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Choose an orthonormal basis of eigenvectors of the positive-definite operator $\alpha$ using the proved spectral theorem. Its eigenvalues ${\lambda }_j$ are strictly positive, because $⟨\alpha e_j,e_j⟩={\lambda }_j>0$. Define $\beta e_j=\sqrt{{\lambda }_j}{e}_j$. It is a positive-definite Hermitian operator and ${\beta }^2=\alpha$, establishing existence of the positive square root of an operator.

For uniqueness, let $B$ be any positive-definite Hermitian square root. The identity $B^2=\alpha$ implies $B\alpha =B^3=\alpha B$, so $B$ preserves each eigenspace $E_{\lambda }$ of $\alpha$. Its restriction to that space is Hermitian and positive-definite. Diagonalize that restriction by the same spectral theorem. Every resulting eigenvalue $\mu$ satisfies $\mu >0$ and ${\mu }^2=\lambda$, hence $\mu =\sqrt{\lambda }$. Therefore $B$ acts as $\sqrt{\lambda }I$ on the entire eigenspace, even when $\lambda$ is repeated. These eigenspaces span the space, so the positive-definite Hermitian square root is unique, with eigenvalues $\sqrt{{\lambda }_j}$.

Work first in a real Hilbert space. A bounded linear operator $L$ is self-adjoint whenFor the variational interpretation, elliptic means uniformly coercive: some $\gamma >0$ satisfiesThis is ellipticity of a bounded Hilbert-space operator, expressed through its quadratic form, rather than the principal-symbol definition for an elliptic differential operator. In the nonsymmetric case this condition concerns the symmetric part of $L$ and does not itself imply self-adjointness.

A positive-definite operator is self-adjoint and strictly positive:Some variational conventions use “positive definite” for the stronger combination of self-adjointness and ellipticity; we distinguish strict from uniform positivity explicitly. A uniformly positive definite symmetric operator is coercive, whereas strict positivity alone in infinite dimension need not give a bounded inverse or solvability for every forcing. If positivity is instead defined only by the displayed real quadratic inequality, symmetry must be added separately for the next part. In a complex Hilbert space, use a self-adjoint operator, the real quadratic form, and $\mathrm{Re}⟨f,v⟩$ in the linear term.

Assume the self-adjoint interpretation of positive-definite operator from the preceding part. For $v,w\in H$ and real $\epsilon$, expand the quadratic functional:The first variation therefore vanishes in every direction precisely when $⟨Lv-f,w⟩=0$ for every $w$, or $Lv=f$. This is its Euler-Lagrange equation and its weak formulation. For any weak solution $u$, write $v=u+w$. Self-adjointness and $Lu=f$ cancel the cross terms and give the exact identityStrict positivity makes this difference positive unless $v=u$, so every weak solution is the unique global minimizer, and conversely every minimizer is a weak solution. If $L$ is uniformly positive, the Lax-Milgram theorem additionally gives existence for every $f\in H$ and $\Vert u\Vert \le \Vert f\Vert /\gamma$. The identity above then gives the quantitative gap $I(v)-I(u)\ge \gamma \Vert v-u{\Vert }^2$.

The hypotheses matter. On ${\mathbb{R}}^2$, letThen $⟨Lv,v⟩=\Vert v{\Vert }^2>0$ for nonzero $v$, but $I$ is minimized at $v=f$, whereas $L^{-1}f=(1/2,-1/2{)}^T$. Thus real quadratic positivity without symmetry does not imply the requested variational assertion: the symmetric part determines a real quadratic functional.

Also, strict self-adjoint positivity does not imply existence for arbitrary $f$. On $H={\ell }^2$, take $(Lv{)}_j=v_j/j$ and $f_j=1/j$. The operator is bounded, self-adjoint and strictly positive, and $f\in {\ell }^2$; a solution would have $u_j=1$, which is not in ${\ell }^2$. In fact the trial vectors with their first $N$ coordinates equal to one give $I(v)=-{\sum }_{j=1}^{N}1/j\to -\infty$. The printed conclusion about a weak solution is valid whenever that solution exists; an unconditional existence assertion needs uniform positivity.

The adjoint operator ${\alpha }^{\ast }:W\to V$ is uniquely characterized byLet $A=\beta {\beta }^{\ast }+{\gamma }^{\ast }\gamma :V\to V$. Its quadratic form is nonnegative:If $Av=0$, both norms vanish. In particular $v\in \ker \gamma =\mathrm{im}\beta$, while ${\beta }^{\ast }v=0$ means $v$ is orthogonal to $\mathrm{im}\beta$, by the defining adjoint operator identity. A vector belonging to a vector subspace and its orthogonal complement must be zero. Hence $A$ is injective, and finite dimensionality makes it surjective.

ThusThe same calculation shows it is a positive-definite operator when $V\ne \{0\}$.

On the zero-boundary Sobolev space $H_0^1(U)$, the associated bilinear form isThe homogeneous Dirichlet boundary condition removes the boundary term, so $(Lu,v{)}_{L^2}=B[u,v]$ for sufficiently regular $u$ and $v\in H_0^1$.

Formal self-adjointness means that $L$ equals its formal adjoint, or equivalently $(L\varphi ,\chi )=(\varphi ,L\chi )$ for compactly supported smooth functions. HereWith the symmetric real principal coefficients in the question, $L=L^{\ast }$ is equivalent to $b=0$. It makes $B$ a symmetric bilinear form.

For the assertions in part (b), positivity must mean strict Dirichlet positivity:This is the positive-definite operator convention. The weaker positive semidefinite operator convention $B[u,u]\ge 0$ does not suffice: $L=-\Delta -{\lambda }_1$ has zero energy on a first Dirichlet eigenfunction. Under strict positivity, the compactness argument in part (b)(i) gives a uniform lower bound as well.

A real linear operator $\mathcal{L}$ on an inner product space is a positive-definite operator when it is self-adjoint andfor every nonzero $u$ in its domain. In the variational setting one normally requires the stronger uniform estimate $⟨\mathcal{L}u,u⟩\ge c\Vert u{\Vert }_V^2$ for some $c>0$, which is coercivity.