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Math Help - Real Analysis

  1. #1
    MHF Contributor chiph588@'s Avatar
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    Real Analysis

    Show that if  f is a continuous real-valued function on the interval  [a,b] and  f(x) \geq 0 \; \forall x \in [a,b] , then
     \lim_{n \to \infty} \left( \int_a^b f^{n}(x) dx \right)^{\frac{1}{n}} = \max \{f(x) : x \in [a,b]\}.
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    Quote Originally Posted by chiph588@ View Post
    Show that if  f is a continuous real-valued function on the interval  [a,b] and  f(x) \geq 0 \; \forall x \in [a,b] , then
     \lim_{n \to \infty} \left( \int_a^b f^{n}(x) dx \right)^{\frac{1}{n}} = \max \{f(x) : x \in [a,b]\}.
    Let m = \max \{f(x) : x \in [a,b]\} . It should be fairly clear that \int_a^b f^{n}(x)\, dx \leqslant m^n. Thus  \lim_{n \to \infty} \left( \int_a^b f^{n}(x) dx \right)^{\!\!1/n} (if it exists) is less than or equal to m.

    To get the reverse inequality, let \varepsilon>0. The continuity of f ensures that there is some interval of length \delta>0 on which f(x)\geqslant m-\varepsilon. Then \int_a^b f^{n}(x)\, dx \geqslant \delta(m-\varepsilon)^n. Take the n'th root and let n\to\infty to see that  \lim_{n \to \infty} \left( \int_a^b f^{n}(x) dx \right)^{\!\!1/n} \geqslant m-\varepsilon. Finally, take the limit as \varepsilon\to0.
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  3. #3
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by Opalg View Post
    Let m = \max \{f(x) : x \in [a,b]\} . It should be fairly clear that \int_a^b f^{n}(x)\, dx \leqslant m^n. Thus  \lim_{n \to \infty} \left( \int_a^b f^{n}(x) dx \right)^{\!\!1/n} (if it exists) is less than or equal to m.
    Shouldn't we say  \int_a^b f^{n}(x)\, dx \leq m^n(b-a) , because what if the length of the interval could be really big. For example,  \int_0^2 m^n\, dx = 2m^n \geq m^n.

    So taking the nth root of both sides would yield  m (b-a)^{\frac{1}{n}} , and as  n \to \infty, \; (b-a)^{\frac{1}{n}} \to 1.
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    Quote Originally Posted by chiph588@ View Post
    Shouldn't we say  \int_a^b f^{n}(x)\, dx \leq m^n(b-a) , because what if the length of the interval could be really big. For example,  \int_0^2 m^n\, dx = 2m^n \geq m^n.

    So taking the nth root of both sides would yield  m (b-a)^{\frac{1}{n}} , and as  n \to \infty, \; (b-a)^{\frac{1}{n}} \to 1.
    Absolutely correct! The length of the interval should be included, and its n'th root goes to 1 as n→∞. I overlooked that.
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  5. #5
    MHF Contributor chiph588@'s Avatar
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    This is one of my favorite theorems.
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    it's not bad to know, if you haven't noticed yet, that the limit is some sort of "continuous version" of this: \lim_{n\to\infty} \sqrt[n]{a_1^n + \cdots + a_k^n} = \max_{1 \leq j \leq k}{a_j}, where a_j are postive numbers.
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    MHF Contributor chiph588@'s Avatar
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    Ah yes, because an integral is just a Riemann sum.
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    I do not want to argue against Opalg because I am a noob compared to him in analysis , but I offer a suggestion.
    I think this problem is better solved using the concepts of limsup's and liminf's.

    Let \ell be the length of the interval. Let I_n = \left( \smallint_a^b f^n \right)^{1/n}.
    Opalg hath shown us that \delta^{1/n} \cdot \ell^{1/n} \cdot (m-\epsilon) \leq I_n \leq m \cdot \ell^{1/n}.
    Then he says m-\epsilon \leq \lim I_n \leq m \implies I_n = m by \epsilon \to 0^+.
    But we need to know the limit exists.

    So a way around this problem is to perhaps say first take \epsilon \to 0^+ and then n\to \infty.
    But the problem is \delta depends on \epsilon and so how we know that \delta^{1/n} \not \to  0 as n\to \infty?

    Maybe I am missing something but I think that if we use limsup's and liminf's the solution is much smoother. Because \delta^{1/n}\cdot \ell^{1/n}\cdot (m-\epsilon) \leq I_n and so m-\epsilon \leq \liminf I_n \implies m \leq \liminf I_n. However, I_n \leq m \implies \limsup I_n  \leq m. Thus, we have shown \liminf I_n \geq \limsup I_n but \liminf I_n \leq \limsup I_n. This means that I_n is a convergent sequence with \lim I_n = \liminf I_n = \limsup I_n = m.
    ---
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    Quote Originally Posted by ThePerfectHacker View Post
    I do not want to argue against Opalg because I am a noob compared to him in analysis , but I offer a suggestion.
    I think this problem is better solved using the concepts of limsup's and liminf's.
    That is quite correct. The efficient way to explain the solution is to use limsup in connection with the upper estimate, and liminf for the lower estimate.

    I didn't do that, because I didn't want to deter anyone who might be unfamiliar with limsup and liminf. So I just included a parenthetical comment to the effect that you need to establish that the limit exists.
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  10. #10
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    I do not want to argue against Opalg because I am a noob compared to him in analysis , but I offer a suggestion.
    I think this problem is better solved using the concepts of limsup's and liminf's.

    Let \ell be the length of the interval. Let I_n = \left( \smallint_a^b f^n \right)^{1/n}.
    Opalg hath shown us that \delta^{1/n} \cdot \ell^{1/n} \cdot (m-\epsilon) \leq I_n \leq m \cdot \ell^{1/n}.
    Then he says m-\epsilon \leq \lim I_n \leq m \implies I_n = m by \epsilon \to 0^+.
    But we need to know the limit exists.

    So a way around this problem is to perhaps say first take \epsilon \to 0^+ and then n\to \infty.
    But the problem is \delta depends on \epsilon and so how we know that \delta^{1/n} \not \to  0 as n\to \infty?

    Maybe I am missing something but I think that if we use limsup's and liminf's the solution is much smoother. Because \delta^{1/n}\cdot \ell^{1/n}\cdot (m-\epsilon) \leq I_n and so m-\epsilon \leq \liminf I_n \implies m \leq \liminf I_n. However, I_n \leq m \implies \limsup I_n  \leq m. Thus, we have shown \liminf I_n \geq \limsup I_n but \liminf I_n \leq \limsup I_n. This means that I_n is a convergent sequence with \lim I_n = \liminf I_n = \limsup I_n = m.
    ---
    how do we know that m-\epsilon \leq \liminf I_n and \limsup I_n  \leq m?
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    Quote Originally Posted by chiph588@ View Post
    how do we know that m-\epsilon \leq \liminf I_n and \limsup I_n  \leq m?
    It was shown that, \delta^{1/n} \cdot \ell^{1/n} \cdot (m-\epsilon) \leq I_n \leq m \cdot \ell^{1/n}.
    Therefore, \liminf \delta^{1/n} \cdot \ell^{1/n} \cdot (m-\epsilon) \leq \liminf I_n \implies m-\epsilon \leq I_n.
    Similary with \limsup.
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    Quote Originally Posted by Opalg View Post
    Let m = \max \{f(x) : x \in [a,b]\} . It should be fairly clear that \int_a^b f^{n}(x)\, dx \leqslant m^n. Thus  \lim_{n \to \infty} \left( \int_a^b f^{n}(x) dx \right)^{\!\!1/n} (if it exists) is less than or equal to m.

    To get the reverse inequality, let \varepsilon>0. The continuity of f ensures that there is some interval of length \delta>0 on which f(x)\geqslant m-\varepsilon. Then \int_a^b f^{n}(x)\, dx \geqslant \delta(m-\varepsilon)^n. Take the n'th root and let n\to\infty to see that  \lim_{n \to \infty} \left( \int_a^b f^{n}(x) dx \right)^{\!\!1/n} \geqslant m-\varepsilon. Finally, take the limit as \varepsilon\to0.
    How did you conclude that  \int_{a}^{b} f^{n} \ dx \leq m^{n} (I know there should be a b-a also)? Are not you using what you are trying to prove?
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    Quote Originally Posted by heathrowjohnny View Post
    How did you conclude that  \int_{a}^{b} f^{n} \ dx \leq m^{n} (I know there should be a b-a also)? Are not you using what you are trying to prove?
    Because 0 \leq f \leq m \implies 0 \leq f^n \leq m^n \implies 0 \leq \int_a^b f^n \leq \int_a^b m^n.
    Thus, we get, \int_a^b f^n \leq m^n(b-a).
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