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The real Lax-Milgram theorem applies to a Hilbert space $H$ and a bounded bilinear form $a$ satisfyingFor every bounded linear functional $L$ there is a unique $u\in H$ withSymmetry of the bilinear form is not required.

By the Riesz representation theorem, write $a(u,v)=⟨Au,v⟩$ and $L(v)=⟨g,v⟩$. The operator $A$ is linear and bounded, with $\Vert A\Vert \le M$. The coercive bilinear form bound and the Cauchy-Schwarz inequality implyHence $A$ is injective. Its range is closed: if $A{u}_n$ converges, this last inequality applied to differences makes $u_n$ a Cauchy sequence, and its limit maps to the proposed range limit. If $w$ is in the orthogonal complement of the range, then $a(u,w)=0$ for all $u$; taking $u=w$ and using coercivity gives $w=0$. The range is thus dense as well as closed, so it is all of $H$. Solve $Au=g$ uniquely; the displayed lower bound gives the asserted estimate.

For complex Hilbert spaces the same proof works for a bounded sesquilinear form, linear in the first argument, with $\mathrm{Re}a(v,v)\ge \alpha \Vert v{\Vert }^2$. In the convention $a(u,v)=L(v)$, $L$ must then be a bounded conjugate-linear functional represented as $⟨g,v⟩$.

The finite element method replaces an infinite-dimensional variational problem by a finite-dimensional one, usually using functions that are polynomial on small mesh elements. Ritz method and Galerkin method describe how the discrete equations are selected; neither intrinsically requires piecewise polynomials, though those spaces make assembly local and sparse.

For a concrete realization of the two-point problem, choose homogeneous Dirichlet boundary conditions at both endpoints of $(0,1)$. The differential expression alone is not a complete boundary value problem, so this boundary choice is an explicit assumption. Let $p\in W^{1,\infty }$ with $p\ge p_0>0$, $q\in L^{\infty }$ with $q\ge 0$, and $f\in L^2$. Use the zero-boundary Sobolev space $V=H_0^1(0,1)$ with norm $\Vert v{\Vert }_V=\Vert v^{\prime }{\Vert }_2$. By the Poincaré inequality this is equivalent to its full first-order Sobolev norm. Multiplying by a test function and applying integration by parts givesThe endpoint terms vanish by the essential boundary condition. The form is a symmetric bilinear form, bounded byand it is a coercive bilinear form:The load is a bounded linear functional. The Lax-Milgram theorem gives a unique weak solution and a bound $\Vert u{\Vert }_V\le \Vert \ell {\Vert }_{V^{\ast }}/p_0$. Under the stated coefficient regularity it solves the differential equation in the usual weak sense.

The Ritz method minimizes the energyover $V$, or over a conforming finite-dimensional subspace $V_h\subset V$. Its first variation is $a(v,w)-\ell (w)$; symmetry and coercivity make the functional strictly convex. In fact, if $u$ is the weak solution,Thus the energy minimizer is exactly the weak solution. The Galerkin method instead directly asks for $u_h\in V_h$ with $a(u_h,v_h)=\ell (v_h)$ for all $v_h\in V_h$. For this symmetric coercive problem, Ritz-Galerkin equivalence for a symmetric coercive form shows that the discrete minimizer and Galerkin solution coincide. For a nonsymmetric problem, a Galerkin formulation can still apply but the same quadratic minimization generally cannot represent its full bilinear form; appropriate coercivity or inf-sup conditions must then justify the chosen spaces.

Subtract the continuous and discrete weak equations to obtain Galerkin orthogonality $a(u-u_h,v_h)=0$. Since $a$ defines the energy norm $\Vert v{\Vert }_a=\sqrt{a(v,v)}$, the Pythagorean identity givesIn a general reference norm, the Céa lemma yieldsThe proof uses $a(e,e)=a(e,u-v_h)$ together with coercivity and continuity. Dense approximation spaces therefore imply convergence; the estimate reduces a numerical error problem to an approximation problem.

For implementation, partition the interval by nodes $0=x_0<\cdots <x_N=1$. Choose continuous piecewise-linear functions and the interior piecewise-linear hat functions ${\varphi }_j$ as a basis, with the two boundary coefficients prescribed. Write $u_h={\sum }_j{U}_j{\varphi }_j$. The coefficients satisfyThe stiffness matrix is symmetric positive definite: $U^{T}KU=a(u_h,u_h)>0$ for nonzero interior coefficients. The local supports make it tridiagonal in this one-dimensional linear-element case.

On an element of length $h_e$, for constant element coefficients $p_e,q_e$, the two-node matrix and constant-load vector areFor variable coefficients, integrate $p,q,f$ against the same shape functions rather than silently replacing them by constants. Element contributions are added at shared nodes, and boundary values are eliminated or lifted into the right side. A sparse direct solve or a suitable iterative method gives the nodal coefficients. Numerical quadrature should preserve the required accuracy and, with positive coefficients and weights, the positive energy structure.

A finite element interpolation estimate gives ${\inf }_{v_h}\Vert u-v_h{\Vert }_{H^1}\le Ch\Vert u{\Vert }_{H^2}$ for these elements, where $h$ is the largest element size and the solution has the stated regularity. The Céa lemma then gives first-order energy error. If the dual elliptic problem has $H^2$ regularity, the Aubin–Nitsche duality argument gives second-order $L^2$ error: for $e=u-u_h$, solve $a(v,z)=(e,v{)}_{L^2}$, then use $a(e,z)=a(e,z-I_{h}z)$ to gain one further factor of $h$. Higher-order elements improve rates when the solution is sufficiently smooth.

Nonzero Dirichlet data are handled by a boundary lifting and an affine trial space. Neumann boundary conditions enter naturally through the integrated boundary term; Robin terms modify both the form and the load. With pure Neumann conditions and $q=0$, constants lie in the kernel: existence needs the load/flux compatibility condition, uniqueness needs a mean constraint, and the preceding Dirichlet coercivity argument cannot simply be reused. Conforming approximation, boundary conditions and coercivity are the ingredients connecting the finite-element construction to a justified Ritz or Galerkin solution.

Use the real Hilbert space $V=H_0^1(0,1)$, the zero-boundary Sobolev space, with norm $\Vert v{\Vert }_V=\Vert v^{\prime }{\Vert }_{L^2}$. The Poincaré inequality makes this equivalent to its usual Sobolev space norm. Define the symmetric bilinear form and continuous linear functionalThe Cauchy-Schwarz inequality and the sharp interval Poincaré inequality implyThus $a$ is a bounded bilinear form and a coercive bilinear form, and $\ell$ is bounded because $f$ is square-integrable. The Lax-Milgram theorem supplies a unique weak solution:This is obtained from the differential equation by integration by parts; the test functions have zero boundary trace. Conversely, the weak solution satisfies $y^{\prime \prime }=xy-f$ in the sense of distributions. Since $xy-f\in L^2$, it belongs to $H^2(0,1)\cap H_0^1(0,1)$ and solves the equation almost everywhere.

Equivalently, the Ritz method minimizesover $V$. Indeed $J(y+w)-J(y)=a(w,w)/2$, since the mixed term is $a(y,w)-\ell (w)=0$. This proves existence and uniqueness of the minimizer directly from the weak solution, as well as its equivalence to the variational problem.

The Lax-Milgram theorem assumes that $H$ is a real Hilbert space and $B:H\times H\to \mathbb{R}$ is a bounded bilinear form and a coercive bilinear form: there are $M,\alpha >0$ such thatThen for every continuous linear functional $\ell \in H^{\prime }$ there is a unique $u\in H$ satisfyingSymmetry of $B$ is not required.

For the proof, the Riesz representation theorem provides a bounded linear operator $T:H\to H$ and $g\in H$ with $B[u,v]=(Tu,v{)}_H$ and $\ell (v)=(g,v{)}_H$. Coercivity and the Cauchy-Schwarz inequality implyThus $T$ is injective and its range is closed: if $T{u}_j$ converges, the displayed inequality makes $u_j$ a Cauchy sequence; completeness gives $u_j\to u$ and $T{u}_j\to Tu$. If $w$ lies in the orthogonal complement of the range, then $B[v,w]=0$ for all $v$. Taking $v=w$ gives $\alpha \Vert w{\Vert }^2\le B[w,w]=0$, so $w=0$. A closed subspace with zero orthogonal complement is all of $H$, hence $T$ is surjective. The unique solution is $T^{-1}g$, and the same inequality gives the asserted norm bound.

Work in the real zero-boundary Sobolev space $X=H_0^1(0,1)$, whose members have square-integrable weak derivatives and zero endpoint trace. Define the bounded bilinear form and linear functionalThe weak formulation is to find $u\in X$ with $a(u,v)=\ell (v)$ for every $v\in X$. Integration by parts recovers the differential equation for a smooth solution; conversely, this identity is the definition of its weak solution.

Use the full $H^1$ norm. The Cauchy-Schwarz inequality gives $|a(u,v)|\le \Vert u{\Vert }_{H^1}\Vert v{\Vert }_{H^1}$ and $|\ell (v)|\le \Vert v{\Vert }_{H^1}/\sqrt{3}$. Also $a(v,v)=\Vert v{\Vert }_{H^1}^2$, so $a$ is a coercive bilinear form with constant one. The Lax-Milgram theorem states that a bounded, coercive bilinear form on a real Hilbert space determines a unique weak solution for every bounded linear functional. It therefore applies here.

The equivalent minimization problem isIf $u$ is the weak solution, symmetry and $a(u,v-u)=\ell (v-u)$ giveThis is positive unless $v=u$, proving a unique global minimizer. Conversely, differentiating $\mathcal{E}(u+tv)$ at $t=0$ gives the weak identity. Thus the unique energy minimizer is exactly the unique weak solution. As a check, the classical representative isIt satisfies both endpoint conditions and $-u^{\prime \prime }+u=x$.

The starting point of a conforming finite element method is a weak formulation in a Hilbert space $X$. For example, a homogeneous Dirichlet boundary condition on a bounded Lipschitz domain gives $X=H_0^1(\Omega )$. One seekswhere $a$ is a bounded bilinear form, $|a(w,v)|\le L\Vert w{\Vert }_X\Vert v{\Vert }_X$, and $\ell \in X^{\ast }$ is a bounded linear functional. A conforming finite element space $X_h\subset X$ is finite dimensional and is typically built from functions that are polynomial on each element of a finite element mesh. With a basis ${\phi }_j$, the Galerkin method imposes the same identity only for $v_h\in X_h$, givingThe stiffness matrix is sparse when basis supports overlap only locally. Local element integrals assemble its entries; the mass matrix represents an $L^2$ term. This separates the choice of trial functions from the variational principle determining their coefficients.

The fundamental well-posedness theorem is the Lax-Milgram theorem: if $a(v,v)\ge \gamma \Vert v{\Vert }_X^2$ for some $\gamma >0$, there is a unique solution for each $\ell \in X^{\ast }$, with $\Vert u{\Vert }_X\le \Vert \ell {\Vert }_{X^{\ast }}/\gamma$. The same theorem on $X_h$ gives a unique discrete solution. The constants must be independent of the mesh for uniform stability of a numerical method. Boundary conditions are built into the trial space when they are essential constraints; flux conditions arise through boundary terms in integration by parts and are natural constraints in the weak formulation.

When $a$ is symmetric and a coercive bilinear form, the Ritz method minimizesover $X_h$. Its first variation is exactly the Galerkin method, so Ritz minimization and Galerkin testing coincide for symmetric coercive problems. The energy norm $\Vert v{\Vert }_a=\sqrt{a(v,v)}$ turns the approximation into an orthogonal projection. Indeed, subtracting the exact and discrete weak equations gives Galerkin orthogonality, $a(u-u_h,v_h)=0$. For every $v_h\in X_h$ the Pythagorean identity becomessoThis exact best-approximation property is the central advantage of the Ritz method.

Symmetry is unnecessary for the Galerkin method. For any bounded, coercive form, the Céa lemma givesTo prove it, put $e=u-u_h$. Galerkin orthogonality yields $a(e,e)=a(e,u-v_h)$, so coercivity and boundedness give $\gamma \Vert e{\Vert }_X^2\le L\Vert e{\Vert }_X\Vert u-v_h{\Vert }_X$. Divide when $e\ne 0$ and take the infimum. Thus approximation capability plus uniform coercivity gives convergence. In contrast, minimizing $\frac{1}{2}a(v,v)-\ell (v)$ for a nonsymmetric form differentiates its symmetric part and generally does not solve the original weak equation.

A concrete symmetric example is the one-dimensional reaction-diffusion problem from Question 5. Continuous piecewise-linear hat functions give the diffusion stiffness matrix $S$ with diagonal $2/h$ and neighbouring entries $-1/h$, and the mass matrix $M$ with diagonal $2h/3$ and neighbouring entries $h/6$. Therefore $K=S+M$ and $F_i=i{h}^2$. Its positive-definite matrix is the coordinate form of the continuous energy. The exact smooth solution $u(x)=x-\sinh (x)/\sinh 1$ makes this an explicit test of the method, not just an abstract existence result.

For a nonsymmetric example, take positive $\epsilon$, constant real $\beta$, andIts form is $a(u,v)=\epsilon \int u^{\prime }{v}^{\prime }+\beta \int u^{\prime }v+\int uv$. Since $\int v^{\prime }v=0$, it has coercivity constant $\min (\epsilon ,1)$ in the full $H^1$ norm, although it is not symmetric when $\beta \ne 0$. The Galerkin method remains well posed. On the uniform hat basis, $K=\epsilon S+\beta C+M$, where $C_{i,i+1}=1/2$, $C_{i,i-1}=-1/2$ and $C_{ii}=0$. The skew-symmetric matrix $C$ cancels from $c^{T}Kc$, proving nonsingularity. A Ritz energy would omit precisely this advection contribution. Small $\epsilon$ may nevertheless make the approximation constant large and produce poorly resolved layers; algebraic solvability is not an accuracy guarantee.

Quantitative convergence uses a finite element interpolation estimate. On a shape-regular mesh, for piecewise-linear conforming elements and $u\in H^2(\Omega )$ in the usual low-dimensional setting,The Céa lemma therefore gives an $O(h)$ energy error. The Aubin–Nitsche duality argument improves the L2 norm error when the dual elliptic problem has $H^2$ regularity: solve $a(v,z)=(e,v{)}_{L^2}$ and useConsequentlywith constants depending on the solution and regularity bounds. The second estimate is conditional on dual regularity; corners or insufficient data regularity can reduce the rate.

The same Galerkin method handles evolution equations. For the homogeneous heat equation, the semidiscrete finite element heat equation is $M\dot{c}+Sc=0$. Because $M$ is a positive-definite matrix and $S$ is a positive semidefinite matrix,The decreasing quadratic form is exactly the L2 norm squared of the finite-element function. This separates spatial stability of a numerical method from the subsequent time integrator, just as in the Schrödinger equation example. Conformity, coercivity and approximation estimates together explain convergence.

Consider a bounded Lipschitz domain $\Omega$ and homogeneous Dirichlet boundary conditions forAssume $A$ is a bounded real symmetric matrix field with ${\xi }^{T}A(x)\xi \ge a_0|\xi {|}^2$ uniformly, $c\ge 0$ is bounded, and $f$ defines a bounded linear functional on $V=H_0^1(\Omega )$, the zero-boundary Sobolev space. The weak formulation isThe Cauchy-Schwarz inequality gives continuity $|a(u,v)|\le M\Vert u{\Vert }_V\Vert v{\Vert }_V$. Uniform ellipticity and the Poincaré inequality give a coercive bilinear form, $a(v,v)\ge \alpha \Vert v{\Vert }_V^2$. The Lax-Milgram theorem therefore supplies a unique weak solution, with $\Vert u{\Vert }_V\le \Vert \ell {\Vert }_{V^{\prime }}/\alpha$. Its proof represents $a$ by a bounded linear operator $T$ using the Riesz representation theorem; coercivity gives an injective operator with closed image, and a zero orthogonal complement makes that image all of $V$. This yields existence, uniqueness and the displayed estimate.

For a conforming finite element space $V_h\subseteq V$, the Galerkin method finds $u_h\in V_h$ such that $a(u_h,v_h)=\ell (v_h)$ for every $v_h\in V_h$. With a basis ${\varphi }_1,\dots ,{\varphi }_N$, this is the linear system $KU=F$, where the stiffness matrix is $K_{ij}=a({\varphi }_j,{\varphi }_i)$ and $F_i=\ell ({\varphi }_i)$. Its positive-definite matrix property proves unique discrete solvability. Local support gives a sparse matrix through elementwise assembly.

For the symmetric problem, the Ritz method minimizes $J(v)=a(v,v)/2-\ell (v)$ over $V_h$. Differentiating in every direction $v_h$ gives precisely the Galerkin method equations. Conversely, if $u_h$ solves them, $J(u_h+w_h)-J(u_h)=a(w_h,w_h)/2$, proving it is the unique minimum. The same argument over $V$ gives $J(v)-J(u)=\Vert v-u{\Vert }_a^2/2$, where $\Vert v{\Vert }_a=\sqrt{a(v,v)}$ is the energy norm. Thus the Ritz method chooses the best trial function in that norm.

Subtracting the continuous and discrete equations gives Galerkin orthogonality, $a(u-u_h,v_h)=0$. For arbitrary $v_h\in V_h$ and $e=u-u_h$, it follows that $a(e,e)=a(e,u-v_h)$. Coercivity and continuity prove the Céa lemma:In the symmetric energy norm, the Cauchy-Schwarz inequality improves the constant to one. Equivalently, Galerkin orthogonality gives the Pythagorean identity $\Vert u-v_h{\Vert }_a^2=\Vert u-u_h{\Vert }_a^2+\Vert u_h-v_h{\Vert }_a^2$. These estimates establish stability and show that approximation properties of the trial space determine convergence.

For continuous piecewise-linear elements on a shape-regular mesh in dimensions at most three, a finite element interpolation estimate gives $\Vert u-I_{h}u{\Vert }_{H^1}\le Ch\Vert u{\Vert }_{H^2}$ when $u\in H^2(\Omega )$. Combining it with the Céa lemma yields $O(h)$ energy norm error. More generally, degree-$p$ elements give $O(h^p)$ error in $H^1$ under the corresponding regularity and approximation assumptions. Mere mesh refinement cannot supply an order whose required solution regularity is absent.

An additional order in the L2 norm follows under a dual elliptic regularity assumption. Solve $a(v,z)=(e,v{)}_{L^2}$ and assume $\Vert z{\Vert }_{H^2}\le C\Vert e{\Vert }_{L^2}$. The Aubin–Nitsche duality argument uses Galerkin orthogonality to obtainTherefore $\Vert e{\Vert }_{L^2}\le Ch\Vert e{\Vert }_{H^1}$, giving $O(h^2)$ for the piecewise-linear case with $u\in H^2$. This improvement explicitly requires regularity of the dual problem.

As a concrete example, solve $-u^{\prime \prime }=1$ on $(0,1)$ with zero endpoint values. On a uniform mesh of spacing $h$, the piecewise-linear hat functions give $K_{jj}=2/h$, $K_{j,j\pm 1}=-1/h$ and $F_j=h$. The discrete equations are $2U_j-U_{j-1}-U_{j+1}=h^2$, whose solution is $U_j=x_j(1-x_j)/2$. The exact solution is $u(x)=x(1-x)/2$, so $u_h$ is its piecewise-linear nodal interpolant. On each interval $[x_j,x_{j+1}]$,Integrating the squared error and squared derivative error gives the explicit rateswhere the $H_0^1$ norm here is $\Vert v^{\prime }{\Vert }_{L^2}$.

For nonsymmetric coercive problems, the Galerkin method, Galerkin orthogonality and the Céa lemma still apply, but the symmetric quadratic minimization interpretation of the Ritz method generally does not. Nonconforming trial spaces, inexact integration and noncoercive equations need additional arguments beyond the conforming coercive theory proved here.

By part (a), $B$ is a bounded bilinear form: $|B(u,v)|\le M\Vert u\Vert \Vert v\Vert$. For each fixed $v$, the map $u\mapsto B(u,v)$ is consequently a bounded linear functional. The Riesz representation theorem supplies a unique vector $Lv$ such thatUniqueness of the representing vector makes $L$ linear, and $\Vert Lv\Vert \le M\Vert v\Vert$ makes it a bounded linear operator.

The coercivity assumption implieshence $\Vert Lv\Vert \ge C\Vert v\Vert$. Moreover, using the adjoint operator,so the same argument gives $\Vert L^{\ast }v\Vert \ge C\Vert v\Vert$. The invertibility from lower bounds on an operator and its adjoint therefore shows that $L$ is invertible and $L^{-1}$ is bounded.

Set $v_f=L^{-1}f$. Then $B(u,v_f)=⟨u,L{v}_f⟩=⟨u,f⟩$ for every $u$. If another $v$ has this property, then $L(v-v_f)=0$, and the injectivity of $L$ gives $v=v_f$. Thus the unique vector is $v_f=L^{-1}f$.

Put $H=H_0^1(\Omega )$ with its usual Hilbert space inner product, and defineThe Cauchy-Schwarz inequality shows that this is a bounded bilinear form. For $u\in H$, Young inequality givesand thereforeBecause $|A{|}_2<2$, this is a coercive bilinear form.

The map $u\mapsto {\int }_{\Omega }uf$ is a bounded linear functional on $H$. By the Riesz representation theorem, it equals $⟨u,F{⟩}_H$ for some $F\in H$. Applying part (b) produces one and only one $v\in H$ such thatThis is precisely the weak formulation of the stated Dirichlet problem, obtained by integration by parts in the Laplace operator term. Thus $v$ is its unique weak solution; this is the constant-drift massive-Laplacian Dirichlet problem.

Define on $H$ the bilinear formand the linear functionalThe Hardy inequality on an interval, Cauchy-Schwarz inequality, and the one-sided Poincaré inequality show that $a$ is a bounded bilinear form and that

For $v\in H$, $v(x)={\int }_0^x{v}^{\prime }(t)dt$, so Cauchy--Schwarz and Fubini's theorem giveHenceThus $a$ is a coercive bilinear form. The Lax-Milgram theorem supplies a unique $u\in H$ satisfying $a(u,v)=\ell (v)$ for every $v\in H$. This is precisely the unique weak solution described by the weak boundary value problem with an inverse-square potential.

Testing the weak formulation with the constant function $1$ proves the necessary compatibility conditionAssume first that $U$ is connected and this condition holds. On the mean-zero Sobolev spaceuse the norm $\Vert v{\Vert }_{†}=\Vert Dv{\Vert }_{L^2(U)}$. To prove the needed Poincare-Wirtinger inequality, suppose it failed. There would be $v_k\in H_{†}^1(U)$ with $\Vert v_k{\Vert }_2=1$ and $\Vert D{v}_k{\Vert }_2\to 0$. The Rellich-Kondrachov compactness theorem gives a subsequence converging strongly in $L^2$ and weakly in $H^1$ to a function $v$. Its weak derivative vanishes, so connectedness makes $v$ constant; its zero mean makes it zero. This contradicts $\Vert v{\Vert }_2=1$. Hence $\Vert Dv{\Vert }_2$ is an equivalent Hilbert space norm on $H_{†}^1(U)$.

DefineBoundedness of $a^{ij}$ makes $B$ a bounded bilinear form, while uniform ellipticity givesso it is a coercive bilinear form. The Cauchy-Schwarz inequality and the Poincare-Wirtinger inequality make $\ell$ a bounded linear functional. The Lax-Milgram theorem supplies a unique $u\in H_{†}^1(U)$ satisfying $B(u,v)=\ell (v)$ for every mean-zero $v$. For arbitrary $v\in H^1(U)$, subtract its mean; the omitted constant contributes zero on both sides because $D1=0$ and ${\int }_{U}f=0$. Thus $u$ solves the original problem.

If two solutions exist, their difference $w$ satisfies $B(w,w)=0$, so uniform ellipticity gives $Dw=0$. It is therefore constant on $U$. The solution is unique up to an additive constant, and its mean-zero representative is unique. If $U$ is disconnected, the precise condition is ${\int }_{U_j}f=0$ on every connected component $U_j$, and one independent additive constant remains on each component.

On the Hilbert space $V$, the form $B$ from part b obeysso it is a bounded bilinear form and a coercive bilinear form. The Cauchy-Schwarz inequality and the Poincare inequality with a partial Dirichlet boundary giveso $\ell$ is a bounded linear functional. The Lax-Milgram theorem now gives a unique weak solution $u\in V$. Taking $v=u$ in the weak identity yieldsand therefore

For a symmetric bounded bilinear form $B$ that is strictly positive on nonzero vectors, and a bounded linear functional $\ell$, set $I(v)=B(v,v)-2\ell (v)$. The first variation is $DI(v)[w]=2(B(v,w)-\ell (w))$. A weak solution $u$ of $B(u,w)=\ell (w)$ satisfies $I(u+w)-I(u)=B(w,w)$, so it is the unique minimizer. Conversely, any minimizer solves the weak equation. A coercive bilinear form gives existence by the Lax-Milgram theorem; strict positivity alone does not.

For a bounded symmetric coercive bilinear form on a Hilbert space, a solution of $a(u,w)=\ell (w)$ minimizes $J(v)=a(v,v)/2-\ell (v)$. The displayed identity proves strict minimality and uniqueness. Restricting to a conforming finite element space makes the discrete Ritz method equivalent to the Galerkin method. Galerkin orthogonality gives exact best approximation in the energy norm; the Céa lemma translates it into a reference-norm estimate.

The stiffness matrix represents a variational bilinear form $a$ in a finite element basis. A symmetric coercive bilinear form gives a positive-definite matrix. Local support of the piecewise-linear hat functions produces a sparse matrix, allowing efficient numerical solution.

On l2 sequence space ${\ell }^2$, let $(Lu{)}_n=u_n/n$ and $f_n=1/n$. This bounded self-adjoint operator has $⟨Lu,u⟩>0$ for nonzero $u$, but the equation $Lu=f$ requires $u_n=1$, which is not in ${\ell }^2$. Taking the first $N$ coordinates equal to one makes $I(u)=-{\sum }_{n=1}^{N}1/n\to -\infty$. Thus strict positivity in isolation does not give existence in that Hilbert space; a coercive bilinear form and a continuous forcing functional in its norm are essential hypotheses for the usual minimization proof.

A continuous symmetric coercive bilinear form $a$ on a real Hilbert space and a continuous linear functional $\ell$ give a unique minimum of $I$. The parallelogram identity makes any minimizing sequence Cauchy in the energy norm; coercivity and completeness give its limit. Differentiation gives $a(u,w)=\ell (w)$, and $I(v)-I(u)=a(v-u,v-u)$ proves uniqueness. A nonsymmetric operator instead differentiates to its symmetric part. Mere pointwise positivity without a lower bound in the chosen norm need not yield a minimizing vector.

A variational problem seeks a stationary point or minimizer of a functional on a specified function space. For a symmetric coercive bilinear form $a$ and bounded linear functional $\ell$ on a real Hilbert space, minimizing $J(v)=a(v,v)/2-\ell (v)$ is equivalent to the weak formulation $a(u,v)=\ell (v)$ for every test vector. The Lax-Milgram theorem gives a unique solution, and $J(u+w)-J(u)=a(w,w)/2$ proves unique minimization. Specifying the admissible space and its boundary constraints is part of the problem.