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For a bounded linear operator $T:H\to K$ between real or complex Hilbert spaces, its adjoint is the unique bounded linear operator $T^{\ast }:K\to H$ satisfyingThe inner product here is linear in its first argument. The adjoint operator transfers an output measurement back to the input: measuring $Tx$ against the vector $y$ gives exactly the same scalar as measuring $x$ against $T^{\ast }y$. It depends on both inner products.

In this real example,Both inner products equal $3$. The dashed lines collect all vectors giving that same measurement: the output line $u_1=3$ becomes the input line $2x_1+x_2=3$. The adjoint operator therefore moves the direction that describes the measurement. It generally differs from the inverse matrix: here $A^{-1}y=(1/2,0{)}^{\mathrm{T}}$.

For an explicitly complex example, use the standard inner product $⟨u,v⟩={\sum }_j{u}_j\overline{v_j}$ on ${\mathbb{C}}^2$. In the standard orthonormal basis, the adjoint operator is represented by the conjugate transpose:For $x=(1+i,2{)}^{\mathrm{T}}$ and $y=(1,i{)}^{\mathrm{T}}$, direct matrix multiplication givesand henceThe complex conjugate is essential to this agreement.

Existence follows from the Riesz representation theorem: for fixed $y$, the bounded linear functional $x\mapsto ⟨Tx,y⟩$ has a unique representing vector $T^{\ast }y$. The resulting adjoint operator has the same operator norm as $T$, andMIT's lecture 18 develops the representation argument and the conjugate transpose example.^[ref]

This transfer explains the normal equation for a linear inverse problem: a least-squares solution has residual $Tx-b$ orthogonal to every attainable direction $Th$, so $⟨h,T^{\ast }(Tx-b)⟩=0$ for every $h$, equivalently $T^{\ast }Tx=T^{\ast }b$. An output residual becomes an input condition through the same adjoint operator. All identities in this section concern bounded operators defined on the whole source Hilbert space; for a densely defined operator that is not bounded, use the adjoint of a densely defined operator and check its operator domain.

If finite Borel measures $\mu$ and $\nu$ on $\mathbb{R}$ satisfyfor every compactly supported continuous function $f$, then $\mu =\nu$. This follows from the uniqueness part of the Riesz representation theorem; alternatively, continuous functions that approximate indicators from above and below recover the measure of every interval and hence every Borel set.

On a compact metric space, every sequence of Borel probability measures has a subsequence with weak convergence of probability measures to a Borel probability measure. To see this, choose a countable uniformly dense subset of continuous functions, extract a diagonal subsequence of their bounded integrals, and extend the resulting positive normalized functional to all continuous functions. The Riesz representation theorem supplies its Borel probability measure. This is useful for constructing an invariant measure from orbit empirical measures.

The duality mapping takes a vector to its supporting dual functionals for half its squared norm. Equivalently $p\in \mathcal{J}(x)$ when $\Vert p{\Vert }_{\ast }=\Vert x\Vert$ and $⟨x,p⟩=\Vert x{\Vert }^2$. This follows from the dual-norm inequality and equality in the conjugate relation for squared norms. In a Hilbert space, the Riesz representation theorem identifies it with the identity; in a general Banach space it can be set-valued.

For a differentiable real-valued functional $F$ on a real Hilbert space $\mathcal{H}$, its functional gradient satisfies $DF(g)[h]=⟨{\nabla }_{\mathcal{H}}F(g),h{⟩}_{\mathcal{H}}$ for every $h\in \mathcal{H}$. A bounded linear derivative has a unique such representative by the Riesz representation theorem. The inner product matters: a formal variational derivative need not equal the functional gradient in another Hilbert space. For $F(g)=\frac{1}{2}{\sum }_i(y_i-g(x_i){)}^2$ in a Reproducing kernel Hilbert space, the reproducing property gives ${\nabla }_{\mathcal{H}}F(g)={\sum }_i(g(x_i)-y_i)k(x_i,\cdot )$. Substitution in the defining inner product proves this formula directly.

For finite positive measures $\mu ,\nu$, put $\rho =\mu +\nu$. The linear functional $g\mapsto \int gd\nu$ is bounded on real $L^2(\rho )$ by the Cauchy-Schwarz inequality. The Riesz representation theorem gives a measurable $h$ with $\nu (A)={\int }_{A}hd\rho$. Testing indicators of its negative and greater-than-one level sets shows $0\le h\le 1$ almost everywhere. Subtraction gives $\mu (A)={\int }_A(1-h)d\rho$. This constructs the densities without assuming the Radon-Nikodym theorem as a prior result.

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 $\mathrm{Re}a(u,u)\ge \alpha \Vert u{\Vert }^2$.

For orthonormal Legendre polynomials on $[-1,1]$, the polynomial $q_n$ represents evaluation at zero on polynomials of degree at most $n$: $⟨r,q_n⟩=r(0)$. This follows by expanding $r$ in the orthonormal basis, or from the Riesz representation theorem. With $M=\lfloor n/2\rfloor$ and $c_m=\left(\genfrac{}{}{0ex}{}{2m}{m}\right)/4^m$, orthonormality givesThe last identity telescopes using $c_m/c_{m-1}=(2m-1)/(2m)$. The norms are unbounded: $c_m\ge 1/(2m+1)$ makes the sum dominate a divergent harmonic series. Thus the evaluation identities hold eventually for every fixed polynomial but cannot extend to convergence for every $L^2$ vector, by the Uniform boundedness principle. Point evaluation is also not defined on $L^2$ equivalence classes.

On a smooth bounded connected domain, the homogeneous Neumann problem seeks $u\in H^1(U)$ such that $\int \nabla u\cdot \nabla v=\int fv$ for every $v\in H^1(U)$. For $f\in L^2(U)$ it is solvable exactly when $\int f=0$, and the weak solution is unique up to constants. The Poincare-Wirtinger inequality and the Riesz representation theorem give existence on the mean-zero Sobolev space. The test function one gives necessity of the compatibility condition.

Use the convention that the inner product is linear in its first argument. If $v\ne 0$, nonnegativity of $\Vert u-⟨u,v⟩v/\Vert v{\Vert }^2{\Vert }^2$ givesThe case $v=0$ is immediate. This Cauchy-Schwarz inequality shows that $f_v$ is bounded linear with norm at most $\Vert v\Vert$. For $v\ne 0$, evaluation at $v/\Vert v\Vert$ gives equality; for $v=0$ both norms vanish. The Banach dual is the space of all bounded linear functionals with the operator norm.

The Riesz representation theorem asserts that every $f\in V^{\prime }$ is uniquely $f(u)=⟨u,v⟩$ for some $v\in V$, with $\Vert f\Vert =\Vert v\Vert$. If $f=0$, take $v=0$. Otherwise put $W=\ker f$, a closed subspace, and use the projection theorem to obtain $w\in W^{\perp }$ with $f(w)=1$. Since $u-f(u)w\in W$, $⟨u,w⟩=f(u)\Vert w{\Vert }^2$. Thus $v=w/\Vert w{\Vert }^2$ represents $f$. If two vectors represent it, their difference has inner product zero with every vector, in particular itself, giving uniqueness. The norm statement was already proved.

The space $P_n$ is finite-dimensional and hence complete; evaluation at zero is bounded there by equivalence of finite-dimensional norms. Riesz therefore supplies exactly one $q_n\in P_n$. In any orthonormal polynomial basis,For the real Legendre functions the bars may be omitted.

The printed even-degree normalization has an error. Orthogonality forces a degree-two polynomial to be a scalar multiple of $3t^2-1$, whose squared integral on $[-1,1]$ is $8/5$. Hence any orthonormal choice has $|p_2(0){|}^2=5/8$, whereas the printed formula gives $3/8$. The factor under its square root must be $4m+1$ instead of $2m+1$. Signs of the individual polynomials can be chosen so that the values at zero are positive; these signs do not affect $q_n$ or its norm.

Put $M=\lfloor n/2\rfloor$ and $c_m=\left(\genfrac{}{}{0ex}{}{2m}{m}\right)/4^m$. With the consistent orthonormal normalization the norm isThe finite sum telescopes because $c_m=(2m-1)c_{m-1}/(2m)$, so $(2m+1{)}^2{c}_m^2-(2m-1{)}^2{c}_{m-1}^2=(4m+1)c_m^2$. If one substitutes the erroneous printed values formally instead, one obtains $\frac{1}{2}{\sum }_{m=0}^M(2m+1)c_m^2$; that is not the norm of the actual orthonormal Legendre evaluation representer.

For a polynomial $r$ of degree $d$, the defining identity gives $⟨r,q_n⟩=r(0)$ for every $n\ge d$, so the requested limit is eventually constant. It does not extend to all of $L^2$. First, evaluation at a point is not well-defined on equivalence classes. More strongly, the functional norms $\Vert q_n\Vert$ are unbounded. Indeed the largest binomial coefficient in row $2m$ is at least the average, so $c_m\ge 1/(2m+1)$, and the displayed sum dominates a divergent constant multiple of the harmonic series. The Uniform boundedness principle then provides an $L^2$ function for which ${\sup }_n|⟨r,q_n⟩|=\infty$. Thus no choice of point representative can restore the assertion for all $L^2$ functions.

The Hilbert projection theorem states that every vector $x$ has a unique decomposition $x=m+n$ relative to a closed linear subspace $M$, with $m\in M$ and $n\in M^{\perp }$. The vector $m$ is the unique nearest point in $M$.

The Riesz representation theorem states that every bounded complex-linear functional $F$ on a Hilbert space has a unique representation $F(x)=⟨x,y⟩$, with $\Vert F\Vert =\Vert y\Vert$, using the convention of linearity in the first argument. If $F=0$, use $y=0$. Otherwise $M=\ker F$ is closed. Project a vector outside $M$ to obtain $z\in M^{\perp }$ with $F(z)\ne 0$, and rescale it so $F(z)=1$. Then $x-F(x)z\in M$, soThis gives the representation. Uniqueness follows because a vector orthogonal to every $x$ is zero. The Cauchy-Schwarz inequality gives $\Vert F\Vert \le \Vert y\Vert$, and evaluation at $y/\Vert y\Vert$ gives equality when $y\ne 0$.

For the adjoint, fix $v$. The functional $u\mapsto ⟨Tu,v⟩$ is bounded with norm at most $\Vert T\Vert \Vert v\Vert$. By the Riesz representation theorem, there is a unique vector $T^{\ast }v$ satisfyingFor scalars $a,b$, conjugate-linearity in the second argument shows that $T^{\ast }(av+bw)=a{T}^{\ast }v+b{T}^{\ast }w$, by uniqueness of the representing vector. The same norm estimate gives $\Vert T^{\ast }v\Vert \le \Vert T\Vert \Vert v\Vert$. Thus $T^{\ast }$ is a unique bounded linear operator, the adjoint operator.

An infinite-dimensional reflexive example is ${\ell }^2$, by the Riesz representation theorem.

The weak topology on $X$ is the coarsest topology making every $f\in X^{\ast }$ continuous. The weak-star topology on $X^{\ast }$ is the coarsest making every evaluation $f\mapsto f(x)$, $x\in X$, continuous. In each case basic neighborhoods impose finitely many evaluation inequalities. The Banach-Alaoglu theorem states that the closed unit ball of a dual space is compact for its weak-star topology. The separation theorem needed here says: if $K$ is a nonempty weak-star compact convex subset of a dual space $E^{\ast }$ and $z\notin K$, then some $e\in E$ strictly separates them,The real-part convention covers complex spaces; for real spaces omit it.

Let $K$ be the weak-star closure of $J_X(B_X)$ inside $B_{X^{\ast \ast }}$. This is compact by Banach-Alaoglu, convex, and nonempty. If $z\in B_{X^{\ast \ast }}\setminus K$, separation in the dual pair $(X^{\ast \ast },X^{\ast })$ gives $f\in X^{\ast }$ withBut $|z(f)|\le \Vert z\Vert \Vert f\Vert \le \Vert f\Vert$, a contradiction. Therefore $K=B_{X^{\ast \ast }}$, proving the required Goldstine theorem density.

If $X$ is reflexive, $J_X$ identifies its weak topology with the weak-star topology of $X^{\ast \ast }$ and carries $B_X$ onto $B_{X^{\ast \ast }}$, so Banach-Alaoglu gives weak compactness. Conversely, if $B_X$ is weakly compact, its image is weak-star compact, hence closed in the Hausdorff bidual topology. Density then forces $J_X(B_X)=B_{X^{\ast \ast }}$. Scaling gives surjectivity on the entire bidual. Thus

In a real Hilbert space, weak convergence in a Hilbert space meansBy the Riesz representation theorem, every continuous linear functional on $X$ has the form $u\mapsto ⟨u,z⟩$. Thus this is exactly weak convergence, defined by convergence under every continuous linear functional.

For a bounded linear operator $T:E\to F$, write the weak square norm of a finite family asIt is a 2-summing operator if $({\sum }_j\Vert T{x}_j{\Vert }^2{)}^{1/2}\le C{w}_2(x_1,\dots ,x_m)$ for all finite families, and ${\pi }_2(T)$ is the least such $C$. This specializes the definition of an absolutely p-summing operator.

First let $E=H_1$, $F=H_2$ be Hilbert spaces. For the column operator $A:{\ell }_2^m\to H_1$, $A{e}_j=x_j$, the Riesz representation theorem gives $w_2(x_1,\dots ,x_m)=\Vert A^{\ast }\Vert =\Vert A\Vert$. If $T$ is a Hilbert-Schmidt operator, the Parseval identity applied to orthonormal bases givesThus ${\pi }_2(T)\le \Vert T{\Vert }_{\mathrm{HS}}$. Conversely test the summing inequality on any finite orthonormal family in $H_1$. Its weak square norm is $1$, by the Bessel inequality, with equality by choosing one of its vectors. Hence ${\sum }_j\Vert T{e}_j{\Vert }^2\le {\pi }_2(T{)}^2$. Taking the supremum over finite subsets of an orthonormal basis proves the Hilbert-Schmidt operator condition and the reverse norm inequality. These basis sums work even for nonseparable Hilbert spaces, being understood as suprema of finite subsums. Therefore

For an $n$-dimensional normed vector space $E$, again associate a family $(x_j)$ with $A:{\ell }_2^m\to E$. By the Hahn-Banach theorem and scalar Euclidean duality,Let $Q$ be the orthogonal projection onto $(\ker A{)}^{\perp }$, whose dimension is at most $n$. Since $A=AQ$, the trace of $Q$ givesThis proves ${\pi }_2(I_E)\le \sqrt{n}$ for every norm, without comparing that norm to an arbitrary Euclidean norm.

To obtain equality and the projection bound, choose a maximum-volume centered ellipsoid inside the unit ball $B_E$ and use coordinates in which it is the Euclidean ball $B_2$. In the real case this is the John ellipsoid; in the complex case maximize among Hermitian ellipsoids. There are unit contact vectors $u_r\in \partial B_E\cap \partial B_2$ and positive weights $c_r$ with the John contact decompositionFor completeness, the maximum-volume argument gives this identity. If $I_E$ were outside the closed convex cone of the contact matrices, finite-dimensional separation would supply a symmetric, or Hermitian, matrix $D$ with $\mathrm{tr}D>0$ and $u^{\ast }Du\le 0$ at every contact. The cone is closed because its compact generators have trace one. Subtract a sufficiently small multiple of the identity to make all contact inequalities strict while retaining positive trace. The support function of $(I+tD)B_2$ decreases near the contact directions for small $t>0$; away from them the strict gap between the support functions of $B_E$ and $B_2$ permits the same small deformation. Thus this deformed ball still lies in $B_E$. Its volume increases, since the real determinant is $1+t\mathrm{tr}D+O(t^2)$ in the real case and $|{\det }_{\mathbb{C}}(I+tD){|}^2=1+2t\mathrm{tr}D+O(t^2)$ in the complex case. This contradicts maximality. The finite-dimensional convex-cone representation gives finitely many contacts, and taking traces gives the sum of the weights.

At a contact $u_r$, a supporting hyperplane to $B_E$ also supports $B_2$. The Euclidean ball has a unique supporting direction there, so the corresponding functional is ${\varphi }_r(x)=u_r^{\ast }x$ and has dual norm one on $E$. In the complex case balancedness of $B_E$ turns the real support bound into $|{\varphi }_r(x)|\le \Vert x{\Vert }_E$. Also $B_2\subseteq B_E$ gives $\Vert x{\Vert }_E\le |x{|}_2$. For any dual-unit functional $\varphi$, this containment implies that its Euclidean norm is at most one. Apply the summing definition to $x_r=\sqrt{c_r}{u}_r$:The weak square norm is exactly one, since any contact functional attains that Euclidean norm. Hence ${\pi }_2(I_E)\ge \sqrt{n}$. Together with the upper estimate, the norm isThis proves the 2-summing norm of a finite-dimensional identity formula.

Finally retain the preceding finite-dimensional hypothesis when $E\subseteq X$. Extend each ${\varphi }_r$ to a dual-unit functional ${\tilde{\varphi }}_r$ on $X$ using the real or complex Hahn-Banach theorem. DefineThe contact identity gives $P{|}_E=I_E$, hence $P^2=P$ and its range is $E$. The synthesis operator $(a_r)\mapsto {\sum }_r\sqrt{c_r}{a}_r{u}_r$ has Euclidean operator norm one, because its product with its adjoint is the identity. ThereforeThus the linear projection satisfies $\boxed{\Vert P\Vert \le \sqrt{n}}$, the Kadec-Snobar projection bound. If $n=0$, the zero projection gives the same conclusion. No assertion about arbitrary infinite-dimensional subspaces is required.

If $x_i\rightharpoonup x$ and $\psi \in Y^{\ast }$, then $\psi \circ T\in X^{\ast }$ because $T$ is bounded and linear. Therefore $\psi (T{x}_i)\to \psi (Tx)$, proving weak continuity of $T$.

For a bounded sequence in ${\ell }^2$, extract successive subsequences on which the first, second, and subsequent coordinates converge, then take a diagonal subsequence. Call the coordinate limits $x_j$. Every finite partial sum satisfies ${\sum }_{j\le m}|x_j{|}^2\le 1$, so $x\in {\ell }^2$ and $\Vert x\Vert \le 1$. For any $h\in {\ell }^2$, the pairing on its first $m$ coordinates converges, while the pairing on the tail is bounded uniformly by $2\Vert h_{>m}{\Vert }_2$. First choose $m$ large, then take the subsequence limit. The Riesz representation theorem identifies these pairings with every continuous linear functional, proving the requested weak convergence.

For ${\ell }^1$, norm convergence implies weak convergence immediately. Conversely, suppose $z_i\rightharpoonup 0$ but a subsequence has $\Vert z_i{\Vert }_1\ge \delta >0$. Coordinate convergence to zero allows us to select indices $i_k$ and cutoffs $0=m_0<m_1<\cdots$ such that the mass of $z_{i_k}$ before or at $m_{k-1}$ is below $\delta /8$, and its mass after $m_k$ is below $\delta /8$. Its mass on the intervening block is then at least $3\delta /4$. On that block set $a_j=\overline{z_{i_k,j}}/|z_{i_k,j}|$ when nonzero and zero otherwise. These prescriptions define $a\in {\ell }^{\infty }$. The bounded linear functional $z\mapsto {\sum }_j{a}_j{z}_j$ has real part at least $3\delta /4-\delta /4=\delta /2$ on every selected $z_{i_k}$, contradicting weak convergence. Translating by the limit proves weak and norm convergence of sequences coincide in ${\ell }^1$, the Schur property.

A compact operator sends bounded sets to relatively norm-compact sets. Every bounded sequence in ${\ell }^2$ has a weakly convergent subsequence by the argument above, after scaling into the unit ball. Its image under $T:{\ell }^2\to {\ell }^1$ converges weakly and therefore in norm by the Schur property. Sequential compactness in a metric space gives relative compactness of the image of the unit ball. Every bounded linear operator from ${\ell }^2$ to ${\ell }^1$ is compact.

We construct the density and singular part, rather than assuming the desired Lebesgue decomposition theorem. Put $\rho =\mu +\nu$, a finite dominating measure, and use the real Hilbert space $L^2(\rho )$. The linear functionalis well defined on its equivalence classes, since a $\rho$-null set is $\nu$-null. The Cauchy-Schwarz inequality and $\nu \le \rho$ giveBy the Riesz representation theorem, there is $h\in L^2(\rho )$ with $T(g)=\int ghd\rho$. Indicators of all measurable sets belong to this Hilbert space, soThe nonnegativity of these two measures forces $0\le h\le 1$ almost everywhere. For instance, if $\rho (\{h\le -1/n\})>0$, its $\nu$-measure would be negative; the analogous sets $\{h\ge 1+1/n\}$ contradict positivity of $\mu$. Their countable union excludes every violation. Modify $h$ on its measurable $\rho$-null exceptional set so that these inequalities hold everywhere. This is the Hilbert-space construction of dominated measure densities.

DefineThen $B\in \Sigma$, $f$ is a nonnegative measurable function, and $\mu (B)={\int }_B(1-h)d\rho =0$. The identity for integration against $\mu$ follows first for indicators, then for nonnegative simple functions and finally for all nonnegative functions by the monotone convergence theorem. Applying it to $f{1}_A$ givesIn particular ${\int }_{\Omega }fd\mu =\nu (\Omega \setminus B)<\infty$, so $f\in L^1(\mu )$. Additivity on the disjoint measurable sets $A\setminus B$ and $A\cap B$ now givesThis is the Lebesgue decomposition from a sum-measure density: the first measure has a density, and the second is concentrated on a $\mu$-null set, so is mutually singular with $\mu$.

The absolute continuity of measures condition $\nu \ll \mu$ means that $\mu (A)=0$ implies $\nu (A)=0$ for every $A\in \Sigma$. If this holds, then $\nu (B)=0$, eliminating the singular part above. Conversely, a measure of the form $\nu (A)={\int }_{A}fd\mu$ vanishes on every $\mu$-null set. ThereforeThus absolute continuity is exactly the absence of the singular part in this construction.

For a finite-rank operator $T$, define $|T|=(T^{\ast }T{)}^{1/2}$ and its trace norm bywhere $s_j>0$ are its nonzero singular values, counted with multiplicity. The finite-rank spectral theorem gives orthonormal systems $(u_j),(v_j)$ with $T={\sum }_j{s}_j{R}_{u_j,v_j}$, where $R_{x,y}z=⟨z,y⟩x$.

For every bounded operator $B$,Choose the contraction $B$ taking $u_j$ to $v_j$ and vanishing on their orthogonal complement. It attains equality, soThis formula proves the triangle inequality and absolute homogeneity. Also $\Vert T\Vert \le \Vert T{\Vert }_1$, so vanishing of the trace norm forces $T=0$. Thus it is a norm on the vector space of finite-rank operators.

The map $B\mapsto F_B$, where $F_B(T)=\mathrm{Tr}(BT)$, is a bounded linear map into the dual of this normed space. Its norm is at most $\Vert B\Vert$. Rank-one tests give the reverse inequality: $\Vert R_{x,y}{\Vert }_1=\Vert x\Vert \Vert y\Vert$ and $F_B(R_{x,y})=⟨Bx,y⟩$, so taking unit vectors $x,y$ gives $\Vert F_B\Vert =\Vert B\Vert$.

Conversely, let $F$ be a bounded linear functional for the trace norm. The sesquilinear form $\beta (x,y)=F(R_{x,y})$ satisfiesThe Riesz representation theorem gives a unique bounded operator $B$ with $\beta (x,y)=⟨Bx,y⟩$, and $\Vert B\Vert \le \Vert F\Vert$. Since every finite-rank operator is a finite sum of rank-one operators, linearity implies $F(T)=\mathrm{Tr}(BT)$ for every $T$. HenceThe same identification extends to the completion, the trace-class operators. This is trace-class duality; completeness of the original finite-rank space is not required to identify its bounded dual.

Choose an upper ellipticity bound $\Lambda <\infty$ for the symmetric coefficient matrix, possible because its entries are bounded. ThenThe Poincaré inequality on a bounded open set for zero-boundary functions states $\Vert u{\Vert }_2\le C_P\Vert Du{\Vert }_2$. ConsequentlyThe symmetric positive bilinear form $Q$ is therefore an inner product, and its square-root norm is equivalent to the usual Sobolev norm. Its triangle inequality follows from the Cauchy-Schwarz inequality for that inner product. completeness of $W_0^{1,2}$ in its original norm makes it a Hilbert space in the $Q$ norm as well.

The linear functional $\ell (v)=-\int fv$ is bounded, sinceThe Hilbert-space Riesz representation theorem states that every bounded linear functional is the inner product with a unique element. Applied in the $Q$ inner product it gives a unique $u$ with $Q(u,v)=\ell (v)$ for all $v$, which is precisely the weak-solution identity. Alternatively uniqueness follows directly by testing the difference of two solutions with itself: $Q(u_1-u_2,u_1-u_2)=0$. Thus $\boxed{\text{the zero-boundary problem has exactly one weak solution}}$.

Construct the densities using the Riesz representation theorem for a Hilbert space, rather than assuming the measure-decomposition conclusion. Put $\rho =\mu +\nu$, a finite dominating measure, and work in the real L2 space $L^2(\rho )$. The linear functionalis well-defined on its equivalence classes because $\nu \le \rho$. By the Cauchy-Schwarz inequality,Thus the Riesz representation theorem supplies $h\in L^2(\rho )$ such that $L(g)=\int ghd\rho$. Since $\rho$ is finite, $h$ is also integrable. Testing $g=1_A$ givesThe first identity and nonnegativity of $\nu$ rule out a positive-measure set where $h<0$; for example, test $\{h\le -1/n\}$. The second identity and nonnegativity of $\mu$ similarly rule out $\{h\ge 1+1/n\}$. Modify a measurable representative on a $\rho$-null set so that $0\le h\le 1$ everywhere. This is the Hilbert-space construction of dominated measure densities.

SetThen $B\in \Sigma$ and $\mu (B)=0$. The indicator identities for $\mu$ extend to integrals of nonnegative measurable functions by approximation with simple functions and the monotone convergence theorem. Consequently,and, for every $A$,HenceThis proves Lebesgue decomposition from a sum-measure density. The argument includes the cases of zero measures; there is no division on $B$, where the denominator vanishes.

Absolute continuity of measures, written $\nu \ll \mu$, means that $\mu (A)=0$ implies $\nu (A)=0$ for every $A\in \Sigma$. If $\nu \ll \mu$, then $\nu (B)=0$ in the construction above, so the singular term vanishes. Conversely, if $\nu (A)={\int }_{A}fd\mu$ with a nonnegative integrable $f$, that integral is zero on every $\mu$-null set. ThusThis is the finite-measure Radon-Nikodym theorem as a consequence of the proved construction.

Finally restrict the base measure to the sub-sigma-algebra ${\Sigma }_0$. For a real $f\in L^1(\mu )$, write $f=f^{+}-f^{-}$, with both parts nonnegative and integrable. On ${\Sigma }_0$ define finite positive measuresThey are absolutely continuous with respect to $\mu {|}_{{\Sigma }_0}$. The result just proved gives nonnegative, ${\Sigma }_0$-measurable densities $g_{\pm }$ satisfyingSet $f_0=g_{+}-g_{-}$. Since $\int (g_{+}+g_{-})d\mu =\int |f|d\mu <\infty$, this is in $L^1(\Omega ,{\Sigma }_0,\mu )$, and subtraction givesFor complex $f$, apply the real construction separately to its real and imaginary parts and combine the two results. This is conditional expectation from finite-measure densities. The resulting conditional expectation is unique almost everywhere: for two real versions, their difference has integral zero on its positive and negative level sets in ${\Sigma }_0$, forcing it to vanish; apply this to both components in the complex case. No probability normalization of $\mu$ is required.

The appropriate bounded-operator formulation uses the zero-boundary Sobolev spaceThe Poincaré inequality makes this a Hilbert space norm equivalent to the usual $H^1$ norm, and the zero endpoint traces encode the Dirichlet boundary conditions. Under the usual regular-coefficient assumptions, for example continuous $p,q$ on the closed interval with $p>0$ and $q\ge 0$, setIf $p_{\ast }=\min p>0$, then Cauchy-Schwarz inequality and the Poincaré inequality giveBy the Riesz representation theorem, a unique bounded linear operator $T:H\to H$ satisfies $(Tu,v{)}_H=a(u,v)$. Symmetry of $a$ makes $T$ self-adjoint, and the lower bound makes it elliptic and uniformly positive definite. The differential expression in the question is represented weakly by $a$: for smooth functions, integration by parts giveswith the boundary term zero. A forcing $f\in L^2$ becomes the bounded functional $v\mapsto \int fv$, or its Riesz representative in $H$.

Equivalently, for sufficiently smooth coefficients one may realize the differential operator itself on $L^2(-1,1)$ with domain $H^2\cap H_0^1$. Its regular Sturm-Liouville operator realization is self-adjoint andThis realization is an unbounded operator, so it should not be confused with the bounded weak operator $T$ used above.

The PDF states sign conditions without coefficient regularity. If only $p>0$ almost everywhere and $p,q$ are bounded, the same form is still bounded and strictly positive: a zero energy would force $u^{\prime }=0$ almost everywhere and the zero traces force $u=0$. Uniform ellipticity, however, requires a positive lower bound. For instance $p(x)=x^2$ for $x\ne 0$, $p(0)=1$, and $q=0$ satisfy literal pointwise positivity. Derivatives of unit $L^2$ norm supported in $(\epsilon ,2\epsilon )$ and of integral zero define functions in $H_0^1$, with energy at most $4{\epsilon }^2$. Thus pointwise positivity without regularity does not imply coercivity in this $H$ norm. These distinctions supply the precise regularity and operator domain behind the intended positive-definiteness proof.