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A bilinear form $B:X\times X\to \mathbb{R}$ on a real normed vector space is bounded when one constant $0\le C<\infty$ satisfiesThis is an upper bound on the magnitude of every pairing. It says nothing about the sign of $B(v,v)$ and does not require a symmetric bilinear form. The chosen norm is part of the hypothesis. Boundedness is equivalent to joint continuity, as shown by the boundedness criterion for a bilinear form.^[ref]

For example, on ${\mathbb{R}}^2$ with the Euclidean norm, the bilinear formsboth satisfy the upper bound with $C=1$, by the Cauchy-Schwarz inequality. Yet their quadratic forms are $B_{-}(v,v)=x^2-y^2$ and $B_0(v,v)=x^2$: the former takes negative values and the latter vanishes at the nonzero vector $(0,1)$. Thus neither is a coercive bilinear form. The latter is also a degenerate bilinear form, whereas $B_{-}$ is nondegenerate. Boundedness controls size, not positivity or loss of directions.

Every bilinear form on a real or complex finite-dimensional vector space equipped with any norm is bounded. In coordinates, write $B(u,v)=u^{T}Av$ using its matrix of a bilinear form. The Cauchy-Schwarz inequality gives $|u^{T}Av|\le \Vert A{\Vert }_2\Vert u{\Vert }_2\Vert v{\Vert }_2$, with $\Vert A{\Vert }_2$ the spectral norm; finite-dimensional equivalence of norms transfers this to the chosen norm. The conclusion concerns each fixed norm and fixed bilinear form: it does not supply a single constant for an arbitrary family of matrices or changing dimensions.

On an infinite-dimensional vector space, the upper bound is an additional condition. Equip the finitely supported sequences $c_{00}$ with the norm inherited from the l2 sequence space and setEach pairing is a finite sum, so this is a well-defined bilinear form. For the basis vector $e_n$, however, $\Vert e_n{\Vert }_2=1$ and $B(e_n,e_n)=n$; no finite $C$ works. It is even a coercive bilinear form with $\alpha =1$, since $\sum n{v}_n^2\ge \sum v_n^2$. Consequently coercivity alone does not imply boundedness on a normed vector space. This example uses $c_{00}$, which is not a Hilbert space; it does not define this pairing on all of the l2 sequence space.

A bounded bilinear form defines a bounded linear operator $T:X\to X^{\prime }$ by $(Tu)(v)=B(u,v)$, where $X^{\prime }$ is the continuous dual space. Its operator norm is the smallest admissible $C$:This converts an upper bound for pairings into an upper bound for a linear map taking an unknown vector to a linear functional.

A real bilinear form $B$ on a normed vector space $X$ is coercive when there is one constant $\alpha >0$ such thatThe lower bound uniformly controls every direction in the chosen norm. Unlike the upper bound defining a bounded bilinear form, it forces positive diagonal values at every nonzero vector. No symmetry is assumed. For complex Hilbert spaces, the usual analogue uses a sesquilinear form and $\mathrm{Re}B(v,v)\ge \alpha \Vert v{\Vert }^2$; it is not the same assertion for a complex bilinear form.

A concrete bounded and coercive example on ${\mathbb{R}}^2$, with the Euclidean norm, isIts quadratic form isThe best lower constant is $\alpha =2$, attained at $(1,0)$. Nevertheless, $B(e_1,e_2)=1$ and $B(e_2,e_1)=-1$, so coercivity does not imply a symmetric bilinear form. The symmetric part of a matrix $S=(A+A^T)/2=\mathrm{diag}(2,3)$ determines $q$, while the skew-symmetric matrix part changes mixed pairings without changing $q$. This is the symmetric and alternating parts of a bilinear form decomposition.^[ref]

For the upper constant, direct multiplication gives $A^{T}A=\left(\begin{array}{cc}5& -1\\ -1& 10\end{array}\right)$. Its eigenvalues are $(15\pm \sqrt{29})/2$, so the optimal upper constant, the spectral norm, is $C=\sqrt{(15+\sqrt{29})/2}$. The two constants answer different questions: how large a mixed pairing can be, and how small a self-pairing can be relative to squared norm.

The plotted quadratic forms show the mechanism: $x^2-y^2$ descends in one direction, $x^2$ leaves one direction uncontrolled, and $2x^2+3y^2$ controls both. These are self-pairings; their level sets cannot reveal the skew-symmetric matrix part. Calling $q$ a quadratic energy here refers to this diagnostic quantity, not to a claim that a nonsymmetric equation comes from minimizing an energy functional.

On a real finite-dimensional vector space with any norm, strict positivity $B(v,v)>0$ for every nonzero $v$ does imply coercivity. Indeed the quadratic form is continuous, and its positive minimum on the compact unit sphere supplies $\alpha$. If $B$ is also a symmetric bilinear form, this is positive definiteness. For general $B$, it is the symmetric and alternating parts of a bilinear form decomposition that isolates the positive part. Switching between equivalent norms preserves coercivity, but usually changes its constant.

The infinite-dimensional vector space distinction is explicit on the real l2 sequence space:The Cauchy-Schwarz inequality gives boundedness with $C=1$, and $B(v,v)>0$ for every nonzero vector. But $B(e_n,e_n)=1/n$ with $\Vert e_n{\Vert }_2=1$, so no positive $\alpha$ works. On the span of $e_1,\dots ,e_N$, the best lower constant is $1/N$: each finite restriction is coercive, but the control disappears as $N$ grows. Strict positivity is therefore weaker than a uniform lower bound.

The practical payoff is a stability estimate for a weak formulation. Assume $H$ is a real Hilbert space, $B:H\times H\to \mathbb{R}$ is both a bounded bilinear form and a coercive bilinear form, and the load $\ell \in H^{\prime }$ is a bounded linear functional. The Lax-Milgram theorem gives a unique $u\in H$ satisfying $B(u,v)=\ell (v)$ for all $v\in H$, including nonsymmetric $B$.^[ref] The dual norm $\Vert \ell {\Vert }_{H^{\prime }}={\sup }_{v\ne 0}|\ell (v)|/\Vert v\Vert$ measures load size. Testing with $v=u$ gives the short energy estimateFor $u=0$ the bound holds directly; otherwise divide by $\Vert u\Vert$. Subtracting the equations for two loads and using the same argument gives $\Vert u_1-u_2\Vert \le \Vert {\ell }_1-{\ell }_2{\Vert }_{H^{\prime }}/\alpha$. Thus a positive lower constant limits the response to load errors. The upper bound and completeness of the Hilbert space enter the existence theorem; the estimate itself uses the lower bound and the load's dual norm.

For the displayed matrix $A$, take $\ell (v)=f^{T}v$. Then $B(u,v)=\ell (v)$ for every $v$ means $A^{T}u=f$, not $Au=f$, with this ordering of the bilinear form. Since $\det A=7$,For $f=t(1,0{)}^T$, the exact response is $u=t(3,-1{)}^T/7$ and $\Vert u{\Vert }_2=\sqrt{10}|t|/7<|t|/2$. For $f=t(0,1{)}^T$, it is $u=t(1,2{)}^T/7$, with norm $\sqrt{5}|t|/7$. The actual maximal amplification is the operator norm $\Vert (A^T{)}^{-1}{\Vert }_2=\sqrt{2/(15-\sqrt{29})}<1/2$: the coercivity estimate is valid but need not be sharp for a nonsymmetric bilinear form.

On the zero-boundary Sobolev space $H_0^1(0,1)$ with norm $\Vert u^{\prime }{\Vert }_{L^2}$, assume measurable coefficients $0<p_0\le p\le p_1$ and $0\le q\le q_1$. The Poincaré inequality and Cauchy-Schwarz inequality give $|a(u,v)|\le (p_1+q_1/{\pi }^2)\Vert u^{\prime }{\Vert }_2\Vert v^{\prime }{\Vert }_2$, while nonnegativity of $q$ gives $a(u,u)\ge p_0\Vert u^{\prime }{\Vert }_2^2$. Thus the form is a bounded bilinear form and a coercive bilinear form. For a source in $H^{-1}$, the Lax-Milgram theorem yields a unique weak solution and every conforming Galerkin method approximation. No derivative of the bounded coefficient $p$ is needed for this weak formulation.

An elliptic boundary value problem couples an elliptic differential equation in a domain to boundary conditions. Its weak formulation is often an equation for a bounded bilinear form on a Sobolev space.

The theorem supplies solvability for a coercive weak formulation.

A bounded bilinear form $a$ on a real Hilbert space $H$ that is a coercive bilinear form represents every bounded linear functional uniquely: for each $\ell \in H^{\prime }$, there is a unique $u\in H$ with $a(u,v)=\ell (v)$ for every $v\in H$.

For a proof, the Riesz representation theorem gives a bounded linear operator $T$ and a vector $g$ with $a(u,v)=⟨Tu,v⟩$ and $\ell (v)=⟨g,v⟩$. Coercivity and the Cauchy-Schwarz inequality give $\Vert Tu\Vert \ge \alpha \Vert u\Vert$, so $T$ is injective and its image is closed: convergence of $T{u}_n$ makes $u_n$ a Cauchy sequence, whose limit supplies the image limit. If $v$ belongs to the orthogonal complement of the image, then $a(u,v)=0$ for every $u$; setting $u=v$ and using coercivity forces $v=0$. Thus the image is also dense and equals $H$. There is consequently a unique $u$ with $Tu=g$, and $\Vert u\Vert \le \Vert \ell \Vert /\alpha$. The complex version uses a bounded sesquilinear form with $\mathrm{Re}a(u,u)\ge \alpha \Vert u{\Vert }^2$.

Let $V$ be a real Hilbert space, $a:V\times V\to \mathbb{R}$ a bounded bilinear form, and $\ell \in V^{\prime }$ a bounded linear functional. Assume constants $C<\infty$, $\alpha >0$ satisfyThe second hypothesis says that $a$ is a coercive bilinear form. The Lax-Milgram theorem then supplies a unique weak solution $u\in V$ with $a(u,v)=\ell (v)$ for every $v\in V$, and $\Vert u{\Vert }_V\le \Vert \ell {\Vert }_{V^{\prime }}/\alpha$.

For a finite-dimensional trial linear subspace $V_h\subset V$, the same boundedness and coercivity constants apply to the restricted form. Since a finite-dimensional subspace is closed, $V_h$ is itself a Hilbert space. Thus there is a unique Galerkin method solution $u_h\in V_h$ satisfying $a(u_h,v_h)=\ell (v_h)$ for every $v_h\in V_h$. The Galerkin orthogonality $a(u-u_h,v_h)=0$ gives the Céa lemma:In particular, approximation by the trial spaces implies numerical convergence. For a complex Hilbert space, replace bilinearity by a sesquilinear form, take $\ell$ antilinear in the test variable, and require $\mathrm{Re}a(v,v)\ge \alpha \Vert v{\Vert }_V^2$.

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 $⟨g,v⟩$.

First let $L$ be a bounded linear operator on $H$ satisfying the symmetric positive-definiteness convention in part (a). For any direction $w\in H$, expansion of the quadratic functional givesThus vanishing of the first variation in every direction is precisely the weak equation $⟨Lu,w⟩=⟨f,w⟩$ for every $w$. In this whole-space bounded-operator setting it is equivalent to $Lu=f$, the Euler-Lagrange equation.

If $u$ solves that equation, set $v=u+w$. The linear terms cancel:with equality only when $v=u$. Hence the weak solution is the unique global minimizer. Conversely, any minimizer has zero first variation, and therefore solves the weak equation. For a symmetric bounded bilinear form $B$ on a form space $V$, exactly the same calculation gives $I(v)=B(v,v)-2\ell (v)$ and $B(u,w)=\ell (w)$ for every $w\in V$; it does not require an unbounded differential operator to map every $v\in V$ into $H$.

Existence for every $f$ requires an extra hypothesis if “positive definite” means only strict positivity. The coercive operator bound makes the form coercive, so the Lax-Milgram theorem supplies existence and uniqueness. Without that bound, the diagonal operator on sequence space $L(x_n)=(x_n/n)$ on ${\ell }^2$ is symmetric and strictly positive, but $f=(1/n)$ belongs to ${\ell }^2$ and its formal inverse $(1,1,\dots )$ does not. Thus strict positivity alone proves uniqueness and the minimizing property of a solution when one exists, not existence for all $f$.

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.

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.

The Lax-Milgram theorem states that if $H$ is a real Hilbert space, $B:H\times H\to \mathbb{R}$ is a bounded bilinear form, and there is an $\alpha >0$ such thatthen for every bounded linear functional $F\in H^{\prime }$ there is a unique $u\in H$ satisfyingMoreover, $\boxed{\Vert u{\Vert }_H\le {\alpha }^{-1}\Vert F{\Vert }_{H^{\prime }}}$. Symmetry of $B$ is not required.

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