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Math Help - Moments of a function

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

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

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

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

    Lemma: J[h] is continuous.
    Proof: Note that if \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 \int_0^1 |f(x)-g(x)|dx\ne0

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

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

    The conclusion follows. \blacksquare.

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

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

    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= 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 \mathcal{C}[0,1] we have that J[h] agrees with the zero function on a dense subset of it's domain. It follows that J[h]=0 for every h. The conclusion follows from this. \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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  3. #3
    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 J:\mathcal{C}[0,1]\to\mathbb{R} by h(x)\mapsto \int_0^1(f(x)-g(x))h(x) dx.

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

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

    Thus, let p(x)\in \mathcal{C}[0,1] be a polynomial, then 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 \mathbb{P} is the set of all polynomials in \mathcal{C}[0,1] that J^{-1}(\{0\})\supseteq \mathbb{P}. But, it follows from Weierstrass's approximation theorem that \mathbb{P} is dense. And since J^{-1}(\{0\}) is closed we have that X\supseteq J^{-1}(\{0\})=\overline{J^{-1}(\{0\})}\supseteq\overline{\mathbb{P}}=X. It follows that J[h]=0. In particular 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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  4. #4
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    Quote Originally Posted by Drexel28 View Post
    If f\in\mathcal{C}[0,1] we call \int_0^1 f(x)x^n dx a moment of f.

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

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

    Lemma: J[h] is continuous.
    Proof: Note that if \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 \int_0^1 |f(x)-g(x)|dx\ne0

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

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

    The conclusion follows. \blacksquare.

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

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

    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= 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 \mathcal{C}[0,1] we have that J[h] agrees with the zero function on a dense subset of it's domain. It follows that J[h]=0 for every h. The conclusion follows from this. \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 f(x)\in C[a,b] and \int_{a}^{b} f(x)g^{(n)} (x)dx=0 for all g\in C_0^{\infty } [a,b] then f is a polynomial of degree at most 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 f(x)\in C[a,b] and \int_{a}^{b} f(x)g^{(n)} (x)dx=0 for all g\in C_0^{\infty } [a,b] then f is a polynomial of degree at most n-1. Try to prove it!
    What is \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 \mathcal{C}^{\infty}_{\color{red}{0}}[a,b]?
    It's the space of smooth functions with compact support in [a,b], ie.  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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