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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 Poincare 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 $f/x\in L^2(0,1)$.

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 $f\in L^2(U)$.