $u$ solves that equation, set $v=u+w$. The linear terms cancel:with equality only when $v=u$. Hence the weak solution is the unique global minimizer. Conversely, any minimizer has zero first variation, and therefore solves the weak equation. For a symmetric bounded bilinear form $B$ on a form space $V$, exactly the same calculation gives $I(v)=B(v,v)-2\ell (v)$ and $B(u,w)=\ell (w)$ for every $w\in V$; it does not require an unbounded differential operator to map every $v\in V$ into $H$.

Existence for every $f$ requires an extra hypothesis if “positive definite” means only strict positivity. The coercive operator bound makes the form coercive, so the Lax-Milgram theorem supplies existence and uniqueness. Without that bound, the diagonal operator on sequence space $L(x_n)=(x_n/n)$ on ${\ell }^2$ is symmetric and strictly positive, but $f=(1/n)$ belongs to ${\ell }^2$ and its formal inverse $(1,1,\dots )$ does not. Thus strict positivity alone proves uniqueness and the minimizing property of a solution when one exists, not existence for all $f$.

For the diagonal operator on sequence space,Equality holds at the first coordinate vector, so $\Vert K\Vert =1$ and $K$ is a bounded linear operator. Every diagonal entry is nonzero, giving the null space $\mathcal{N}(K)=\{0\}$. A datum has a preimage exactly when its weighted coordinates belong to the l2 sequence space. HenceFinite sequences lie in the range and are dense in ${\ell }^2$, so $\mathcal{R}(K{)}^{\perp }=\{0\}$ andThe sequence $f^{(n)}=e_n/n$ converges to zero in the data norm, but $K^{†}{f}^{(n)}=e_n$ has norm one. Thus $K^{†}$ is unbounded and discontinuous. In particular the dense range is not closed; the datum $f_j=1/j$ also proves that it is a proper subset of ${\ell }^2$.

Let $e_m$ denote the sequence with a one in coordinate $m$ and zeros elsewhere.

For $T_1$, one has $\Vert e_m{\Vert }_p=1$ but $\Vert T_1{e}_m{\Vert }_p=m$ for every $p$, including $p=\infty$. Thus $T_1$ is unbounded for every $p$.

For $T_2$, the sequence $T_2{e}_m$ has entries $-(m-1)$ and $m$ in coordinates $m-1$ and $m$, respectively. Hence $\Vert T_2{e}_m{\Vert }_p\ge m$, so $T_2$ is also unbounded for every $p$.

For $T_3$, coordinatewise division givesfor every $1\le p\le \infty$. Thus $T_3$ is a bounded diagonal operator on sequence space for every $p$.

For $T_4$, write $a=(n^{-1/2}{)}_{n\ge 1}$. Then $T_{4}x=x_{1}a$, and coordinate evaluation satisfies $|x_1|\le \Vert x{\Vert }_p$. By the p-series criterion, $a\in {\ell }^p$ exactly when $p>2$, while $a\in {\ell }^{\infty }$. Therefore $T_4$ is bounded precisely forFor $p\le 2$, even $T_4{e}_1$ does not belong to ${\ell }^p$.

For $T_5$, let $q$ be the Holder conjugate exponent of $p$. Holder inequality gives, for finite $p$,and consequentlyFor $p=\infty$,Thus $T_5$ is bounded for every $1\le p\le \infty$. Altogether,

If $B={\sup }_j|d_j|<\infty$, then for every $f\in {\ell }^2$,so the diagonal operator on sequence space is bounded. Conversely, for the standard unit vector $e_j$,so boundedness of $D$ implies ${\sup }_j|d_j|\le \Vert D\Vert <\infty$.

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.