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A beautiful classical example from Functional Analysis is the Hausdorff moment problem: characterize the sequences $m:=(m_0,m_1,\dots)$ of real numbers that are moments of some positive, finite Borel measure on the unit interval $I:=[0,1]$: $$m_k=\int_I x^kd\mu(x).$$ A necessary condition immediately comes from $\int_I x^{\ j}(1-x)^{\ k} d\mu(x)\geq0$, and is expressed saying that $m$ has to be a "completely monotone" sequence, that is $$(I-S)^k m\ge0,$$ where $S$ is the shift operator acting on sequences (in other words, the $k$-th discrete difference of $m$ has the sign of $(-1)^k$: $m$ is positive, decreasing, convex,...). The nontrivial fact is that this is also a sufficient condition, thus caracterizing the sequences of moments. Moreover, the measure is then unique.

I'll quote two proofs, both very nice. The first is close to the original one by Hausdorff; the second is a consequence of the Choquet's theorem.

Proof I, with computation (skipped). Bernstein polynomials $$B_nf(x):=\sum_{k=0}^n f\big(\frac{k}{n}\big)\Big( {n \atop k}\Big)x^k(1-x)^{n-k}, \qquad n\in\mathbb{N},\; f\in C^0(I), \; x\in I ,$$ define a sequence of linear positive operators strongly convergent to the identity $$B_n:C^0(I)\to C^0(I)\ .$$

Therefore the transpose operators $$B_n ^*:C^0(I)^ *\to C^0(I)^ *$$ give a sequence of operators weakly convergent to the identity. If you write down what is $B_n^ *(\mu)$ for a Radon measure $\mu\in C^0(I)^ *$ you'll observe that it is a linear combinations of Dirac measures located at the points $\{k/n\}_{0\leq k\leq n}$, and with coefficients only depending on the moments of $\mu$. This gives a uniqueness result and a heuristic argument: if $m$ is a sequence of moments for some measure $\mu$, then $\mu$ can be reconstructed by its moments as a weak* limit of discrete measures $\mu_n:=B_n^*(\mu)$. This observation leads to a constructive solution of the problem. Indeed, given a completely monotone sequence $m$, consider the corresponding sequence of measures $\mu_n$ suggested by the experssion of $B_n^*(\mu)$ in terms of the $(m_k)$. Due to the assumption of complete monotoniticy they turns out to be positive measures, and with some more computations one shows that they converges weakly* to a measure $\mu$ with moment's sequence $m$.

Proof II, no or little computation. Completely monotone sequences with $m_0=1$ are a closed convex, thus weakly* compact and metrizable subset $M$ of $l^\infty$. A one-line, smart computation shows that the extremal points of $M$ are exactly the exponential sequences, $m^{ (t)}:=(1,t,t^2,\dots)$, for $0\leq t \leq1$ (these turn out to be the moments of Dirac measures in points of $I$, of course). By the Choquet's theorem, for any given $m\in M$ there exists a probability measure on $\mathrm{ex}(M),$ that we identify with $I,$ such that $m=\int_I m^{ (t) } d\mu(t).$ But this exactly means $m_k=\int_I t^{\ k} d \mu(t)$ for all $k\in\mathbb{N}.$

A beautiful classical example from Functional Analysis is the Hausdorff moment problem: characterize the sequences $m:=(m_0,m_1,\dots)$ of real numbers that are moments of some positive, finite Borel measure on the unit interval $I:=[0,1]$: $$m_k=\int_I x^kd\mu(x).$$ A necessary condition immediately comes from $\int_I x^{\ j}(1-x)^{\ k} d\mu(x)\geq0$, and is expressed saying that $m$ has to be a "completely monotone" sequence, that is $$(I-S)^k m\ge0,$$ where $S$ is the shift operator acting on sequences (in other words, the $k$-th discrete difference of $m$ has the sign of $(-1)^k$: $m$ is positive, decreasing, convex,...). The nontrivial fact is that this is also a sufficient condition, thus caracterizing the sequences of moments. Moreover, the measure is then unique.

I'll quote two proofs, both very nice. The first is close to the original one by Hausdorff; the second is a consequence of the Choquet's theorem.

Proof I, with computation (skipped). Bernstein polynomials $$B_nf(x):=\sum_{k=0}^n f\big(\frac{k}{n}\big)\Big( {n \atop k}\Big)x^k(1-x)^{n-k}, \qquad n\in\mathbb{N},\; f\in C^0(I), \; x\in I ,$$ define a sequence of linear positive operators strongly convergent to the identity $$B_n:C^0(I)\to C^0(I)\ .$$

Therefore the transpose operators $$B_n ^*:C^0(I)^ *\to C^0(I)^ *$$ give a sequence of operators weakly convergent to the identity. If you write down what is $B_n^ *(\mu)$ for a Radon measure $\mu\in C^0(I)^ *$ you'll observe that it is a linear combinations of Dirac measures located at the points $\{k/n\}_{0\leq k\leq n}$, and with coefficients only depending on the moments of $\mu$. This gives a uniqueness result and a heuristic argument: if $m$ is a sequence of moments for some measure $\mu$, then $\mu$ can be reconstructed by its moments as a weak* limit of discrete measures $\mu_n:=B_n^*(\mu)$. This observation leads to a constructive solution of the problem. Indeed, given a completely monotone sequence $m$, consider the corresponding sequence of measures $\mu_n$ suggested by the expression of $B_n^*(\mu)$ in terms of the $(m_k)$. Due to the assumption of complete monotoniticy they turns out to be positive measures, and with some more computations one shows that they converges weakly* to a measure $\mu$ with moment's sequence $m$.

Proof II, no or little computation. Completely monotone sequences with $m_0=1$ are a closed convex, thus weakly* compact and metrizable subset $M$ of $l^\infty$. A one-line, smart computation shows that the extremal points of $M$ are exactly the exponential sequences, $m^{ (t)}:=(1,t,t^2,\dots)$, for $0\leq t \leq1$ (these turn out to be the moments of Dirac measures in points of $I$, of course). By the Choquet's theorem, for any given $m\in M$ there exists a probability measure on $\mathrm{ex}(M),$ that we identify with $I,$ such that $m=\int_I m^{ (t) } d\mu(t).$ But this exactly means $m_k=\int_I t^{\ k} d \mu(t)$ for all $k\in\mathbb{N}.$

A beautiful classical example from Functional Analysis is the Hausdorff moment problem: characterize the sequences $m:=(m_0,m_1,\dots)$ of real numbers that are moments of some positive, finite Borel measure on the unit interval $I:=[0,1]$: $$m_k=\int_I x^kd\mu(x).$$ A necessary condition immediately comes from $\int_I x^{\ j}(1-x)^{\ k} d\mu(x)\geq0$, and is expressed saying that $m$ has to be a "completely monotone" sequence, that is $$(I-S)^k m\ge0,$$ where $S$ is the shift operator acting on sequences (in other words, the $k$-th discrete difference of $m$ has the sign of $(-1)^k$: $m$ is positive, decreasing, convex,...). The nontrivial fact is that this is also a sufficient condition, thus caracterizing the sequences of moments. Moreover, the measure is then unique.

I'll quote two proofs, both very nice. The first is close to the original one by Hausdorff; the second is a consequence of the Choquet's theorem.

Proof I, with computation (skipped). Bernstein polynomials $$B_nf(x):=\sum_{k=0}^n f\big(\frac{k}{n}\big)\Big( {n \atop n}\Big)x^k(1-x)^{n-k}, \qquad n\in\mathbb{N},\; f\in C^0(I), \; x\in I ,$$ define a sequence of linear positive operators strongly convergent to the identity $$B_n:C^0(I)\to C^0(I)\ .$$

Therefore the transpose operators $$B_n ^*:C^0(I)^ *\to C^0(I)^ *$$ give a sequence of operators weakly convergent to the identity. If you write down what is $B_n^ *(\mu)$ for a Radon measure $\mu\in C^0(I)^ *$ you'll observe that it is a linear combinations of Dirac measures located at the points $\{k/n\}_{0\leq k\leq n}$, and with coefficients only depending on the moments of $\mu$. This gives a uniqueness result and a heuristic argument: if $m$ is a sequence of moments for some measure $\mu$, then $\mu$ can be reconstructed by its moments as a weak* limit of discrete measures $\mu_n:=B_n^*(\mu)$. This observation leads to a constructive solution of the problem. Indeed, given a completely monotone sequence $m$, consider the corresponding sequence of measures $\mu_n$ suggested by the experssion of $B_n^*(\mu)$ in terms of the $(m_k)$. Due to the assumption of complete monotoniticy they turns out to be positive measures, and with some more computations one shows that they converges weakly* to a measure $\mu$ with moment's sequence $m$.

Proof II, no or little computation. Completely monotone sequences with $m_0=1$ are a closed convex, thus weakly* compact and metrizable subset $M$ of $l^\infty$. A one-line, smart computation shows that the extremal points of $M$ are exactly the exponential sequences, $m^{ (t)}:=(1,t,t^2,\dots)$, for $0\leq t \leq1$ (these turn out to be the moments of Dirac measures in points of $I$, of course). By the Choquet's theorem, for any given $m\in M$ there exists a probability measure on $\mathrm{ex}(M),$ that we identify with $I,$ such that $m=\int_I m^{ (t) } d\mu(t).$ But this exactly means $m_k=\int_I t^{\ k} d \mu(t)$ for all $k\in\mathbb{N}.$

A beautiful classical example from Functional Analysis is the Hausdorff moment problem: characterize the sequences $m:=(m_0,m_1,\dots)$ of real numbers that are moments of some positive, finite Borel measure on the unit interval $I:=[0,1]$: $$m_k=\int_I x^kd\mu(x).$$ A necessary condition immediately comes from $\int_I x^{\ j}(1-x)^{\ k} d\mu(x)\geq0$, and is expressed saying that $m$ has to be a "completely monotone" sequence, that is $$(I-S)^k m\ge0,$$ where $S$ is the shift operator acting on sequences (in other words, the $k$-th discrete difference of $m$ has the sign of $(-1)^k$: $m$ is positive, decreasing, convex,...). The nontrivial fact is that this is also a sufficient condition, thus caracterizing the sequences of moments. Moreover, the measure is then unique.

I'll quote two proofs, both very nice. The first is close to the original one by Hausdorff; the second is a consequence of the Choquet's theorem.

Proof I, with computation (skipped). Bernstein polynomials $$B_nf(x):=\sum_{k=0}^n f\big(\frac{k}{n}\big)\Big( {n \atop k}\Big)x^k(1-x)^{n-k}, \qquad n\in\mathbb{N},\; f\in C^0(I), \; x\in I ,$$ define a sequence of linear positive operators strongly convergent to the identity $$B_n:C^0(I)\to C^0(I)\ .$$

Therefore the transpose operators $$B_n ^*:C^0(I)^ *\to C^0(I)^ *$$ give a sequence of operators weakly convergent to the identity. If you write down what is $B_n^ *(\mu)$ for a Radon measure $\mu\in C^0(I)^ *$ you'll observe that it is a linear combinations of Dirac measures located at the points $\{k/n\}_{0\leq k\leq n}$, and with coefficients only depending on the moments of $\mu$. This gives a uniqueness result and a heuristic argument: if $m$ is a sequence of moments for some measure $\mu$, then $\mu$ can be reconstructed by its moments as a weak* limit of discrete measures $\mu_n:=B_n^*(\mu)$. This observation leads to a constructive solution of the problem. Indeed, given a completely monotone sequence $m$, consider the corresponding sequence of measures $\mu_n$ suggested by the experssion of $B_n^*(\mu)$ in terms of the $(m_k)$. Due to the assumption of complete monotoniticy they turns out to be positive measures, and with some more computations one shows that they converges weakly* to a measure $\mu$ with moment's sequence $m$.

Proof II, no or little computation. Completely monotone sequences with $m_0=1$ are a closed convex, thus weakly* compact and metrizable subset $M$ of $l^\infty$. A one-line, smart computation shows that the extremal points of $M$ are exactly the exponential sequences, $m^{ (t)}:=(1,t,t^2,\dots)$, for $0\leq t \leq1$ (these turn out to be the moments of Dirac measures in points of $I$, of course). By the Choquet's theorem, for any given $m\in M$ there exists a probability measure on $\mathrm{ex}(M),$ that we identify with $I,$ such that $m=\int_I m^{ (t) } d\mu(t).$ But this exactly means $m_k=\int_I t^{\ k} d \mu(t)$ for all $k\in\mathbb{N}.$

A beautiful classical example from Functional Analysis is the Hausdorff moment problem: characterize the sequences $m:=(m_0,m_1,\dots)$ of real numbers that are moments of some positive, finite Borel measure on the unit interval $I:=[0,1]$: $$m_k=\int_I x^kd\mu(x).$$ A necessary condition immediately comes from $\int_I x^{\ j}(1-x)^{\ k} d\mu(x)\geq0$, and is expressed saying that $m$ has to be a "completely monotone" sequence, that is $$(I-S)^k m\ge0,$$ where $S$ is the shift operator acting on sequences (in other words, the $k$-th discrete difference of $m$ has the sign of $(-1)^k$: $m$ is positive, decreasing, convex,...). The nontrivial fact is that this is also a sufficient condition, thus caracterizing the sequences of moments. Moreover, the measure is then unique.

I'll quote two proofs, both very nice. The first is close to the original one by Hausdorff; the second is a consequence of the Choquet's theorem.

Proof I, with computation (skipped). Bernstein polynomials give a sequence of linear positive operators strongly convergent to the identity $$B_n:C^0(I)\to C^0(I).$$ Therefore the transpose operators $$B_n ^*:C^0(I)^ *\to C^0(I)^ *$$ give a sequence of operators weakly convergent to the identity. If you write down what is $B_n^ *(\mu)$ for a Radon measure $\mu\in C^0(I)^ *$ you'll observe that it is a linear combinations of Dirac measures located at the points $\{k/n\}_{0\leq k\leq n}$, and with coefficients only depending on the moments of $\mu$. This gives a uniqueness result and a heuristic argument: if $m$ is a sequence of moments for some measure $\mu$, then $\mu$ can be reconstructed by its moments as a weak* limit of discrete measures $\mu_n:=B_n^*(\mu)$. This observation leads to a constructive solution of the problem. Indeed, given a completely monotone sequence $m$, consider the corresponding sequence of measures $\mu_n$ suggested by the experssion of $B_n^*(\mu)$ in terms of the $(m_k)$. Due to the assumption of complete monotoniticy they turns out to be positive measures, and with some more computations one shows that they converges weakly* to a measure $\mu$ with moment's sequence $m$.

Proof II, no or little computation. Completely monotone sequences with $m_0=1$ are a closed convex, thus weakly* compact and metrizable subset $M$ of $l^\infty$. A one-line, smart computation shows that the extremal points of $M$ are exactly the exponential sequences, $m^{ (t)}:=(1,t,t^2,\dots)$, for $0\leq t \leq1$ (these turn out to be the moments of Dirac measures in points of $I$, of course). By the Choquet's theorem, for any given $m\in M$ there exists a probability measure on $\mathrm{ex}(M),$ that we identify with $I,$ such that $m=\int_I m^{ (t) } d\mu(t).$ But this exactly means $m_k=\int_I t^{\ k} d \mu(t)$ for all $k\in\mathbb{N}.$

A beautiful classical example from Functional Analysis is the Hausdorff moment problem: characterize the sequences $m:=(m_0,m_1,\dots)$ of real numbers that are moments of some positive, finite Borel measure on the unit interval $I:=[0,1]$: $$m_k=\int_I x^kd\mu(x).$$ A necessary condition immediately comes from $\int_I x^{\ j}(1-x)^{\ k} d\mu(x)\geq0$, and is expressed saying that $m$ has to be a "completely monotone" sequence, that is $$(I-S)^k m\ge0,$$ where $S$ is the shift operator acting on sequences (in other words, the $k$-th discrete difference of $m$ has the sign of $(-1)^k$: $m$ is positive, decreasing, convex,...). The nontrivial fact is that this is also a sufficient condition, thus caracterizing the sequences of moments. Moreover, the measure is then unique.

I'll quote two proofs, both very nice. The first is close to the original one by Hausdorff; the second is a consequence of the Choquet's theorem.

Proof I, with computation (skipped). Bernstein polynomials $$B_nf(x):=\sum_{k=0}^n f\big(\frac{k}{n}\big)\Big( {n \atop n}\Big)x^k(1-x)^{n-k}, \qquad n\in\mathbb{N},\; f\in C^0(I), \; x\in I ,$$ define a sequence of linear positive operators strongly convergent to the identity $$B_n:C^0(I)\to C^0(I)\ .$$

Therefore the transpose operators $$B_n ^*:C^0(I)^ *\to C^0(I)^ *$$ give a sequence of operators weakly convergent to the identity. If you write down what is $B_n^ *(\mu)$ for a Radon measure $\mu\in C^0(I)^ *$ you'll observe that it is a linear combinations of Dirac measures located at the points $\{k/n\}_{0\leq k\leq n}$, and with coefficients only depending on the moments of $\mu$. This gives a uniqueness result and a heuristic argument: if $m$ is a sequence of moments for some measure $\mu$, then $\mu$ can be reconstructed by its moments as a weak* limit of discrete measures $\mu_n:=B_n^*(\mu)$. This observation leads to a constructive solution of the problem. Indeed, given a completely monotone sequence $m$, consider the corresponding sequence of measures $\mu_n$ suggested by the experssion of $B_n^*(\mu)$ in terms of the $(m_k)$. Due to the assumption of complete monotoniticy they turns out to be positive measures, and with some more computations one shows that they converges weakly* to a measure $\mu$ with moment's sequence $m$.

Proof II, no or little computation. Completely monotone sequences with $m_0=1$ are a closed convex, thus weakly* compact and metrizable subset $M$ of $l^\infty$. A one-line, smart computation shows that the extremal points of $M$ are exactly the exponential sequences, $m^{ (t)}:=(1,t,t^2,\dots)$, for $0\leq t \leq1$ (these turn out to be the moments of Dirac measures in points of $I$, of course). By the Choquet's theorem, for any given $m\in M$ there exists a probability measure on $\mathrm{ex}(M),$ that we identify with $I,$ such that $m=\int_I m^{ (t) } d\mu(t).$ But this exactly means $m_k=\int_I t^{\ k} d \mu(t)$ for all $k\in\mathbb{N}.$