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For a real positive-definite matrix $A$ in finite dimension, write $A=Q\Lambda Q^T$ by the spectral theorem for real symmetric matrices. In the coordinates in an orthonormal basis $w=Q^{T}x$, the positive level set $q(x)=k$ becomesThis is an ellipsoid, with semi-major and semi-minor axes in two dimensions forming an ellipse; each semi-axis has length $r_j=\sqrt{k/{\lambda }_j}$ and direction given by the corresponding eigenvector. In the opening example, the shorter semi-axis is $\sqrt{k}/2$ along $e_{+}$, while the longer is $\sqrt{k/2}$ along $e_{-}$. A stiffer direction, meaning a larger eigenvalue, reaches the same energy at a smaller displacement. The ratio of longest to shortest semi-axis is $\sqrt{{\lambda }_{\max }/{\lambda }_{\min }}$, the square root of the spectral condition number of a positive-definite matrix.^[ref]

Strict positivity matters because ${\lambda }_{\min }>0$ givesHence a fixed positive level set cannot extend arbitrarily far. The Cholesky decomposition $A=L{L}^T$ gives another view: $q(x)=\Vert L^{T}x{\Vert }^2$, so the ellipsoid is the inverse image of a sphere under an invertible linear map. A singular positive semidefinite matrix has a nonzero kernel; adding a kernel vector preserves its quadratic form, so any nonempty positive level set is unbounded. These statements concern finite dimension; the positive definite symmetric operator entry explains why strict positivity alone need not give a uniform lower bound on an infinite-dimensional Hilbert space.

A Mehrstellen method gains accuracy by exploiting the differential equation in the local truncation error of a compact finite difference method stencil. The extra nearby values alone do not guarantee higher order: one must also modify the source term. Consider $\Delta u=f$ on a square with prescribed Dirichlet boundary data, using the same spacing $h$ in both directions.

For comparison, the five-point Laplacian has expansionIts leading error is not a multiple of ${\Delta }^{2}u$, because the mixed fourth derivative is missing. To obtain an isotropic leading error, consider a symmetric nine-point finite-difference stencil with weight $a$ on each axial neighbor, $b$ on each diagonal neighbor, and $c$ on the center, all divided by $h^2$. Vanishing on constants and consistency of a numerical method require $4a+4b+c=0$ and $a+2b=1$. Taylor expansion then gives the fourth-derivative contributionFor this to equal $h^2{\Delta }^{2}u/12$ we must choose $b=1/6$, hence $a=2/3$, $c=-10/3$. The resulting nine-point finite-difference stencil operator isIts expansion, retaining the next term for later use, isThus $D_9$ by itself is second order on a general function. On a Poisson equation solution, however, ${\Delta }^{2}u=\Delta f$. Approximating this known source derivative by $D_{5}f$ gives the fourth-order Mehrstellen method equationEquivalently, its two stencils areThe normalized local truncation error is $O(h^4)$, since $D_{5}f-\Delta f=O(h^2)$. If the equation is multiplied by $6h^2$, the row residual is $O(h^6)$; this change of normalization does not make the solution sixth order. The fourth-order correction of the nine-point Poisson stencil retains only the closest axial and diagonal unknowns. A fourth-order one-dimensional second derivative instead uses $(-U_{i+2}+16U_{i+1}-30U_i+16U_{i-1}-U_{i-2})/(12h^2)$, widening the unknown stencil and introducing negative outer weights. Mehrstellen method accuracy is obtained by correcting known source values rather than widening that unknown stencil.

The compact scheme has useful stability and solvability properties. With homogeneous Dirichlet boundary data, $-D_9$ is symmetric, has positive diagonal and nonpositive off-diagonal entries, and is a positive definite symmetric operator. Indeed its quadratic form is a sum of positive edge weights times squared differences, with boundary values set to zero. Vanishing of this sum would force a constant on the connected grid and then zero by connection to the boundary. Consequently the Dirichlet boundary condition linear system has a unique solution. The same weights give a discrete maximum principle: if $D_{9}v\ge 0$ at every interior node and $v\le 0$ on the boundary, a positive maximum inside would make every weighted difference to its neighbors nonpositive. Equality forces the maximum to propagate to the boundary, which is impossible. Thus $v\le 0$ everywhere.

This also turns consistency of a numerical method into a quantitative convergence result, rather than merely suggesting an order. Let $B_{h}f=f+h^2{D}_{5}f/12$, let $\tau =D_{9}u-B_{h}f$ be the normalized residual of the exact solution, and set $e=U-u$. Exact Dirichlet boundary data imply $e=0$ on the boundary and $D_{9}e=-\tau$. On the unit square the quadratic barrierhas $D_{9}q=-1$ exactly, is nonnegative on the boundary, and has maximum $1/8$. If $\Vert \tau {\Vert }_{\infty }\le \epsilon$, then $D_9(e-\epsilon q)\ge 0$ and $D_9(-e-\epsilon q)\ge 0$, while both comparison functions are nonpositive on the boundary. The discrete maximum principle provesThis is maximum-norm convergence of the corrected nine-point Poisson scheme. It requires sufficient smoothness of the solution up to the boundary for the uniform truncation bound. On other domains the source and boundary closures must retain the required accuracy; boundary singularities can invalidate a smooth-solution order estimate.

The same expansion reveals how to add still more accuracy. Differentiating $\Delta u=f$ givesThus the sixth derivatives appearing in the $h^4$ term can also be expressed solely through the source. With exact source derivatives,is the sixth-order source correction of the nine-point Poisson stencil, with normalized residual $O(h^6)$. If source derivatives are replaced by differences, $\Delta f$ must be approximated through fourth order and the fourth derivatives through second order to retain that accuracy. The unknowns still occupy the same nine-point finite-difference stencil; only evaluation of known source data becomes more elaborate. The same inverse bound yields sixth-order convergence when these approximations and the boundary data have matching accuracy.

A particularly instructive case is the Laplace equation, $f=0$. Both displayed error corrections automatically vanish: ${\Delta }^{2}u=0$ and $u_{xxxxyy}+u_{xxyyyy}=\Delta u_{xxyy}=0$, which also makes the pure sixth derivatives sum to zero. Expanding the stencil two orders further gives, for a sufficiently smooth harmonic function,For verification, the three eighth-derivative contributions are $(u_{xxxxxxxx}+u_{yyyyyyyy})/20160$, $(u_{xxxxxxyy}+u_{xxyyyyyy})/2160$ and $u_{xxxxyyyy}/864$. Because $u$ is a harmonic function, their sum is $u_{xxxxxxxx}(1/10080-1/1080+1/864)=u_{xxxxxxxx}/3024$. The example $u=\mathrm{Re}(x+iy{)}^8$ has $D_{9}u(0,0)=40h^6/3$, so this term is genuinely present. The unmodified harmonic nine-point scheme therefore has sixth-order nodal convergence under the same smoothness and boundary assumptions. This harmonic superconvergence of the nine-point stencil is special to harmonic functions; it should not be confused with the fourth-order source-corrected scheme for general Poisson equation data.

The appeal of Mehrstellenverfahren is this combination of a compact sparse matrix, favorable discrete maximum principle properties, and high accuracy obtained by using the governing equation. The derivation also shows precisely which accuracy belongs to the operator on arbitrary functions, which belongs to its action on solutions, and which survives in the computed boundary-value solution.

In the variational convention needed here, a positive definite symmetric operator on a real Hilbert space satisfiesFor a bounded operator on the whole space, symmetry is self-adjointness. On a complex Hilbert space, the corresponding convention is Hermitian positivity; the quadratic expression is real. Strict positivity should be distinguished from uniformly positive definite symmetric operator coercivity, which requires a fixed $\gamma >0$ such that $⟨Lv,v⟩\ge \gamma \Vert v{\Vert }^2$.

Symmetry matters in a real space. If “positive definite” means only the inequality for the quadratic form, the next variational assertion is false: $L=\left(\begin{array}{cc}1& -1\\ 1& 1\end{array}\right)$ has $⟨Lv,v⟩=\Vert v{\Vert }^2>0$, but its skew part disappears from that form. Its Euler-Lagrange equation is $v=f$, not $Lv=f$. Thus the usual symmetric/Hermitian convention, or an explicit symmetry assumption, is required.

For a real Hilbert space, differentiate along an arbitrary direction $w$:The last equality uses the symmetry in the definition of a positive definite symmetric operator. Vanishing of this first variation for every $w$ is equivalent to $Lv=f$, givingIf $u$ solves it, expand the functional about $u$ to findThis is strictly positive for nonzero $w$, so $u$ is the unique minimizer. Conversely a minimizer has zero first variation and solves the equation. On a complex Hilbert space, use the real functional $⟨Lv,v⟩-2\mathrm{Re}⟨f,v⟩$ and real directional derivatives; varying $w$ and $iw$ yields the same result.

This proves the quadratic variational principle for a symmetric positive operator, conditional on existence. Strict positivity in infinite dimension does not by itself guarantee a minimizer for every $f$: on ${\ell }^2$, the operator $(Lv{)}_j=v_j/j$ is strictly positive, but $f_j=1/j$ would require the non-square-summable solution $v_j=1$. A coercive bilinear form supplies existence through the Lax-Milgram theorem under its usual boundedness assumptions.

For the quadratic variational principle for a symmetric positive operator below, use a symmetric operator $L$ on the real Hilbert space $H$. Strict positive definiteness meansA uniformly positive definite symmetric operator satisfies the stronger coercive operator conditionIn this variational setting, “positive definite” is often used for a symmetric operator with this uniform bound. We will state explicitly where the coercive operator bound is needed. For a bounded linear operator defined on all of $H$, symmetry means that the operator is self-adjoint. For an unbounded operator, positivity is imposed on its operator domain, and the variational formulation is made on its form domain.

Symmetry is essential in a real Hilbert space: positivity of the quadratic expression alone does not imply symmetry. For example, $L=I+K$ with nonzero real skew-symmetric matrix $K$ satisfies $⟨Lv,v⟩=\Vert v{\Vert }^2$, but its quadratic functional has derivative involving $I$, not $I+K$. A positive definite symmetric operator supplies both the positivity and symmetry needed in part (b).

There is a genuine conflict in the printed coefficient assumptions: no uniformly positive coefficient in a zero-boundary Sobolev space exists. A function $a\in H_0^1(\Omega )$ cannot also obey $a\ge a_{-}>0$ almost everywhere. Indeed, the Lipschitz truncation $g(s)=\min (\max (s,0),a_{-})$ has $g(0)=0$, so Lipschitz truncation preserves zero-boundary Sobolev spaces, giving $g(a)\in H_0^1(\Omega )$. But $g(a)$ would be the nonzero constant $a_{-}$. Its zero gradient contradicts the Poincaré inequality in the zero-boundary Sobolev space. Thus the literal coefficient class is empty.

For the meaningful uniformly elliptic problem, take $a\in L^{\infty }(\Omega )$ with $0<a_{-}\le a\le a_{+}$, and impose the Dirichlet boundary condition on the unknown and test functions: $V=H_0^1((0,1{)}^2)$. Additional $H^1$ regularity of $a$ is harmless, but a zero trace for $a$ must be removed. The divergence-form elliptic operator is $L=-\mathrm{div}(a\nabla )$. Integration by parts defines the symmetric bounded bilinear formIn particular,Here $2{\pi }^2$ is the first Dirichlet Laplacian eigenvalue on the unit square; the Poincaré inequality follows, for example, by applying the one-dimensional inequality in each coordinate and adding. Thus $B$ is coercive in the gradient norm on $V$, and the Dirichlet realization of $L$ is a positive definite symmetric operator. On its operator domain, $⟨Lv,v⟩=B(v,v)$.

The required functional and weak equation areBecause $f\in L^2$, the right side is a bounded linear functional by the Cauchy-Schwarz inequality and Poincaré inequality. The Lax-Milgram theorem gives a unique weak solution, and part (b) proves that it uniquely minimizes $I$. These formulas prove the intended conclusion under the repaired coefficient hypothesis; under the literal hypothesis there is no coefficient to which the conclusion can be applied.

A bounded Hermitian operator $T$ on a Hilbert space is positive, also called a positive semidefinite operator, whenThe value must be real as well as nonnegative. This permits zero values on nonzero vectors: the zero linear operator is positive. For a bounded Hermitian operator, positivity is equivalent to its spectrum being contained in $[0,\infty )$; in finite dimension, this is the condition that all eigenvalues are nonnegative.^[ref]

The Hermitian operator diagram separates three cases. $D=\mathrm{diag}(1,-1)$ is self-adjoint but not positive, because $q_D((0,1{)}^{\mathrm{T}})=-1$. The orthogonal projection $P=\mathrm{diag}(1,0)$ is positive semidefinite, with $q_P(x)=|x_1{|}^2$, but is not strictly positive on every nonzero vector. The complex Hermitian matrix $H=\left(\begin{array}{cc}2& i\\ -i& 2\end{array}\right)$ is positive definite because $q_H(a{e}_1+b{e}_3)=|a{|}^2+3|b{|}^2>0$ whenever $(a,b)\ne (0,0)$. For real symmetric matrices, MIT's lecture 27 relates these tests to the geometry of quadratic forms and their minima.^[ref]

A fundamental source of positive operators is $T^{\ast }T$ for a bounded linear operator $T:H\to K$:For the complex matrix $T=\left(\begin{array}{cc}1& i\\ 2& 1-i\end{array}\right)$ used above,Its eigenvalues are $4-\sqrt{6}$ and $4+\sqrt{6}$, both positive. Positivity organizes the squared error minimized by a least-squares solution: the same $T^{\ast }T$ occurs in the normal equation for a linear inverse problem. The positive square root of an operator gives the converse squared-length description, $q_T(x)=\Vert T^{1/2}x{\Vert }^2$ for every bounded positive operator $T$.

On a real Hilbert space, nonnegativity of $⟨Tx,x⟩$ alone does not imply self-adjointness, so the Hermitian operator hypothesis is part of the definition. On a complex Hilbert space, real-valuedness of all these expressions already forces self-adjointness by the polarization identity.

Strict positivity and a uniform positive lower bound also differ in infinite dimension. On the l2 sequence space, the bounded Hermitian operator $(Tx{)}_n=x_n/n$ satisfiesConsequently no $\gamma >0$ satisfies $q_T(x)\ge \gamma \Vert x{\Vert }^2$ for every $x$. Compare the positive definite symmetric operator and uniformly positive definite symmetric operator conditions.

A positive definite symmetric operator is uniformly positive definite if $⟨Lv,v⟩\ge \gamma \Vert v{\Vert }^2$ for some $\gamma >0$. For a bounded linear operator defined on a whole Hilbert space, this is a symmetric coercive operator, and the Lax-Milgram theorem supplies a unique solution to $Lu=f$ for every $f$. For a differential operator, the corresponding bounded coercive form is used on its form space.