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Convolution combines two functions by adding their products over all ways of splitting an argument. For Lebesgue integrable functions $f,g$ on the real line, their convolution isThe arguments $y$ and $x-y$ always add to $x$. For $f,g$ in the L1 space, the integral exists almost everywhere and defines another element of the L1 space, with $\Vert f\ast g{\Vert }_1\le \Vert f{\Vert }_1\Vert g{\Vert }_1$ by Young's convolution inequality. Particular examples below have a value at every $x$.

To see what the integral counts, take the indicator function $q=1_{[0,1]}$. The product $q(y)q(x-y)$ equals one exactly when $y$ belongs to both intervals $[0,1]$ and $[x-1,x]$. Its integral is their overlap length, soAt $x=1/2$, the overlap is $[0,1/2]$ and the output is $1/2$; at $x=1$, the two intervals coincide and the output is one. The result is the shifted triangular function $\mathrm{tri}(x-1)$. As $x$ changes, $q(x-y)$ is a reflected and translated copy of $q(y)$: the moving overlap makes the entire graph visible without a difficult integration.

The same calculation gives the probability distribution of a sum of two independent random variables. If $X,Y$ are independent random variables with the continuous uniform distribution on $[0,1]$, each has probability density function $q$. Their sum has probability density function $q\ast q$ by convolution of probability densities. The probability density function at one is larger than near zero or two because there are more combinations of the two inputs giving that sum. Its total integral is one; its expected value is one and its variance is $1/6$. These follow equally from integrating the triangular function or from linearity of expectation and variance additivity for independent random variables.

Convolution also gives a useful averaging operation. For $\epsilon >0$, let $k_{\epsilon }(y)=1_{[-\epsilon ,\epsilon ]}(y)/(2\epsilon )$. Thenthe mean of $f$ in a neighborhood of $x$. A smaller $\epsilon$ examines a smaller neighborhood. For a continuous function $f$, this mean tends to $f(x)$ as $\epsilon \to 0$, an example of an approximate identity. Integrating a sharp input against a distributed convolution kernel can produce a smoother output; the triangular example turns two discontinuous inputs into a continuous output.

In signal processing, a linear time-invariant system combines shifted copies of its impulse response by the same rule. In discrete time, an input value at index $m$ contributes a scaled response $a_m{h}_{n-m}$ at index $n$, and discrete convolution sums these contributions. This is the superposition interpretation developed in Oppenheim, MIT Signals and Systems, Lecture 4, convolution sums and integrals. The convolution theorem then gives another description: transform both inputs with a Fourier transform and multiply their transforms. Combining shifts, combining independent quantities and multiplying frequency descriptions are three faces of the same operation.

An integrable kernel is uniquely determined by its convolution operator on $L^2$. Convolve it with a smooth compactly supported approximate identity ${\eta }_{\epsilon }$, whose members also lie in $L^2$. A zero operator makes $h\ast {\eta }_{\epsilon }=0$, while translation continuity in Lp gives $h\ast {\eta }_{\epsilon }\to h$ in $L^1$, proving the assertion. Since the adjoint operator of convolution by $f$ is convolution by $f^{♯}(s)=\overline{f(-s)}$, this also proves the exact self-adjointness criterion $f=f^{♯}$ almost everywhere, including necessity at the level of kernels.

The conformal map $w=z^2$ sends $x,y>0$ to the complex upper half-plane. For bounded continuous boundary values $g_2$ on the horizontal axis and $g_1$ on the vertical axis with equal corner values, set $f(s)=g_2(\sqrt{s})$ for $s>0$ and $f(s)=g_1(\sqrt{-s})$ for $s<0$. The bounded harmonic function with those Dirichlet boundary conditions isEquivalently, if $z^2=a+ic$ with $c>0$, this is $\frac{2c}{\pi }{\int }_0^{\infty }r\left[g_2(r)/((r^2-a{)}^2+c^2)+g_1(r)/((r^2+a{)}^2+c^2)\right]dr$. The Poisson kernel is a positive approximate identity, so the prescribed edge values are attained. Boundedness is essential for uniqueness on this unbounded domain.

Let $I_C[F]=(2\pi {)}^{-1}{\int }_C{e}^{ikx-(k^2-i\alpha k)t}F(k)dk$. The drift-diffusion half-line global relation and the half-line drift null contour identity give the formula in the secondary title. Its real-space kernel is the difference of two drifted Gaussian heat kernels, and its boundary-forcing kernel is $x(2\sqrt{\pi }{t}^{3/2}{)}^{-1}\exp [-(x+\alpha t{)}^2/(4t)]$. This kernel is an approximate identity in time as $x\downarrow 0$, which verifies the Dirichlet boundary condition.

The Fourier inversion theorem makes decomposition into frequencies reversible: a complete Fourier transform, including its wave phases, determines the original function up to equality almost everywhere. With the angular-frequency conventionif $f$ and $\hat{f}$ belong to the L1 space, thenalmost everywhere. The right side is a continuous representative of $f$. If $f$ itself is continuous, equality holds everywhere, as explained by the continuous version of Fourier inversion. The integrability hypotheses ensure that both displayed integrals are ordinary absolutely convergent integrals. For arbitrary elements of the L2 space, the Plancherel theorem instead supplies inversion in the L2 space norm.

Each factor $e^{ix\cdot \xi }$ is a Fourier mode; the inverse integral adds them with the complex weights $\hat{f}(\xi )/(2\pi {)}^n$. The normalization is part of the recipe. In one dimension, writing ordinary frequency as $\nu =\xi /(2\pi )$ gives the equivalent pairwhere $F(\nu )=\hat{f}(2\pi \nu )$. Reversing both exponential signs gives another equivalent convention, but the forward and inverse formulas must use opposite signs consistently.

For example, the Fourier transform of a Gaussian givesFourier inversion says that a weighted continuum of oscillations reconstructs this smooth Gaussian function. Narrowing the original Gaussian function by increasing $a$ broadens its frequency spectrum. This is a concrete demonstration that a sharply localized shape needs a wider range of frequencies.

Inversion is what makes a frequency calculation useful: transform a function, solve the simpler problem there, then return to the original variables. A brief proof outline is to multiply $\hat{f}(\xi )$ by $e^{-ϵ|\xi {|}^2}$, $ϵ>0$. The inverse integral becomes convolution with a normalized Gaussian function, an approximate identity. As $ϵ$ tends to zero, dominated convergence gives the unregularized inverse integral, while the approximate identity recovers $f$ almost everywhere. Joseph Fourier's Théorie analytique de la chaleur (1822), Chapter IX, article 417, contains a cosine double-integral representation antecedent to this modern formula; the hypotheses above are the modern statement.^[ref]

The centered normal densitiesform an approximate identity: for every $f\in L^1(\mathbb{R})$, $f\ast g_t\to f$ at almost every point where the Lebesgue differentiation theorem applies. Their Fourier transforms are $e^{-t{\xi }^2/2}$ in the angular-frequency convention.

The normalized Gaussian mass of an interval isIt is the convolution of an interval indicator with the heat kernel, and has an error function representation. As the variance tends to zero it converges to the interval indicator away from its endpoints, with value one half at an endpoint. This is a concrete approximate identity.

Convolution of this kernel with Dirichlet boundary data supplies the boundary forcing for $u_t=u_{xx}+\alpha u_x$ on $x>0$. It is $-(2{\partial }_x+\alpha )H(x+\alpha t,t)$. For $\alpha \ge 0$, its total time mass is $e^{-\alpha x}$, tending to one as $x\downarrow 0$, and its mass away from zero time tends to zero. Thus the boundary value is recovered through an approximate identity, not by pointwise substitution $x=0$ in the kernel.

For the heat equation on ${\mathbb{R}}^n$ with diffusion coefficient one, the heat kernel is $K_t(x)=(4\pi t{)}^{-n/2}{e}^{-|x{|}^2/(4t)}$. Its convolution with bounded initial data is smooth for positive time, solves the equation by differentiation under the integral, and returns continuous initial data locally uniformly through the approximate identity property. It is Gaussian filtering with per-coordinate variance $2t$.

For a bounded Fourier multiplier operator $T$ on $L^2$ and a bump function $\varphi$, the displayed convolution identity follows by multiplying Fourier transforms. The convergence holds in $L^2$ and almost everywhere by the approximate identity theorem, after accounting for the integral of the kernel. The limit is $T(f)$ when $\int \varphi =1$. Smoothness and compact support alone do not imply this normalization.

The inverse Fourier transform of $(1-|\xi |{)}_{+}$ in three dimensions, with $a=2\pi |x|$, is $-4\pi \sin a/a^3+8\pi (1-\cos a)/a^4$. Its leading oscillatory $|x{|}^{-3}$ tail is not absolutely integrable against the radial volume element $4\pi |x{|}^{2}d|x|$. Multiplying by a Gaussian Fourier transform retains a nonzero boundary-derivative tail, so the radial cutoff need not map even Schwartz functions into $L^1$. A tensor product of one-dimensional triangular cutoffs is a valid approximate identity, and must not be confused with this Euclidean radial cutoff.

If $L>0$ and $f$ is bounded and continuous at zero, the positive kernel $n{e}^{-nx}$ has mass $1-e^{-nL}$ on $[0,L]$, tending to one, and mass at most $e^{-n\delta }$ outside $[0,\delta ]$. Split the integral of $f(x)-f(0)$ at $\delta$: continuity makes the near part arbitrarily small, and boundedness times the exponentially small tail controls the far part. This proves the displayed endpoint approximate identity. A shrinking cutoff $L_n=n^{-1/2}$ still captures mass $1-e^{-\sqrt{n}}$ and therefore gives the same limit.

Define $f^{♯}(s)=\overline{f(-s)}$. Interchanging the integrals and substituting variables givesMoreover Cauchy-Schwarz in the inner integral givesTaking the supremum over unit $g_2$ proves $\Vert T_f{g}_1{\Vert }_2\le \Vert f{\Vert }_1\Vert g_1{\Vert }_2$, so $T_f$ is bounded with norm at most $\Vert f{\Vert }_1$, and the adjoint identity is justified.

It remains to ensure necessity is a condition on the kernel itself. If $T_h=0$ for some $h\in L^1$, convolve $h$ with a smooth compactly supported approximate identity ${\eta }_{\epsilon }\in L^1\cap L^2$. Since $T_h{\eta }_{\epsilon }=0$, all these convolutions vanish. But $h\ast {\eta }_{\epsilon }\to h$ in $L^1$, by continuity of translations in $L^1$, so $h=0$ almost everywhere. Thus the map from kernels to convolution operators is injective. Applying this to $h=f-f^{♯}$ yields the exact criterionFor real-valued kernels this reduces to evenness; real-valuedness by itself is insufficient.

For $\epsilon >0$, insert a Gaussian function into the proposed inverse Fourier transform:The Fourier bounds for Schwartz functions show that $\hat{\varphi }$ is an integrable function, so dominated convergence givesOn the other hand, Fubini's theorem applies because $\Vert \varphi {\Vert }_1\int e^{-\epsilon {\xi }^2/2}d\xi <\infty$. The given Fourier transform of a Gaussian, rescaled by $\sqrt{\epsilon }$, yieldsThe last integrand is bounded by $\Vert \varphi {\Vert }_{\infty }{e}^{-z^2/2}/\sqrt{2\pi }$. Since $\varphi$ is continuous, another application of dominated convergence makes this tend to $\varphi (x)$. Equating the limits proves the Fourier inversion theorem:The Gaussian functions above form an approximate identity; no Fourier inversion was used in evaluating their integrals.

Choose a nonnegative mollifier $\rho \in C_c^{\infty }({\mathbb{R}}^3)$ with $\int \rho =1$, and put ${\rho }_{\epsilon }(x)={\epsilon }^{-3}\rho (x/\epsilon )$. Sobolev approximation by mollifiers in the Sobolev space under consideration statesIn particular, it gives density of smooth functions in a Sobolev space. We prove convergence of both the function and all its first weak derivatives.

The usual smoothing properties of convolution give $u_{\epsilon }\in C^{\infty }$. For a test function $\varphi \in C_c^{\infty }({\mathbb{R}}^3)$ and $i\in \{1,2,3\}$, Fubini's theorem and the definition of a weak derivative giveThe integrals are absolutely convergent: the compact supports of ${\rho }_{\epsilon }$ and $\varphi$ restrict $y$ to a compact set, on which $u$ and ${\partial }_{i}u$ are integrable by Holder inequality. ConsequentlyThe approximate identity theorem in the Lp space $L^6$ applies separately to $u$ and ${\partial }_{i}u$. ThusEach convolution belongs to $L^6$ by Young's convolution inequality. With the componentwise Sobolev norm, the sixth power of the required difference is the sum of these four sixth powers, proving the boxed claim. Any equivalent Sobolev norm gives the same convergence.

The Weierstrass approximation theorem asserts that every real continuous function on a compact interval is a uniform limit of polynomials. An affine change of variable reduces the interval to $[0,1]$. There are two complementary routes: construct positive approximation operators, or construct an algebra dense enough to approximate absolute values and hence pointwise maxima and minima.

For the first route, the Korovkin theorem reduces convergence of positive linear operators on continuous functions $L_n$ to the three tests $1,t,t^2$. Here is its proof. Given $f\in C[0,1]$ and $\epsilon >0$, uniform continuity supplies a $\delta >0$ such that $|f(t)-f(x)|<\epsilon$ for $|t-x|<\delta$. For all $t,x$ one then hasApply positivity to the two real inequalities obtained from this bound. At $x$ this givesThe expression in brackets is nonnegative and tends uniformly to zero by the three test assumptions. Also $L_{n}1\to 1$ uniformly. Taking the upper limit of the supremum norm and then letting $\epsilon \downarrow 0$ proves convergence for every $f$. This is the quadratic barrier estimate for positive approximation operators.

The Bernstein polynomial supplies the constructive corollaryThe binomial theorem makes its nonnegative weights sum to one. Differentiating the binomial theorem, or using the first two binomial distribution moments, givesThe Korovkin theorem proves $B_{n}f\to f$ uniformly, and each $B_{n}f$ is a polynomial, proving the Weierstrass approximation theorem. The same criterion gives the Schoenberg spline operator convergence in Solution 3. For periodic functions its counterpart uses $1,\cos t,\sin t$: the nonnegative barrier $1-\cos (t-x)$ is bounded away from zero outside any fixed small arc. This periodic Korovkin test set gives convergence of the normalized Fejér sums and density of trigonometric polynomials.

The Lebesgue convergence criterion for approximation operators addresses stability. Suppose bounded linear operators $L_n:C[a,b]\to C[a,b]$ converge to the identity on every polynomial. Then they converge on every continuous function if and only if ${\sup }_n\Vert L_n\Vert <\infty$. Sufficiency is the extension of approximation operators from a dense subspace: for a polynomial $p$ with $\Vert f-p\Vert <\eta$ and a common bound $M$,First let $n\to \infty$, then $\eta \downarrow 0$. Necessity follows from the Uniform boundedness principle, since a convergent $L_{n}f$ is bounded for every $f$. Applied to approximation projectors, the Lebesgue inequality for approximation projectors isIndeed $P_{n}v=v$ makes $f-P_{n}f=(f-v)-P_n(f-v)$; take the infimum. Thus density of the ranges and a uniform bound on the Lebesgue constants of interpolation imply convergence. Density alone does not control a sequence of interpolation projectors.

There is also a kernel version of this stability criterion, useful for trigonometric polynomial approximation. If integrable periodic kernels $K_n$ have uniformly bounded $L^1$ norms, their integrals tend to one, and ${\int }_{|t|\ge \delta }|K_n(t)|dt\to 0$ for every $\delta >0$, then convolution with $K_n$ converges uniformly to every continuous periodic function. Subtract $f(x)\int K_n$ and split the error into a small arc, bounded by $\omega (f,\delta ){\sup }_n\Vert K_n{\Vert }_1$, and its complement, bounded by $2\Vert f{\Vert }_{\infty }{\int }_{|t|\ge \delta }|K_n|$. The remaining mass error tends to zero. This approximate identity argument shows directly why nonnegative unit-mass Fejér kernels work even though ordinary Fourier partial sums need not converge uniformly for every continuous target.

Lebesgue's constructive approach to the Weierstrass approximation theorem starts instead with the absolute value. On $[-1,1]$ define the polynomial recurrence for the absolute valueInduction gives $0\le p_r(x)\le |x|$, monotonicity, andThe pointwise limit satisfies $p(x{)}^2=x^2$ and is nonnegative, hence equals $|x|$. Dini's theorem makes the convergence uniform. More explicitly, the continuous decreasing errors have, for any positive tolerance, open sublevel sets covering the compact interval; a finite subcover supplies one index valid everywhere. Every continuous piecewise linear function has the formThe $c_j$ are its changes in slope. Scaling and translating the absolute-value approximation shows that $L$ is uniformly approximable by polynomials. Solution 4, or polygonal interpolation and uniform continuity, makes these piecewise linear functions dense in $C[a,b]$. Approximating $f$ first by $L$ and then by a polynomial is another complete proof of the Weierstrass approximation theorem.

The Stone-Weierstrass theorem generalizes from intervals to a compact Hausdorff space $K$: a real subalgebra $A\subseteq C(K,\mathbb{R})$ containing the constants and separating points is uniformly dense. Its proof uses the same absolute-value mechanism. The uniform closure of a function algebra is an algebra. If $a$ belongs to this closure, approximation of $|t|$ by polynomials on a bounded interval containing its range proves that $|a|$ also belongs to the closure. Therefore the closure containsPoint separation and constants allow exact interpolation of any two real values: choose $h\in A$ with $h(s)\ne h(t)$ and solve for $\alpha ,\beta$ in $\alpha h+\beta$. The lattice approximation from two-point approximation proved in Solution 4 now makes the closure all of $C(K,\mathbb{R})$.

The complex version requires a unital, point-separating subalgebra closed under complex conjugation. Its real-valued members separate points using real or imaginary parts, so the real result approximates both parts of every complex target. Conjugation closure is essential: polynomials in $z$ alone on the closed unit disk cannot approximate $\overline{z}$ uniformly, because ${\int }_{|z|=1}p(z)dz=0$ whereas ${\int }_{|z|=1}\overline{z}dz=2\pi i$.

Among the corollaries are density of multivariate polynomials on compact subsets of real Euclidean space, since coordinates separate points, and density of trigonometric polynomials on the circle, since sine and cosine separate points there. On a compact interval, polynomials with rational coefficients are a countable dense family, giving separability of $C[a,b]$. The real and complex approximation results above concern uniform approximation of continuous targets; neither asserts convergence of arbitrary interpolation schemes or arbitrary Fourier series.

Use the Fourier transform convention $\hat{h}(\xi )=\int h(x)e^{-2\pi ix\cdot \xi }dx$. The deleted-ball kernel need not be integrable at infinity, so first work with finite annuli. WriteWe will prove a uniform bound, convergence as $R\to \infty$, and then convergence as $\epsilon \downarrow 0$. This makes all Fourier-transform assertions precise for the annularly cancelling singular convolution kernel.

Fix $\xi \ne 0$ and put $y=\xi /(2|\xi {|}^2)$, so $e^{-2\pi iy\cdot \xi }=-1$. For $R>r\ge 4|y|$, let $\Omega =\{r<|x|<R\}$ and defineThe Hörmander integral kernel condition givesOn changing variables $u=x-y$, the phase factor shows $J_{r,R}=-{\int }_{\Omega -y}K(u)e^{-2\pi iu\cdot \xi }du$. The symmetric difference between $\Omega$ and $\Omega -y$ lies in the two shells whose radii are within $|y|$ of $r$ and $R$. On each such shell, the kernel's size bound gives, by polar integration,Thereforeand adding the two comparisons proves the oscillatory tail estimate for a singular kernel:The integral on the right is at most $B$. Moreover, for fixed $\xi$, it tends to zero as $r\to \infty$ because the translated difference is absolutely integrable outside $2|y|$. The shell term also tends to zero. Thus the far Fourier integral is Cauchy and converges as its outer radius tends to infinity.

Set $r_0=4|y|=2/|\xi |$. If $\epsilon <r_0$, annular cancellation allows subtraction of the constant phase on the near annulus:The same estimate applies to a shorter near annulus when $R<r_0$. The far annulus is bounded by $B/2+C_{d}A$, using the tail estimate; if $\epsilon \ge r_0$, apply that estimate with $r=\epsilon$. At $\xi =0$ the annular integral vanishes by cancellation. We have provedFor each fixed $\epsilon >0$, the size bound makes $K_{\epsilon }\in L^2$, and $K{1}_{\{\epsilon <|x|<R\}}\to K_{\epsilon }$ in $L^2$ as $R\to \infty$. By the Plancherel theorem, their Fourier transforms converge in $L^2$ to ${\hat{K}}_{\epsilon }$. The pointwise improper limits just established identify this transform, up to a null set, withFor example identification follows by selecting an almost-everywhere convergent subsequence from the $L^2$ convergence. HenceThe near-annulus integral with the constant phase subtracted converges absolutely at zero, since its bound is proportional to $|x{|}^{1-d}$. The far integral has already converged. Therefore $m_{\epsilon }(\xi )\to m(\xi )$ for every $\xi \ne 0$, with the same uniform bound; the value at zero may be set arbitrarily.

For $f\in L^2$, the convolution $K_{\epsilon }\ast f$ is absolutely convergent for each $x$ by Cauchy-Schwarz inequality. Its Fourier transform is $m_{\epsilon }\hat{f}$: first prove this for Schwartz functions and then approximate any $L^2$ input. The convolutions converge uniformly under this approximation by the same Cauchy-Schwarz bound, and the multiplier outputs converge in $L^2$ by their uniform operator bound, so these two definitions agree. Define the bounded Fourier multiplier operator $T$ by $\widehat{Tf}=m\hat{f}$. Dominated convergence theorem and the Plancherel theorem now give

For the final request, use a normalized bump function, $\int \varphi =1$, so that its dilates form an approximate identity. Multiplying Fourier transforms givesThese products are integrable by Cauchy-Schwarz inequality, since $T{\varphi }_{\epsilon },f\in L^2$, and also define $L^2$ functions because $m$ and $\hat{\varphi }$ are bounded. Thus the convolution identity is justified even though $T{\varphi }_{\epsilon }$ need not be in $L^1$:The allowed standard approximate identity results, applied to $Tf\in L^2$, prove convergence to $Tf$ in norm and almost everywhere. This is the mass normalization in mollified singular integrals; it does not require proving almost-everywhere convergence of the sharp truncations $T_{\epsilon }f$.

If “bump function” means merely smooth with compact support, the stated limit needs the unit-integral qualification. For arbitrary such $\varphi$, the limit is $(\int \varphi )Tf$ instead. This is also the almost-everywhere limit: at each Lebesgue point of $Tf$, subtract $(\int \varphi )Tf(x)$ inside the convolution and bound the remaining local average by the boundedness and compact support of $\varphi$. As a concrete counterexample without normalization, take $K(x)=1/(\pi x)$ in dimension one and $\varphi =2\rho$ for a unit-integral mollifier $\rho$. This kernel has the required size, odd annular cancellation, and translation-difference integral $\log (3)/\pi$ on $\{|x|>2|y|\}$ for every $y\ne 0$. Its limit operator is the nonzero Hilbert transform; by the Hilbert-transform Fourier multiplier, it is an $L^2$ isometry. For any nonzero $f\in L^2$ the mollified limit is $2Tf$, not $Tf$.

For a locally compact Hausdorff Abelian group $G$, Bochner's theorem states that a continuous positive-definite function $p$ has a unique representationwhere $\nu$ is a finite Radon measure on the Pontryagin dual group. Conversely every such measure gives a continuous positive-definite function. Here positive definiteness means ${\sum }_{i,j}{c}_i\overline{c_j}p(x_i{x}_j^{-1})\ge 0$ for all finite choices. The Abelian and continuity hypotheses are part of the theorem; the representation is not claimed for arbitrary non-Abelian groups or discontinuous functions.

First prove existence without presupposing the representation. The elementary consequences of positive definiteness give $p(x^{-1})=\overline{p(x)}$, $p(e)\ge 0$ and $|p(x)|\le p(e)$. If $p(e)=0$, then $p=0$ and take $\nu =0$. Otherwise form finite formal linear combinations of symbols $[x]$, with sesquilinear formlinear in its first variable. Positive definiteness makes this form positive semidefinite. The Cauchy-Schwarz inequality for such a form shows that null vectors are orthogonal to everything, so quotienting by them gives an inner product space. Complete it to a Hilbert space $H$. Since $G$ is Abelian, translation $U_a[x]=[ax]$ preserves the form and extends to a unitary operator. Its inverse is $U_{a^{-1}}$, so $(U_a)$ is a unitary representation. With $v=[e]$,The translated copies of $v$ span a dense subspace. MoreoverFinite sums and density, together with $\Vert U_a\Vert =1$, give strong continuity on all of $H$. This constructs the cyclic unitary representation of a positive-definite function directly.

For $f\in L^1(G)$ define the integrated unitary representation $\pi (f)=\int f(x)U_{x}dx$, with the integral applied to each vector. Its norm is at most $\Vert f{\Vert }_1$. Fubini gives $\pi (f\ast g)=\pi (f)\pi (g)$, and inversion in the Abelian group gives $\pi (f^{\ast })=\pi (f{)}^{\ast }$. Thusis a commutative C-star algebra. It is nondegenerate: choose nonnegative compactly supported functions $u_V$ of Haar integral one with supports shrinking to the identity element. Strong continuity impliesfor every $w\in H$. These form an approximate identity in the representation. Neighbourhoods may form a net, since no metrizability is being assumed here.

The Commutative Gelfand--Naimark theorem identifies $\mathcal{A}$ with $C_0(X)$, where $X$ is its character space. The positive vector functional $\omega (B)=⟨Bv,v⟩$ becomes a bounded positive functional on $C_0(X)$, so the Riesz-Markov-Kakutani representation theorem supplies a finite positive Radon measure $\rho$ withThese general representation theorems are applied to the explicitly constructed algebra; they are not an assumption of Bochner's conclusion.

For each $\theta \in X$, the functional $f\mapsto \theta (\pi (f))$ is a nonzero algebra character of $L^1(G)$, since $\pi (L^1(G))$ is dense in $\mathcal{A}$. By Question 3 it has the form $\int f(x){\chi }_{\theta }(x)dx$. The resulting map $\iota :X\to \hat{G}$ is continuous: pointwise convergence of $\theta$ on $\mathcal{A}$ gives convergence on all $\pi (f)$, and the topological character identification in Question 3 gives compact-open convergence of ${\chi }_{\theta }$. Push $\rho$ forward under $\iota$ to obtain $\nu$. This finite measure is Radon: compact subsets approximating the measure of any inverse-image Borel set have compact images, giving inner regularity; finiteness and complements then give outer regularity.

To recover $p(x)$, note that $U_x\pi (u_V)=\pi (T_x{u}_V)\in \mathcal{A}$. HenceThe left side tends to $p(x)$. On the right, $|{\Phi }_{\chi }(u_V)|\le 1$ and ${\Phi }_{\chi }(u_V)\to 1$ uniformly on compact subsets of $\hat{G}$. For completeness, the evaluation $(\chi ,x)\mapsto \chi (x)$ is jointly continuous: near $x_0$ restrict $x$ to a compact neighbourhood, control the character uniformly there, and use continuity of the fixed character. A finite cover of a compact set of characters then makes those characters equicontinuous at $e$, proving the stated uniform convergence. Continuity of $\iota$ gives uniform convergence on compact subsets of $X$, and the finite Radon measure $\rho$ makes the remaining tails arbitrarily small. Thus passage to the limit is valid even for a net, andTaking $x=e$ also gives the required total mass $p(e)$.

Conversely, a finite positive Radon measure givesIts transform is continuous. Indeed, choose a compact set of characters with arbitrarily small measure outside it; joint continuity of evaluation gives uniform continuity in $x$ near the specified point on that compact set, while the outside contribution is bounded by twice its measure. This proves continuity without using an unjustified dominated-convergence assertion for arbitrary nets.

Finally suppose two finite Radon measures have the same transform. Multiply by $f\in L^1(G)$ and integrate over $G$. Fubini shows that the measures agree on every function $\chi \mapsto {\Phi }_{\chi }(f)$. Question 3 showed that these functions are uniformly dense in $C_0(\hat{G})$. Boundedness of both measure functionals extends equality to all of $C_0(\hat{G})$; uniqueness in the Riesz representation theorem gives equality of the measures. This proves existence, uniqueness and the converse, completing Bochner's theorem. In particular $p(e)=1$ corresponds exactly to a probability measure on the dual group.

Use the positive Laplace-Beltrami operator $\Delta =-\mathrm{div}\nabla$. For the global discrete spectral statements, take $M$ to be a closed manifold with a smooth Riemannian metric. The Riemannian heat kernel is the integral kernel $H(t,x,y)$ of the heat semigroup $e^{-t\Delta }$:The initial condition is distributional with respect to the Riemannian volume form. Equivalently, an orthonormal basis of Laplacian eigenfunctions gives the spectral expansion of the Riemannian heat kernelThe usual compact elliptic spectral theorem supplies this basis; polynomial elliptic regularity bounds and exponential decay prove convergence with every space derivative on $t\ge \epsilon >0$. These analytic results may be used as clearly stated external ingredients. We will also construct the kernel locally and prove its short-time expansion.

On a sufficiently small neighbourhood of the diagonal, let $r=d(x,y)$ and write $x={\exp }_y\xi$ in geodesic normal coordinates, with $d{V}_x=J(y,\xi )d\xi$. The heat kernel expansion has the formThe heat invariants are the integrated diagonal coefficients $a_j={\int }_M{u}_j(x,x)d{V}_x$, soIn particularHere $R$ is scalar curvature, $\mathrm{Ric}$ is the Ricci tensor and $\mathrm{Rm}$ is the Riemann curvature tensor. The last expression is the integrated second scalar heat coefficient; a local divergence term integrates to zero on a closed manifold. The exponent of the leading heat-trace term determines dimension, and these coefficients determine volume, total scalar curvature, and the displayed integrated curvature combination. On a surface $R=2K$, where $K$ is Gaussian curvature, so $a_1=\frac{1}{3}\int KdA=\frac{2\pi }{3}\chi (M)$ by the Gauss-Bonnet theorem, and $a_2=\frac{1}{15}\int K^{2}dA$. Thus the Euler characteristic, and the genus of a connected orientable closed surface, are determined. The first three coefficients also detect constant Gaussian curvature, since they determine $\int (K-\overline{K}{)}^{2}dA$. These conclusions do not claim that the pointwise curvature function or the metric is determined.

Here is the local construction. Put $g_t=(4\pi t{)}^{-m/2}{e}^{-r^2/(4t)}$. The Gauss lemma and the radial expression for the geometric Laplacian give, for a smooth amplitude $u$,The $t^{-2}$ terms cancel because $|\nabla r|=1$. Define $D=r{\partial }_r+(r/2){\partial }_r\log J$. The singular term disappears with $u_0=J^{-1/2}$, normalized by $u_0(y,y)=1$. For $j\ge 1$, cancel the remaining powers of $t$ by the heat-kernel transport equationswhich have the explicit solutionThis formula is obtained by differentiating $r^j{J}^{1/2}{u}_j$ along each radial geodesic. It proves smoothness jointly in $x,y$, including the diagonal, by induction: the integrand is smooth on a fixed compact integration interval. A homogeneous solution of $(D+j)u=0$ has $r^j{J}^{1/2}u$ constant along a ray, and smoothness at $r=0$ forces this constant to vanish for $j>0$. Thus the smooth transport coefficients are unique. In particular $u_j(y,y)=-({\Delta }_x{u}_{j-1})(y,y)/j$. Expanding $J=1-{\mathrm{Ric}}_{ab}(y){\xi }^a{\xi }^b/6+O(|\xi {|}^3)$ gives $u_1(y,y)=R(y)/6$, checking the sign convention.