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The Lax-Milgram theorem gives, for a bounded coercive bilinear form on a real Hilbert space and every bounded linear functional, a unique vector representing that functional through the form, with the inverse bound supplied by coercivity. The Fredholm alternative for a compact operator $K$ says that $I+\mu K$ is invertible if its kernel is zero; in general its range is closed and consists exactly of vectors orthogonal to $\ker (I+\mu K{)}^{\ast }$ in the Hilbert-space case.

Uniform ellipticity gives $\sum a_{ij}{\xi }_i{\xi }_j\ge \theta |\xi {|}^2$ for some $\theta >0$. Bounded coefficients make $B_{\lambda }$ bounded on $H^1$, and for $\lambda <-\gamma$,The forcing $v\mapsto \int fv$ is bounded on $H^1$. The Lax-Milgram theorem therefore gives a unique weak solution for every $\lambda <-\gamma$.

When $c=0$, choose the shift ${\lambda }_0=-1$ and let $Kf$ be the unique solution of $B[Kf,v]+\int (Kf)v=\int fv$. Thus $K=(L+I{)}^{-1}$ in the weak sense. The equation $Lu=f$ is equivalent to $u=K(f+u)$, orConversely a solution in $L^2$ of this operator equation lies in $H^1$ because it equals $K(f+u)$, and satisfies the variational equation. The assumed boundedness $K:L^2\to H^1$, followed by the compact Rellich embedding $H^1\hookrightarrow L^2$, proves compactness of $K$ as an $L^2$ operator.

By the supplied kernel assumption, $\ker (I-K)$ consists of constants. Also $K1=1$. Since $K$ is a self-adjoint operator, the Fredholm alternative condition is $Kf\perp 1$, andTherefore a weak solution exists exactly when the mean of $f$ is zero, and solutions differ by a constant. On a disconnected domain the actual kernel would consist of constants on each component and the condition would apply separately on each; the question's given kernel assumption specifies the single-constant case.

For a general coefficient matrix, the stated variational definition encodes zero conormal flux $(aDu)\cdot n=0$. It agrees with ordinary zero normal derivative only under additional conditions on $a$. The argument above uses the weak formulation explicitly defined in the paper, so does not require that stronger boundary identification.

The quadratic form of a matrix only sees its symmetric part. Thus $\lambda |\xi {|}^2\le {\xi }^{T}A\xi \le \Lambda |\xi {|}^2$ need not bound the skew part of a measurable coefficient matrix. For $A=\left(\begin{array}{cc}1& k\\ -k& 1\end{array}\right)$ the quadratic form is exactly $|\xi {|}^2$, regardless of $k$. On a square, take $k(x)=1/|x_1|$ off the line $x_1=0$, with a finite arbitrary value on that line. Two compactly supported smooth functions that equal $x_2$ and $x_1$ near the origin have $(A\nabla u)\cdot \nabla v=k$ there; the purported energy pairing is not integrable. A Lax-Milgram theorem proof therefore requires bounded full coefficients or another hypothesis that makes the bilinear form bounded. Symmetry supplies the missing matrix bound from the quadratic upper bound.

Let $V$ be a closed linear subspace of a real Hilbert space, and let $Q(v)=a(v,v)$ for a symmetric bounded bilinear form that is nonnegative on the ambient space and coercive on $V$. Then $Q$ has a unique minimizer on every closed affine subspace $g+V$. If $Q(v_j)\to m={\inf }_{g+V}Q$, the midpoint belongs to $g+V$ and the parallelogram law gives $Q(v_j-v_k)\le 2Q(v_j)+2Q(v_k)-4m\to 0$. Coercivity on $V$ makes the minimizing sequence Cauchy; completeness and continuity give a minimizer. The vanishing first variation is $a(u,v)=0$ for all $v\in V$. Two minimizers differ by $w\in V$ with $Q(w)=0$, so coercivity gives uniqueness. This proves the symmetric existence theorem directly, without invoking the Lax-Milgram theorem.

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.

Suppose the mixed boundary value problem $\Delta w=g$, with zero Dirichlet boundary condition on a nonempty boundary portion and zero Neumann boundary condition on the remainder, has an $H^2$ solution with $\Vert w{\Vert }_{H^2}\le C\Vert g{\Vert }_{L^2}$. Assume the domain is connected and that zero Sobolev trace on the specified portion excludes nonzero constants. Then the same boundary conditions for $-\Delta z+z=e$ give $\Vert z{\Vert }_{H^2}\le 2C\Vert e{\Vert }_{L^2}$.

Indeed, the Lax-Milgram theorem first constructs $z$ in the Sobolev space with a partial Dirichlet condition; testing with $z$ shows $\Vert z{\Vert }_{H^1}\le \Vert e{\Vert }_{L^2}$. Apply the assumed elliptic regularity result to $g=z-e$. Its solution $w$ and $z$ satisfy the same weak Laplace equation, so testing their difference shows $\nabla (w-z)=0$. The Sobolev function with zero weak gradient property and zero Sobolev trace give $w=z$. Finally $\Vert z-e{\Vert }_{L^2}\le 2\Vert e{\Vert }_{L^2}$. This argument supplies the dual regularity needed by the Aubin–Nitsche duality argument for the equation with a positive reaction term.

For $Au=u^{\prime \prime }-(1+x^2)u$ on $D(A)=\{u\in H^2(\mathbb{R}):x^{2}u\in L^2\}$, the form $a_{\lambda }(u,v)=\int u^{\prime }\overline{v^{\prime }}+(1+\lambda +x^2)u\bar{v}$ is coercive on the one-dimensional harmonic oscillator form domain when $\lambda >0$. The Lax-Milgram theorem gives a unique weak solution of $(\lambda -A)u=g$. The graph estimate for the harmonic oscillator puts it in $D(A)$, and $a_{\lambda }(u,u)\ge (1+\lambda )\Vert u{\Vert }_2^2$ gives the displayed resolvent bound. The domain is dense and the bounded resolvent proves closedness, so the Hille-Yosida theorem gives a contraction semigroup. An exponentially shifted semigroup with generator $A+I$ solves the heat equation with oscillator potential.

A skew-coupled elliptic system has zeroth-order couplings with opposite signs. For the displayed example with homogeneous Dirichlet boundary conditions, its weak formulation on $V=H_0^1(\Omega )\times H_0^1(\Omega )$ has the bilinear formThe coupling cancels for equal arguments, so $b((p,q),(p,q))=\Vert \nabla p{\Vert }_0^2+\Vert \nabla q{\Vert }_0^2$. On a domain satisfying the Poincaré inequality, this is a coercive bilinear form. With the product first-order Sobolev norm, the Cauchy-Schwarz inequality applied to $(\nabla p,\nabla q,-q,p)$ and $(\nabla v,\nabla z,v,z)$ gives $|b(P,T)|\le \Vert P{\Vert }_V\Vert T{\Vert }_V$. The Lax-Milgram theorem therefore applies even though the bilinear form is not symmetric. Eliminating $w=-\Delta u$ recovers ${\Delta }^{2}u+u=f$, involving the biharmonic operator.

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.

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.

On $H=\{v\in H^1(0,1):v(0)=0\}$, the formis bounded and coercive. The Hardy inequality on an interval controls its singular term, while the one-sided Poincaré inequality controls the first-order term. Thus the Lax-Milgram theorem gives a unique weak solution for every functional $v\mapsto {\int }_0^1(f/x)(v/x)$ with $f/x\in L^2(0,1)$.

For continuous real coefficients with $p>0$ and $q,r\ge 0$ on a bounded interval, use the clamped second-order Sobolev space and the displayed bilinear form. A forcing in $L^2$ defines $\ell (v)=\int fv$. The weak solution satisfies $a(u,v)=\ell (v)$ for every clamped test vector. Compactly supported test functions give $(p{u}^{\prime \prime }{)}^{\prime \prime }-(q{u}^{\prime }{)}^{\prime }+ru=f$ as a distribution; density extends that equation to the energy space. Continuous coefficients alone do not justify differentiating them classically. The clamped interval derivative norm bound gives coercivity, so Lax-Milgram theorem supplies a unique solution and the quadratic energy has that unique minimum.

Let a positive-measure part ${\Gamma }_D$ of the boundary of a bounded connected domain carry a homogeneous Dirichlet boundary condition, and let the complementary part carry a homogeneous Neumann boundary condition. Onthe weak Poisson problem isThe Poincare inequality with a partial Dirichlet boundary makes the left side coercive, so the Lax-Milgram theorem gives a unique weak solution.

For $Lu=-{\partial }_j(a^{ij}{\partial }_{i}u)+b^i{\partial }_{i}u+cu$ with smooth coefficients on a bounded smooth domain, the weak formulation of $(L+\lambda )u=f$ and conormal Robin boundary condition $a^{ij}{\partial }_{i}u{\nu }_j+\beta u=0$ isThe bilinear form is bounded by the Sobolev trace theorem. If $a^{ij}$ satisfy the positive uniformly elliptic operator inequality, the multiplicative trace inequality on a bounded smooth domain and the Young inequality show that the form is coercive when $\lambda$ is sufficiently large, even if $\beta$ and $c$ have negative parts. The Lax-Milgram theorem then gives a unique weak solution for every $f\in L^2(U)$.