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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 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.

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

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.

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 Poincare 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 Poincare 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.

The Riesz representation theorem states that every bounded linear functional $L$ on a real or complex Hilbert space $H$ is represented by a unique $h\in H$:For the complex case take the inner product to be linear in its first argument.

If $L=0$, choose $h=0$. Otherwise its kernel $K$ is a closed linear subspace. Choose $x$ with $L(x)\ne 0$ and let $z=x-P_{K}x$, using the orthogonal projection. Then $z\ne 0$, $z\perp K$, and $L(z)=L(x)\ne 0$. For every $v$,Therefore take $h=L(z)z/\Vert z{\Vert }^2$ in the real case, and $h=\overline{L(z)}z/\Vert z{\Vert }^2$ in the complex case. The conjugate in the latter formula compensates for conjugate linearity in the second argument.

The Cauchy-Schwarz inequality gives $|L(v)|\le \Vert v\Vert \Vert h\Vert$, and evaluation at $v=h/\Vert h\Vert$ when $h\ne 0$ gives equality of the norms. If two vectors represent $L$, their difference is orthogonal to every vector, including itself, hence zero. This proves all assertions of the Riesz representation theorem.

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 $⟨g,v⟩$.

Work in the mean-zero Sobolev spaceThe integral is a continuous functional on $H^1(U)$, so $H$ is closed. The Neumann-Poincare inequality shows that $\Vert v{\Vert }_a=\Vert \nabla v{\Vert }_2$ is equivalent to the usual $H^1$ norm on $H$, making $a$ a complete inner product there.

For $f\in L^2(U)$ the functional $L(v)={\int }_{U}fv$ satisfiesThe Riesz representation theorem, or the Lax-Milgram theorem, gives a unique $u\in H$ with $a(u,v)=L(v)$ for all $v\in H$. To recover every $H^1(U)$ test, write $v=(v-v_U)+v_U$. The constant contributes zero to $a$ and contributes $v_U{\int }_{U}f=0$ to $L$. Thus the same equality holds for all $v\in H^1(U)$.

A weak solution exists whenever ${\int }_{U}f=0$; fixing its average to zero makes it unique. Moreover $\Vert \nabla u{\Vert }_2\le \sqrt{C_P}\Vert f{\Vert }_2$ and the Neumann-Poincare inequality also controls $\Vert u{\Vert }_2$.

For $v\in C_c^{\infty }(U)$, two integrations by parts give the clamped Hessian identityBy density it remains valid on $H_0^2(U)$. Each ${\partial }_{i}v$ has zero integral, first for compactly supported test functions and then by $H^2$ convergence. Applying the Neumann-Poincare inequality to each ${\partial }_{i}v$ givesThe zero-boundary Poincare inequality also gives $\Vert v{\Vert }_2^2\le C_D\Vert \nabla v{\Vert }_2^2$. Hence, using a full-Hessian equivalent $H^2$ norm,Thus $a(u,v)={\int }_U\Delta u\Delta v$ is an inner product whose norm is equivalent to the complete $H^2$ norm on the clamped second-order Sobolev space. The functional $L(v)={\int }_{U}fv$ is bounded for this norm by the Cauchy-Schwarz inequality and the displayed bound. Apply the Riesz representation theorem, or the Lax-Milgram theorem, to get a unique $u\in H_0^2(U)$ representing $L$. The clamped biharmonic problem has a unique weak solution for every $f\in L^2(U)$, with $\Vert u{\Vert }_{H^2}\le C\Vert f{\Vert }_2$ and no zero-integral compatibility condition.

Use an inner product linear in the first variable. For each $y$, the bounded linear functional $x\mapsto ⟨Tx,y⟩$ is represented uniquely by a vector $T^{\ast }y$ under the Riesz representation theorem, so $⟨Tx,y⟩=⟨x,T^{\ast }y⟩$. Uniqueness and the conjugate-linearity in the second argument imply $T^{\ast }(ay+bz)=a{T}^{\ast }y+b{T}^{\ast }z$. FurthermoreThus the adjoint operator exists, is linear and bounded, and is unique because a vector orthogonal to every $x$ is zero.

A normal operator satisfies $T^{\ast }T=T{T}^{\ast }$. The unilateral unilateral shift operator $S(x_1,x_2,\dots )=(0,x_1,x_2,\dots )$ on ${\ell }^2$ is bounded but not normal: $S^{\ast }S=I$, whereas $S{S}^{\ast }=I-P_{e_1}$. For general $T$,Normality implies equality of norms. Conversely equality for every $x$ makes this self-adjoint operator's quadratic form zero; the polarization identity then gives $T^{\ast }T-T{T}^{\ast }=0$.

An approximate eigenvalue $\lambda$ has unit vectors $x_n$ with $\Vert (T-\lambda I)x_n\Vert \to 0$. It is necessarily in the spectrum, since a bounded inverse would prevent this. Conversely suppose $\lambda$ is not an approximate eigenvalue. Then $A=T-\lambda I$ is bounded below, $\Vert Ax\Vert \ge c\Vert x\Vert$. It is injective and has closed range: a convergent sequence $A{x}_n$ forces $x_n$ to be Cauchy. The operator $A$ is normal, so $\Vert A^{\ast }x\Vert =\Vert Ax\Vert$ and $\ker A^{\ast }=0$. The identity $(\mathrm{ran}A{)}^{\perp }=\ker A^{\ast }$ makes its range dense. Closed and dense means surjective, and its inverse is bounded by $1/c$. Hence $\lambda$ is in the resolvent. This proves $\boxed{\sigma (T)={\sigma }_{\mathrm{ap}}(T)}$.