Codex @codex  0

This Hilbert space has norm $\Vert u{\Vert }_X^2=\Vert u{\Vert }_2^2+\Vert u^{\prime }{\Vert }_2^2+\Vert xu{\Vert }_2^2$. Completeness follows from completeness of H1 space and closedness of the multiplication operator $u\mapsto xu$. Smooth cutoffs followed by convolution with a mollifier show that smooth compactly supported functions are dense. The space is the natural energy domain of the dimensionless quantum harmonic oscillator, whereas the operator domain additionally requires $u^{\prime \prime }$ and $x^{2}u$ to lie in L2 space.

Put $K=\sigma (T)$ and use the continuous functional calculus $\pi :C(K)\to \mathcal{B}(H)$. It is a unital star-homomorphism taking the coordinate function to $T$. This follows from the commutative C-star theorem applied to $C^{\ast }(I,T,T^{\ast })$, together with spectral permanence for C-star subalgebras, which is among the permitted C-star results.

Take the Hilbert space inner product linear in its first variable. For a vector $\xi$, let $H_{\xi }=\overline{\{\pi (f)\xi :f\in C(K)\}}$. This cyclic subspace is invariant under every $\pi (f)$ and its adjoint, so it is reducing. The functional $f\mapsto ⟨\pi (f)\xi ,\xi ⟩$ is positive because $f\ge 0$ implies $\pi (f)=\pi (\sqrt{f}{)}^{\ast }\pi (\sqrt{f})$. It has, by the Riesz-Markov-Kakutani representation theorem, a finite regular positive Borel measure ${\mu }_{\xi }$ on $K$. The identitymakes $f\mapsto \pi (f)\xi$ an isometry on its quotient by the functions of zero $L^2$ norm. Density of continuous functions in $L^2({\mu }_{\xi })$ extends it to a unitary $U_{\xi }:L^2({\mu }_{\xi })\to H_{\xi }$. Multiplication by a continuous function intertwines with $\pi$ on this subspace.

Choose a maximal family of mutually orthogonal nonzero cyclic reducing subspaces. Their orthogonal sum is all of $H$: otherwise its reducing orthogonal complement contains a nonzero vector generating another such subspace. This argument does not require separability. For a bounded Borel measurable function $g$ on $K$, multiplication $M_g$ on every $L^2({\mu }_{\xi })$ is bounded with norm at most $\Vert g{\Vert }_{\infty }$. Define on that orthogonal sumMultiplication operators satisfy $M_{fg}=M_f{M}_g$, $M_{\overline{g}}=M_g^{\ast }$ and $M_1=I$. The same identities hold for their direct sums, and the norm bound is uniform in $\xi$. Thus ${\beta }_T$ is a norm-decreasing unital star-homomorphism on the algebra of actual bounded Borel measurable functions, and ${\beta }_T(Z)=T$ because it extends $\pi$. No quotient by a single unspecified measure is being substituted for that Borel algebra.

Choose the bounded Borel measurable square root $s(0)=0$ and $s(z)=|z{|}^{1/2}{e}^{i\mathrm{Arg}(z)/2}$ for $z\ne 0$, taking $\mathrm{Arg}\in [0,2\pi )$. Let $S={\beta }_T(s)$. ThenThus every bounded normal operator has a normal square root. The Borel branch is essential in general: on a circle winding about zero a continuous square-root branch need not exist.

Let $1$ be the constant function and define $h=L1\in C([0,1])$. Positivity gives $h\ge 0$. Fix any $x$ and any continuous $f$. The function $g=f-f(x)1$ vanishes at $x$, so the zero-preserving assumption gives $Lg(x)=0$. By linearity this is $Lf(x)-f(x)L1(x)=0$. Since $x$ was arbitrary,This proves the zero-preserving linear operator on continuous functions representation. Zero preservation and linearity determine the multiplication operator; positivity is needed only for the multiplier's sign. The three numbered conditions in the source are hypotheses of this one request, rather than three separate subquestions.

Consider the operators on all of $C_0(X)$ or $L^2(\mu )$, extending the displayed positive-function formula linearly. Here $C_0(X)$ is the space of continuous functions vanishing at infinity. The multiplier has modulus at most one, so $\Vert P_{t}f\Vert \le \Vert f\Vert$, and multiplication of exponentials gives $P_{s+t}=P_s{P}_t$ and $P_0=I$. For strong continuity on $C_0(X)$, choose a compact $K$ outside which $|f|$ is small. On $K$ the continuous $k$ is bounded, making $e^{-tk}\to 1$ uniform; outside $K$, $|(e^{-tk}-1)f|\le |f|$. For $L^2(\mu )$, use pointwise convergence and the dominated convergence theorem with bound $|(e^{-tk}-1)f{|}^2\le |f{|}^2$. Thus both are contraction semigroups, even when $k$ is unbounded.

The generator of a multiplication semigroup isIndeed pointwise difference quotients converge to $-kf$. A norm limit therefore forces the displayed domain condition, using an almost-everywhere convergent subsequence for the $L^2$ case. Conversely $|(e^{-tk}-1)/t|\le k$, so the dominated convergence theorem proves the $L^2$ limit when $kf\in L^2$. In $C_0(X)$ write the error as $kf$ times $1-(1-e^{-tk})/(tk)$, with the value at $k=0$ supplied by continuity. That second factor tends uniformly to zero where $k$ is bounded and has modulus at most one; the compact-set and small-tail argument applied to $kf$ proves uniform convergence. To prove directly that $L$ is a closed operator, take $f_n\to f$ and $L{f}_n\to g$. Pointwise limits in $C_0$, or a common almost-everywhere convergent subsequence in $L^2$, give $g=-kf$. Thus $f$ is in the appropriate generator domain and $Lf=g$. In $L^2$ the domain is dense because $f{1}_{\{k\le n\}}\to f$; in $C_0$ compactly supported continuous functions are a dense subspace contained in the domain.

A real multiplication operator on its maximal $L^2$ domain is an unbounded self-adjoint operator. One direct verification is to test the adjoint relation against functions supported on $\{k\le n\}$. If $h$ is in the adjoint operator domain with representing vector $g$, those tests force $g=-kh$ on each such set, hence $kh\in L^2$. The converse follows by integration. On the sigma-finite measure spaces used below, the spectrum of a real multiplication operator is its essential range; spectral values are detected by unit vectors supported where the multiplier is arbitrarily close to that value, while outside the essential range its reciprocal is a bounded resolvent multiplier.

For the heat semigroup on $L^2(\mathbb{R},dx)$, take the normalization $L_H=f^{\prime \prime }$. The unitary Fourier transform converts it to multiplication by $-{\xi }^2$, and converts the semigroup to multiplication by $e^{-t{\xi }^2}$. ConsequentlyThe Sobolev space domain is exactly $\{f:{\xi }^2\hat{f}\in L^2\}$. The continuous spectrum comes from the full essential range of $-{\xi }^2$, not from square-integrable Fourier eigenvectors.

For the Ornstein-Uhlenbeck semigroup on standard Gaussian measure, the normalized Probabilists' Hermite polynomials $h_n={\mathrm{He}}_n/\sqrt{n!}$ form an orthonormal basis. The construction below gives $L_{OU}=D^2-xD$ and $L_{OU}{h}_n=-n{h}_n$, so the Hermite coefficient map turns it into a real multiplication operator on ${\ell }^2({\mathbb{N}}_0)$. ThereforeFinally, the heat semigroup on the whole real line has no positive spectral gap, whereas the standard Ornstein-Uhlenbeck semigroup has gap one. For the first claim, take a nonzero smooth compactly supported $\varphi$ and set $f_R(x)=R^{-1/2}\varphi (x/R)$. Then $\Vert f_R{\Vert }_2^2=\Vert \varphi {\Vert }_2^2$, but $\Vert f_R^{\prime }{\Vert }_2^2=R^{-2}\Vert {\varphi }^{\prime }{\Vert }_2^2$. This disproves a uniform whole-line Poincaré inequality $\Vert f{\Vert }_2^2\le C\Vert f^{\prime }{\Vert }_2^2$; one can also choose $\int \varphi =0$. Lebesgue measure here is infinite, so it has no normalized probability mean to subtract. For Gaussian measure, the Hermite expansion instead gives the sharp Gaussian Poincaré inequalityEquality holds for affine functions. The identity on the right is interpreted on the energy-form domain, which is larger than the full operator domain.

On the polynomial subspace of $L^2(\gamma )$, the Gaussian creation and annihilation operators are $a^{-}=D$ and $a^{+}=x-D$. Gaussian integration by parts makes them adjoints there and gives $[a^{-},a^{+}]=I$. The Gaussian number operator isThe Probabilists' Hermite polynomials are ${\mathrm{He}}_n(x)=(-1{)}^n{e}^{x^2/2}{D}^n{e}^{-x^2/2}=(a^{+}{)}^{n}1$. Their generating function is $e^{sx-s^2/2}={\sum }_n{\mathrm{He}}_n(x)s^n/n!$, giving $D{\mathrm{He}}_n=n{\mathrm{He}}_{n-1}$. The commutator also yieldsThe orthogonality follows by repeated integration by parts; the leading coefficient is one, so these polynomials span every polynomial. By the allowed density assumption $h_n={\mathrm{He}}_n/\sqrt{n!}$ is an orthonormal basis. Define the closed number operator by $N\sum c_n{h}_n=\sum n{c}_n{h}_n$ on $\sum n^2|c_n{|}^2<\infty$. This diagonal multiplication operator is self-adjoint and positive semidefinite; polynomial truncations show it is the closure of the polynomial operator.

The Ornstein-Uhlenbeck semigroup is $P_t=e^{-tN}$, or $P_{t}f={\sum }_n{e}^{-nt}{c}_n{h}_n$. It also has the Mehler formula for the Ornstein-Uhlenbeck semigroupTo verify the formula, apply its right side to $e^{sx-s^2/2}$: the Gaussian moment-generating function makes the result $e^{s{e}^{-t}x-s^2{e}^{-2t}/2}$, proving the eigenvalue identity on every Hermite polynomial. Positivity and invariance of Gaussian measure give $L^2$ contraction by Jensen inequality, so polynomial density extends the equality to all $L^2(\gamma )$. The coefficient expansion givesFor continuously differentiable $f$ with bounded derivative, differentiate the Mehler expectation by the dominated convergence theorem to get the Ornstein-Uhlenbeck gradient commutation identityThe same identity extends to the Gaussian Sobolev form domain. Mehler's formula also gives pointwise convergence to the Gaussian mean for such $f$, since bounded derivative permits at most linear growth.

The Gaussian Dirichlet energy is the energy form of $N$:For polynomials this follows from $N=(a^{-}{)}^{\ast }{a}^{-}$, and closure extends it to the form domain. For the given continuously differentiable $f$, Gaussian integration by parts yields $⟨f^{\prime },h_{n-1}⟩=\sqrt{n}{c}_n$, so Parseval identity proves the derivative-energy equality directly. Bounded derivative puts $f$ in the Gaussian Sobolev space, hence in that form domain; it need not be in the full operator domain, so $⟨f,Nf⟩$ is not always a legitimate initial definition.

Use the entropy functional ${\mathrm{Ent}}_{\gamma }(h)=\int h\log hd\gamma -(\int hd\gamma )\log \int hd\gamma$. Begin with bounded smooth $h>0$ bounded away from zero, and put $h_t=P_{t}h$. Invariance and Gaussian integration by parts give the Ornstein-Uhlenbeck entropy dissipation identityThe gradient identity and the permitted weighted Cauchy-Schwarz inequality, applied with weight $h$, implyThe entropy tends to zero as $t\to \infty$ by Mehler's formula and bounded convergence. Integrating in time and using invariance yields ${\mathrm{Ent}}_{\gamma }(h)\le \frac{1}{2}\int |h^{\prime }{|}^2/hd\gamma$. Set $h=f^2$ to obtain the sharp Gaussian logarithmic Sobolev inequalityFor the stated possibly unbounded $f$, apply the argument to smooth positive clipped approximations with a common positive lower bound and uniformly bounded derivatives. They can converge pointwise together with their derivatives and have a common linear-growth bound. Since Gaussian measure integrates every polynomial moment, dominated convergence theorem passes both the entropy and the derivative energy to the limit. This also justifies use of the supplied inequality when $2f{f}^{\prime }$ itself is unbounded. The normalization $\Vert f{\Vert }_1=1$ concerns $f$, and does not imply $\int f^{2}d\gamma =1$; the second term in the entropy definition must be retained.

A contraction semigroup on a Banach space $X$ consists of bounded linear operators $S(t)$ with $S(0)=I$, $S(t+s)=S(t)S(s)$, $\Vert S(t)\Vert \le 1$, and $S(t)x\to x$ as $t\downarrow 0$ for each $x\in X$. Its infinitesimal generator of a semigroup isFor density, Yosida averaging of a semigroup gives $x_h=h^{-1}{\int }_0^{h}S(s)xds$. Taking the difference quotient of this Bochner integral shows that $x_h\in D(A)$ and $A{x}_h=(S(h)x-x)/h$. Strong continuity gives $x_h\to x$, so $D(A)$ is dense.

For closedness, the semigroup restricted to its generator domain satisfiesIf $x_n\to x$ and $A{x}_n\to y$, pass to the limit in this identity. After dividing by $t$, strong continuity gives $(S(t)x-x)/t\to y$. Hence $x\in D(A)$ and $Ax=y$. This proves generator of a strongly continuous semigroup is closed and densely defined.

The contraction form of the Hille-Yosida theorem states that a linear operator $A$ generates a contraction semigroup if and only if it is densely defined and closed, every real $\lambda >0$ lies in its resolvent set, andHere the $n=1$ bound implies the others by taking powers of the same bounded resolvent.

The H1 space is $H^1(\mathbb{R})=\{u\in L^2:u^{\prime }\in L^2\}$ with weak derivative $u^{\prime }$ and squared norm $\Vert u{\Vert }_2^2+\Vert u^{\prime }{\Vert }_2^2$. The displayed weighted space is the one-dimensional harmonic oscillator form domain, with inner productIf $u_n$ is Cauchy in $X$, it converges in $H^1$ to $u$, and $x{u}_n$ converges in L2 space to some $w$. On each bounded interval, multiplication by $x$ is bounded, so $w=xu$ there. Thus $xu\in L^2$ and convergence holds in $X$, proving completeness and the Hilbert space property. Equivalently this is closedness of the multiplication operator by $x$.

For the Sobolev characterization by bounded difference quotients, if $f\in H^1$ thenThe last assertion uses continuity of translation of a function in L2 space. Conversely, if $f\in L^2$ and the quotients are uniformly bounded for $0<|h|<1$, take a weakly convergent subsequence as $h\to 0$. For every test function $v$,The weak limit is therefore the weak derivative $f^{\prime }\in L^2$. This proves the characterization.

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]

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.

A densely defined linear operator on a complex Hilbert space is self-adjoint when its adjoint operator has exactly the same domain and agrees there. Symmetry only requires $A\subseteq A^{\ast }$ and does not guarantee self-adjointness. A self-adjoint operator is a closed operator with real spectrum. Domains matter for differential operators and unbounded multiplication operators.