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Thread: Moments of a function

  1. #1
    MHF Contributor Drexel28's Avatar
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    Moments of a function

    If $\displaystyle f\in\mathcal{C}[0,1]$ we call $\displaystyle \int_0^1 f(x)x^n dx$ a moment of $\displaystyle f$.

    Problem: Prove that if $\displaystyle f,g\in\mathcal{C}[0,1]$ have the same moments (for $\displaystyle n=0,1,2,\cdots$) then the must be the same function.

    Proof: Define the functional $\displaystyle J:\mathcal{C}[0,1]\to\mathbb{R}$ by $\displaystyle J[h]=\int_0^1 (f(x)-g(x))h(x)\text{ }dx$. We first prove that $\displaystyle J$ is continuous.

    Lemma: $\displaystyle J[h]$ is continuous.
    Proof: Note that if $\displaystyle \int_0^1 |f(x)-g(x)|dx=0$ we're done since the integral of a continuous non-negative function is zero only when the function is identically zero. Thus, we may assume that $\displaystyle \int_0^1 |f(x)-g(x)|dx\ne0$

    Let $\displaystyle \varepsilon>0$ be given. Then choosing $\displaystyle h$ such that $\displaystyle \|h-h_0\|_{\infty}<\left(\int_0^1 |f(x)-g(x)|dx\right)^{-1}\varepsilon$ we see that

    $\displaystyle \left|J[h_0]-J[h]\right|=\left|\int_0^1(f(x)-g(x))(h(x)-h_0(x))dx\right|$$\displaystyle \leqslant \int_0^1|f(x)-g(x)||h(x)-h_0(x)|dx$$\displaystyle <\varepsilon\left(\int_0^1|f(x)-g(x)|\right)^{-1}\int_0^1|f(x)-g(x)|dx=\varepsilon$

    The conclusion follows. $\displaystyle \blacksquare$.

    But notice by our assumption that $\displaystyle \int_0^1 f(x)x^n dx=\int_0^1 g(x)x^n dx$ for every $\displaystyle n\in\mathbb{N}\cup\{0\}$ or equivalently $\displaystyle J[x^n]=\int_0^1(f(x)-g(x))x^n dx=0$.

    But, we in fact see that given any polynomial $\displaystyle p(x)=a_0+a_1x+\cdots a_n x^n$ that

    $\displaystyle J[p(x)]=\int_0^1(f(x)-g(x))p(x)dx=\int_0^1(f(x)-g(x))(a_0+\cdots+a_n x^n)dx=$$\displaystyle a_0\int_0^1(f(x)-g(x))x^0+a_1\int_0^1(f(x)-g(x))x+\cdots+a_n\int_0^1(f(x)-g(x))x^n=0+\cdots+0=0$

    And since by Weirstrass's theorem we have that the polynomials are dense in $\displaystyle \mathcal{C}[0,1]$ we have that $\displaystyle J[h]$ agrees with the zero function on a dense subset of it's domain. It follows that $\displaystyle J[h]=0$ for every $\displaystyle h$. The conclusion follows from this. $\displaystyle \blacksquare$


    Does that look right? I feel like I made this way harder than it need be.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Oops...I just realized that I didn't need to prove that the functional itself was continuous. Oh well...it's late.
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    MHF Contributor Drexel28's Avatar
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    If anyone is interested, now that I'm awake there is a much more clean way to say this.

    Proof: Define, the functional $\displaystyle J:\mathcal{C}[0,1]\to\mathbb{R}$ by $\displaystyle h(x)\mapsto \int_0^1(f(x)-g(x))h(x) dx$.

    Using the same rational as above we may conclude that $\displaystyle J$ is continuous.

    Also, it is apparent that $\displaystyle J$ is functional and by our conditions we have that $\displaystyle J[x^n]=0,\text{ }n=0,1,\cdots$.

    Thus, let $\displaystyle p(x)\in \mathcal{C}[0,1]$ be a polynomial, then $\displaystyle J[p(x)]=J\left[a_0+\cdots+a_n x^n\right]=\sum_{k=0}^{n}a_k J[x^k]=\sum_{k=1}^{0}0=0$.

    It follows that if $\displaystyle \mathbb{P}$ is the set of all polynomials in $\displaystyle \mathcal{C}[0,1]$ that $\displaystyle J^{-1}(\{0\})\supseteq \mathbb{P}$. But, it follows from Weierstrass's approximation theorem that $\displaystyle \mathbb{P}$ is dense. And since $\displaystyle J^{-1}(\{0\})$ is closed we have that $\displaystyle X\supseteq J^{-1}(\{0\})=\overline{J^{-1}(\{0\})}\supseteq\overline{\mathbb{P}}=X$. It follows that $\displaystyle J[h]=0$. In particular $\displaystyle J[f(x)-g(x)]=\int_0^1(f(x)-g(x))^2dx=0$ and thus by prior remarks the conclusion follows.
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    Quote Originally Posted by Drexel28 View Post
    If $\displaystyle f\in\mathcal{C}[0,1]$ we call $\displaystyle \int_0^1 f(x)x^n dx$ a moment of $\displaystyle f$.

    Problem: Prove that if $\displaystyle f,g\in\mathcal{C}[0,1]$ have the same moments (for $\displaystyle n=0,1,2,\cdots$) then the must be the same function.

    Proof: Define the functional $\displaystyle J:\mathcal{C}[0,1]\to\mathbb{R}$ by $\displaystyle J[h]=\int_0^1 (f(x)-g(x))h(x)\text{ }dx$. We first prove that $\displaystyle J$ is continuous.

    Lemma: $\displaystyle J[h]$ is continuous.
    Proof: Note that if $\displaystyle \int_0^1 |f(x)-g(x)|dx=0$ we're done since the integral of a continuous non-negative function is zero only when the function is identically zero. Thus, we may assume that $\displaystyle \int_0^1 |f(x)-g(x)|dx\ne0$

    Let $\displaystyle \varepsilon>0$ be given. Then choosing $\displaystyle h$ such that $\displaystyle \|h-h_0\|_{\infty}<\left(\int_0^1 |f(x)-g(x)|dx\right)^{-1}\varepsilon$ we see that

    $\displaystyle \left|J[h_0]-J[h]\right|=\left|\int_0^1(f(x)-g(x))(h(x)-h_0(x))dx\right|$$\displaystyle \leqslant \int_0^1|f(x)-g(x)||h(x)-h_0(x)|dx$$\displaystyle <\varepsilon\left(\int_0^1|f(x)-g(x)|\right)^{-1}\int_0^1|f(x)-g(x)|dx=\varepsilon$

    The conclusion follows. $\displaystyle \blacksquare$.

    But notice by our assumption that $\displaystyle \int_0^1 f(x)x^n dx=\int_0^1 g(x)x^n dx$ for every $\displaystyle n\in\mathbb{N}\cup\{0\}$ or equivalently $\displaystyle J[x^n]=\int_0^1(f(x)-g(x))x^n dx=0$.

    But, we in fact see that given any polynomial $\displaystyle p(x)=a_0+a_1x+\cdots a_n x^n$ that

    $\displaystyle J[p(x)]=\int_0^1(f(x)-g(x))p(x)dx=\int_0^1(f(x)-g(x))(a_0+\cdots+a_n x^n)dx=$$\displaystyle a_0\int_0^1(f(x)-g(x))x^0+a_1\int_0^1(f(x)-g(x))x+\cdots+a_n\int_0^1(f(x)-g(x))x^n=0+\cdots+0=0$

    And since by Weirstrass's theorem we have that the polynomials are dense in $\displaystyle \mathcal{C}[0,1]$ we have that $\displaystyle J[h]$ agrees with the zero function on a dense subset of it's domain. It follows that $\displaystyle J[h]=0$ for every $\displaystyle h$. The conclusion follows from this. $\displaystyle \blacksquare$


    Does that look right? I feel like I made this way harder than it need be.
    As you noted, this is a rather trivial generalization of the fundamental lemma of calculus of variations. A more interesting (and surprising) one is this: If $\displaystyle f(x)\in C[a,b]$ and $\displaystyle \int_{a}^{b} f(x)g^{(n)} (x)dx=0$ for all $\displaystyle g\in C_0^{\infty } [a,b]$ then $\displaystyle f$ is a polynomial of degree at most $\displaystyle n-1$. Try to prove it!
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  5. #5
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Jose27 View Post
    As you noted, this is a rather trivial generalization of the fundamental lemma of calculus of variations. A more interesting (and surprising) one is this: If $\displaystyle f(x)\in C[a,b]$ and $\displaystyle \int_{a}^{b} f(x)g^{(n)} (x)dx=0$ for all $\displaystyle g\in C_0^{\infty } [a,b]$ then $\displaystyle f$ is a polynomial of degree at most $\displaystyle n-1$. Try to prove it!
    What is $\displaystyle \mathcal{C}^{\infty}_{\color{red}{0}}[a,b]$?
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  6. #6
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    Quote Originally Posted by Drexel28 View Post
    What is $\displaystyle \mathcal{C}^{\infty}_{\color{red}{0}}[a,b]$?
    It's the space of smooth functions with compact support in $\displaystyle [a,b]$, ie.$\displaystyle C_0^{\infty } [a,b]:= \{ f\in C^{\infty }[a,b] : f^{(k)}(a)=f^{(k)}(b)=0 \ \mbox{for all} \ k\geq 0 \}$.
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