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Use $⟨f,g{⟩}_{\pi }={\sum }_x\pi (x)f(x)g(x)$. The Dirichlet form of a Markov chain isThe equality follows by expanding the square and using row sums one and the stationary distribution identity. For the reversible Markov chain, detailed balance makes $P$ a self-adjoint operator, with real orthonormal eigenbasis $f_1\equiv 1,f_2,\dots ,f_m$ and ordered eigenvalues $1={\lambda }_1>{\lambda }_2\ge \cdots \ge {\lambda }_m$.

Adding a constant changes neither the variance nor the Dirichlet form of a Markov chain. Expand the centered function as $f-\pi (f)={\sum }_{j\ge 2}{a}_j{f}_j$. ThenFor every nonconstant $f$ the quotient is at least $1-{\lambda }_2$, and equality is attained at $f=f_2$. This provesEquivalently minimize the Dirichlet form of a Markov chain over centered unit-norm functions. This is the variational characterization underlying the Poincare inequality for a reversible Markov chain, and does not require the transition matrix to have nonnegative eigenvalues. For a one-state chain there is no nonconstant test function; the displayed minimization concerns $m\ge 2$.

A self-adjoint operator $T$ satisfies $⟨Tx,y⟩=⟨x,Ty⟩$. If $Tx=\lambda x$, then $\lambda ⟨x,x⟩=⟨Tx,x⟩=⟨x,Tx⟩=\overline{\lambda }⟨x,x⟩$, so $\lambda \in \mathbb{R}$. An eigenvector exists over $\mathbb{C}$; its orthogonal complement is $T$-invariant. Induction on dimension produces an orthonormal basis of eigenvectors, proving the finite-dimensional spectral theorem.

A self-adjoint $T$ is positive definite exactly when every eigenvalue is positive, since $⟨Tx,x⟩={\sum }_i{\lambda }_i|x_i{|}^2$ in an eigenbasis. The quadratic forms of two positive-definite operators add, so their sum is positive definite.

The same eigenbasis shows that $\mathrm{spec}T\subset [a,b]$ exactly when $T-\lambda I$ is positive definite for every $\lambda <a$ and negative definite for every $\lambda >b$. Finally, if $v$ is a unit eigenvector of $\alpha +\beta$ with eigenvalue $\gamma$, the Rayleigh quotient bounds giveThus every eigenvalue of $\alpha +\beta$ lies in $[a+c,b+d]$.

First construct the Dirichlet eigensystem. By part (i) and the Lax-Milgram theorem, the inverse $S=A_D^{-1}$ exists from $L^2$ into $H_0^1$. Elliptic regularity gives $Sf\in H^2\cap H_0^1$, and the Rellich-Kondrachov compactness theorem makes $S$ a compact operator on $L^2$. Symmetry givesso $S$ is a self-adjoint operator. It is injective, and $(Sf,f)=B[Sf,Sf]>0$ for $f\ne 0$. The spectral theorem for compact Hermitian operators supplies an orthonormal basis $(w_m)$ with $S{w}_m={\mu }_m{w}_m$, ${\mu }_m>0$, and ${\mu }_m\to 0$. Put ${\lambda }_m={\mu }_m^{-1}$. ThenRepeated elliptic regularity makes each $w_m$ smooth up to the boundary.

As with part (ii), the stated characterization omits boundary compatibility. At $k=0$, its right-hand side is finite for every $u\in L^2$ by Parseval identity, whereas its left-hand side would require $u\in H_0^1$. At higher orders, the preceding polynomial example is also decisive for the Dirichlet eigensystem: on $(0,\pi )$,Although this function belongs to every ordinary $H^k$, the weighted sum diverges at $k=3$.

The correct spectral characterization of elliptic Dirichlet domains iswith $X_k$ defined in part (ii). For finite eigenfunction sums the squared norm $((u,u){)}_k$ is exactly the weighted sum: even orders follow from $L^l{w}_m={\lambda }_m^l{w}_m$, and odd orders also use $B[w_m,w_j]={\lambda }_m{\delta }_{mj}$. These sums are dense in $X_1$: if $v$ is $B$-orthogonal to every $w_m$, then $0=B[v,w_m]={\lambda }_m(v,w_m)$, so $v=0$. They are dense in $X_0$ by construction, and the isomorphisms $A_D:X_{k+2}\to X_k$ propagate density to every $X_k$. Completion in the equivalent norms from part (ii) proves both directions of the corrected equivalence. In particular, the printed characterization is correct for $k=1,2$.

Put $B=\Vert f{\Vert }_{\infty }$, $v={\mathrm{Var}}_{\mu }(f)$, and $N_n=n-m_n$. We first obtain a uniform-in-$f$ stationary bound, then compare the burned-in trajectory with a stationary one.

For a reversible Markov chain, geometric ergodicity gives an absolute L2 spectral gap of a reversible Markov chain. Thus, for some $r<1$, the Markov kernel viewed as an operator satisfiesOne can use the rate in the total-variation estimate. To see why, reversibility makes $P$ a self-adjoint operator. For bounded mean-zero $h$ supported on a set where $M$ is bounded,The spectral theorem for normal operators writes the left side as $\int t^{2k}d{\nu }_h(t)$. Any positive spectral mass where $|t|>r$ would contradict the inequality as $k\to \infty$. These test functions are dense in $L_0^2(\mu )$: truncate a function and restrict it to $\{M\le j\}$, then correct its mean using a bounded function supported on a fixed positive-measure set where $M$ is bounded. Hence the spectral restriction extends to the entire mean-zero space.

Let $(W_i)$ be the stationary chain. For $h=f-\mu f$, the stationary covariance bound for reversible Markov chains givesConsequently

The initial law of the burned-in trajectory is $P^{m_n}(x,\cdot )$. Its total variation distance from $\mu$ is at most ${\epsilon }_n=M(x)r^{m_n}$. Attaching the same conditional future evolution cannot increase that distance, so the entire length-$N_n$ path laws differ by at most ${\epsilon }_n$. For either path the sum $S$ has $|S|\le B{N}_n$. The dual bound for total variation distance gives a difference of at most $2B^2{N}_n^2{\epsilon }_n$ in the second moments, and at most $4B^2{N}_n^2{\epsilon }_n$ in the squares of the means. This burn-in total variation comparison yieldsSince $m_n=\lfloor n^{1/3}\rfloor$, $N_n{r}^{m_n}\to 0$. Dividing by $N_n$ givesThe constant depends on the chain, not on $f$.

For a bounded linear operator $T:H\to H$, the spectrum iswhile the point spectrum is the set of eigenvalues, equivalently the $\lambda$ for which $T-\lambda I$ is not injective.

A compact operator maps the unit ball to a relatively compact set. A self-adjoint operator satisfies $T=T^{\ast }$, or $⟨Tx,y⟩=⟨x,Ty⟩$ for all $x,y\in H$. A finite-rank operator has finite-dimensional image.

For an oriented smooth surface $S$, the Gauss map sends $p$ to the chosen unit normal vector $N(p)\in S^2$. Since $|N{|}^2=1$, differentiation shows that $D{N}_p(X)$ is perpendicular to $N(p)$ and hence lies in $T_{p}S$.

In a local parametrization $\varphi (u,v)$ with $n=N\circ \varphi$, differentiatinggivesThus the bilinear form $(X,Y)\mapsto D{N}_p(X)\cdot Y$ is symmetric, so $D{N}_p$ is self-adjoint. The Gaussian curvature isWriting the coefficients of the first fundamental form asand those of the second fundamental form asone obtains

At an umbilic point, the self-adjoint map $D{N}_p$ has a repeated eigenvalue, so it is a scalar map. If every point is umbilic, there is a function $\lambda$ withEquality of mixed partial derivatives givesThe two tangent vectors are linearly independent, hence ${\lambda }_u={\lambda }_v=0$. Since ${\mathbb{R}}^2$ is connected, $\lambda$ is constant.

If $\lambda =0$, then $n$ is constant andso $S$ lies in a plane. If $\lambda \ne 0$, thenThus $n-\lambda \varphi =c$ is constant, andTherefore $S$ lies in a sphere of radius $1/|\lambda |$. This proves that the surface is part of a plane or part of a sphere.

Write $Z=X-\mu$, so the covariance operator isFor $g,h\in L^2[0,1]$,so $C_X$ is self-adjoint. Moreover,so it is a positive operator.

Let $(e_j)$ be any orthonormal basis. Tonelli theorem and Parseval identity giveA positive operator with finite trace is a trace-class operator, completing the proof.

Write $H={\partial }_x^2-V(x)$. The real potential and periodic boundary condition make $H$ a self-adjoint operator on the periodic domain. Since $u_t=-iHu$,because the inner product $⟨u,Hu⟩$ is real. Thus the Schrödinger equation preserves the $L^2$ norm.

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 Schrödinger 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.

For the given weighted inner product, a self-adjoint operator $L$ obeys $L=L^{†}$, including equality of its domain with the adjoint domain. In the differential-equation calculation this requiresfor all functions in a specified common domain, with endpoint conditions that remove the boundary form. Equality of these inner products by itself describes symmetry; the domain conditions complete the self-adjoint realization.

Use the weighted inner product $⟨f,g{⟩}_w={\int }_a^{b}w(x)\overline{f(x)}g(x)dx$ with positive weight $w$. For a self-adjoint operator $L$ and nonzero eigenfunction $Ly=\lambda y$, self-adjointness gives $\lambda ⟨y,y{⟩}_w=⟨y,Ly{⟩}_w=⟨Ly,y{⟩}_w=\overline{\lambda }⟨y,y{⟩}_w$. The positive norm therefore forces $\lambda$ to be real. For two eigenfunctions of distinct eigenvalues, the same calculation gives $({\lambda }_2-{\lambda }_1)⟨y_1,y_2{⟩}_w=0$, proving their orthogonality.

The Chebyshev differential equation takes the Sturm-Liouville formIts operator is $Ly=-(1-x^2)y^{\prime \prime }+x{y}^{\prime }$, so $Lx=x$ as required. For $y=x^3+Bx$, direct substitution gives $Ly=9x^3+(B-6)x$. Comparing coefficients yieldsExplicitly, using $x=\cos \theta$,The endpoints are singular because $p$ vanishes and $w$ diverges there. The polynomial solutions use the natural condition $p{y}^{\prime }\to 0$ at each end; under $x=\cos \theta$ these are the Neumann eigenfunctions of the Chebyshev differential operator, proportional to $\cos (n\theta )$. They differ from the zero-value endpoint realization of the same differential expression.

Use a Lagrange multiplier $\mu$ for the norm constraint. For real variations $u+\delta \eta$ with $\eta (0)=\eta (\pi )=0$, integration by parts givesThus stationarity is equivalent to the Sturm-Liouville problemMultiplying by $u$ and integrating shows $F[u]=\mu G[u]=\mu$, so the constrained stationary values are the eigenvalues, and the normalized stationary functions are their eigenfunctions. Conversely the eigenvalue equation makes the constrained first variation vanish. The self-adjoint operator has weight one.

For $ϵ=0$, zero and negative eigenvalues yield no nonzero solution with both Dirichlet boundary conditions. Positive eigenvalues have solutions $u=A\sin (\sqrt{\lambda }x)$ after the left condition, and the right condition gives $\sqrt{\lambda }=n$, $n\ge 1$. Hence the lowest eigenpair iswhere the amplitude follows from ${\int }_0^{\pi }{\sin }^{2}xdx=\pi /2$.

The Rayleigh-Ritz variational principle bounds the full lowest eigenvalue by $F[Y_1]$. Its derivative contribution is one. For the linear potential, reflection about $\pi /2$ givesThus the sine trial bound for a linear potential on an interval isFor $ϵ>0$ the bound is strict, since $-Y_1^{\prime \prime }+ϵx{Y}_1=(1+ϵx)Y_1$ is not a constant multiple of $Y_1$. Also ${\lambda }_{ϵ}\ge 1$ by the Poincaré inequality and the nonnegative potential. Equality in the upper bound holds at $ϵ=0$.

The time dependence gives $u_{tt}=-{\omega }^{2}u$, and the angular derivative vanishes for these radial normal modes. The Laplacian in polar coordinates therefore yieldsMultiplication by $r$ puts the Sturm-Liouville problem in the formThe relevant self-adjoint operator is $\mathcal{A}R=-r^{-1}(r{R}^{\prime }{)}^{\prime }$ in the weighted inner product with weight $r$. A physical displacement is regular at the origin, excluding the logarithmic singular solution. In particular $R$ is bounded and $r{R}^{\prime }\to 0$ there; regular eigenfunctions also have $R^{\prime }(0)=0$. This origin condition is needed in addition to the fixed outer boundary.

Write $N[\psi ]={\int }_a^b(p{\psi }^{\prime 2}+q{\psi }^2)dx$ and $D[\psi ]={\int }_a^{b}w{\psi }^{2}dx>0$. For a real variation, with $\Lambda =N/D$,Suitable endpoint conditions are homogeneous Dirichlet boundary conditions, homogeneous Neumann boundary conditions $p{\psi }^{\prime }=0$, or one of each. They remove the variation boundary term and the product $[p\psi {\psi }^{\prime }{]}_a^b$. Stationarity for all interior variations is equivalent toConversely, multiplication of an eigenfunction equation by $\psi$ and integration by parts gives $N=\lambda D$, hence $\boxed{\Lambda [\psi ]=\lambda }$. General Robin data require their boundary-energy contributions unless these endpoint products happen to vanish.

Use an eigenbasis orthonormal with respect to the weight: ${\int }_a^{b}w{\psi }_i^{\ast }{\psi }_{j}dx={\delta }_{ij}$. For complex coefficients use the Hermitian version of the quotient, with absolute squares. Self-adjointness givesWhen $c_0\ne 0$, dividing numerator and denominator by $|c_0{|}^2$ yieldsIf $c_0=0$, these ratios are undefined, but the unscaled formula and upper-bound conclusion remain valid. The assumptions are a problem with a self-adjoint operator bounded below and a complete eigenbasis, with the trial function in its energy-form domain. The Rayleigh-Ritz method minimizes this quotient over an admissible trial family, obtaining an upper bound on ${\lambda }_0$. Higher eigenfunctions are stationary directions and need not be unrestricted extrema.

The necessary boundary identity for a self-adjoint operator isAt regular endpoints, a standard full family of separated self-adjoint conditions iswith real coefficients and neither pair $({\alpha }_a,{\beta }_a)$, $({\alpha }_b,{\beta }_b)$ zero. For example, if ${\beta }_a\ne 0$, both functions obey $p{f}^{\prime }=-{\alpha }_{a}f/{\beta }_a$ at $a$, so $p{f}^{\prime \ast }g-p{f}^{\ast }{g}^{\prime }=0$ there. The cases ${\beta }_a=0$ and the endpoint $b$ follow directly. These include Dirichlet boundary conditions, Neumann boundary conditions and Robin boundary conditions.

Separate vanishing at each endpoint is sufficient but not necessary. Coupled conditions such as $y(b)=e^{i\chi }y(a)$ and $p(b)y^{\prime }(b)=e^{i\chi }p(a)y^{\prime }(a)$, with real $\chi$, make the two endpoint forms equal and hence their difference zero. More generally the allowed endpoint data must be a maximal subspace on which the boundary form vanishes; this ensures that the adjoint has the same domain. Merely imposing extra endpoint restrictions until the boundary term vanishes proves symmetry, but can overrestrict the domain and fail to give self-adjointness.