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For $f,g\in L^2(\pi )$, detailed balance says that the measure $\pi (dx)K(x,dy)$ is invariant under exchanging $x$ and $y$. ThereforeThis is precisely the defining identity for a self-adjoint operator.

Subtracting the expected value of $f$ does not alter either side, so suppose $\pi (f)=0$. Since reversibility makes $K$ a self-adjoint operator and stationarity makes it a contraction,The Discrete-time Poincaré inequality for a Markov kernel is therefore equivalent toApplying this inequality successively to $f,Kf,\dots ,K^{t-1}f$ yieldsConversely, the asserted variance contraction with $t=1$ rearranges to the Poincaré inequality. Hence the two statements are equivalent.

Let $\mathbb{T}=\mathbb{R}/(2\pi \mathbb{Z})$ and define the positive self-adjoint operatoron $L^2(\mathbb{T})$. On the Fourier mode $e^{inx}$ it acts by multiplication by $\sqrt{1+n^2}$. ConsequentlyThe energy space is the periodic Sobolev spacewith inner productIf $u=\sum u_n{e}^{inx}$ and $v=\sum v_n{e}^{inx}$, thenwhich is precisely the stated energy norm.

With $v=u_t$, the periodic Klein-Gordon equation becomesFor $AZ$ to belong to $H^1\times L^2$, one needs $v\in H^1$ and $u\in H^2$. Thus

After the pressure enforces incompressibility, let $L$ denote the displayed linear operator. Integration by parts givesThe pressure terms vanish by incompressibility and the boundary conditions. The expression is symmetric under $i\leftrightarrow j$, soThus $L$ is a self-adjoint operator in the energy inner product and hence a normal operator. Its orthogonal eigenmodes cannot generate non-normal transient growth.

Write $H=-{\partial }_x^2+V(x)$. The real potential and periodic boundary conditions make $H$ a self-adjoint operator. Since $u_t=-iHu$,because the expectation of a self-adjoint operator is real. Equivalently, integrating the kinetic term by parts leaves $\int |u_x{|}^{2}dx$ and the periodic boundary term cancels. Thus the continuous Schrodinger equation preserves its $L^2$ norm.

A projection-valued measure on the Borel sets of $\mathbb{C}$ is a map $E$ into the orthogonal projections on a separable Hilbert space $\mathcal{H}$ such thatand for pairwise disjoint $B_j$,for every $v\in \mathcal{H}$, with convergence in norm.

The spectral theorem for normal operators on a separable Hilbert space states that a bounded normal operator $A$ has a unique projection-valued measure supported on $\mathrm{Sp}(A)$ for whichMore generally, the Borel functional calculus for a normal operator is

For $v,w\in \mathcal{H}$, the scalar spectral measures areIf $A$ is self-adjoint, its spectrum and hence the support of $E$ lie in $\mathbb{R}$. Moreover,so ${\mu }_v$ is a positive measure, andThe paper prints total mass $\Vert v\Vert$; with the standard definition it is $\Vert v{\Vert }^2$, so the unsquared norm is a typographical error.

A self-adjoint operator $A$ is positive definite when $⟨Av,v⟩>0$ for every nonzero vector $v$ in its domain. In finite dimension it is invertible and has positive eigenvalues. In infinite dimension, strict positivity of individual quadratic-form values need not give a bounded inverse; the stronger uniform bound $⟨Av,v⟩\ge c\Vert v{\Vert }^2$ with $c>0$ supplies coercivity.

For maps $\beta :U\to V$ and $\gamma :V\to W$ between finite-dimensional inner-product spaces, with $\mathrm{im}\beta =\ker \gamma$, the operator $\beta {\beta }^{\ast }+{\gamma }^{\ast }\gamma$ is positive definite. Its quadratic form is $\Vert {\beta }^{\ast }v{\Vert }^2+\Vert \gamma v{\Vert }^2$, whose zero set is $\mathrm{im}\beta \cap (\mathrm{im}\beta {)}^{\perp }=\{0\}$. This uses the adjoint operator relation $\ker {\beta }^{\ast }=(\mathrm{im}\beta {)}^{\perp }$.

A Schrodinger operator is a differential operator of the formwhere $\Delta$ is the Laplace operator and $V$ is a potential function. Its eigenvalue problem is the Time-independent Schrodinger equation.

For the elastic filament operator $K=A{\partial }_x^4$, two integration by parts operations give boundary form $A[\overline{u}{v}^{\prime \prime \prime }-\overline{u^{\prime }}{v}^{\prime \prime }+\overline{u^{\prime \prime }}{v}^{\prime }-\overline{u^{\prime \prime \prime }}v]$. Four elementary homogeneous choices at each endpoint are $h=h^{\prime }=0$ (clamped), $h=h^{\prime \prime }=0$ (hinged), $h^{\prime \prime }=h^{\prime \prime \prime }=0$ (free), and $h^{\prime }=h^{\prime \prime \prime }=0$ (slope-constrained and transverse-force-free). Each defines a self-adjoint operator on a finite interval when imposed at both ends. More general real Robin boundary conditions also give self-adjointness; the four choices are not an exhaustive classification of all boundary subspaces.

For a bounded self-adjoint operator $L$, every bounded sequence whose images converge has a norm-convergent subsequence exactly when its kernel is finite-dimensional and its range is closed, or equivalently $0\notin {\Sigma }_{\mathrm{e}}(L)$ for the essential spectrum of a bounded self-adjoint operator. Split the sequence into kernel and kernel complement: finite dimensionality gives a subsequence on the first part and the closed-range bound on the kernel complement makes the second part a Cauchy sequence. Infinite kernel or approximate null unit vectors in its complement obstruct the property.

A singular Weyl sequence for a bounded self-adjoint operator $L$ at $\lambda$ is a spectral Weyl sequence that also converges weakly to zero. Such a sequence exists exactly when $\lambda$ is in the essential spectrum of a bounded self-adjoint operator. An infinite-dimensional shifted kernel gives a weakly null orthonormal sequence. If that kernel is finite-dimensional but the shifted range is not closed, choose approximate null vectors in the kernel complement and extract a weakly convergent subsequence; its limit belongs to both kernel and complement, so is zero.

For a finite irreducible Markov chain satisfying detailed balance with $\pi$, its transition matrix is a self-adjoint operator for $⟨f,g{⟩}_{\pi }={\sum }_x\pi (x)f(x)g(x)$. For a real orthonormal eigenbasis $(f_j)$ with eigenvalues $({\lambda }_j)$,Indeed, apply the spectral decomposition of $P^t$ to the function $1_{\{y\}}/\pi (y)$, whose coefficient against $f_j$ is $f_j(y)$. For $t=0$, every eigenvalue multiplier is interpreted as one.

For a bounded self-adjoint operator $L$, a spectral Weyl sequence at $\lambda$ consists of unit vectors $f_n$ with $\Vert (L-\lambda I)f_n\Vert \to 0$. Such a sequence exists exactly at points of the spectrum of a bounded operator. If no sequence exists, the shifted operator is bounded below, with closed range and zero kernel. Self-adjointness then makes its range dense, so it is invertible. This spectral criterion is different from the equidistribution Weyl criterion.

For a nonzero horizontal Fourier mode $k$, solve $(D^2-k^2{)}^{2}w=R{k}^2\theta$, $w=w^{\prime \prime }=0$, to obtain $w={\mathcal{W}}_R\theta$. Under homogeneous temperature Dirichlet boundary conditions, the sine modes diagonalize this self-adjoint operator, with eigenvalues $R{k}^2/(k^2+n^2{\pi }^2{)}^2$. The temperature linear operator is ${\mathcal{L}}_R=D^2-k^2+{\mathcal{W}}_R$. At a marginal mode $g=\sin (n\pi z)$ it has ${\mathcal{W}}_{R}g=(k^2+n^2{\pi }^2)g$.

On ${\ell }^2$, the bounded self-adjoint operator $(Lv{)}_j=v_j/j$ is strictly positive. Yet $f_j=1/j$ lies in ${\ell }^2$ and $Lu=f$ would require the non-square-summable vector $u_j=1$. A positive uniform lower bound, rather than strict positivity alone, guarantees a bounded inverse for a bounded self-adjoint operator.

A real matrix is symmetric when $A^T=A$. It represents a self-adjoint operator in the standard Euclidean inner product.

A densely defined operator $T$ on a Hilbert space is symmetric whenEquivalently, $T$ agrees with its adjoint operator on $D(T)$, while $D(T^{\ast })$ may be larger. A self-adjoint operator additionally requires equality of these operator domains.