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Thread: f integrable => f^2 integrable?

  1. #1
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    f integrable => f^2 integrable?

    Here is one that I don't really know how to approach:

    Prove: if $\displaystyle f(x)$ is integrable on $\displaystyle [a,b],$ then $\displaystyle (f(x)^2)$ is integrable on $\displaystyle [a,b].$
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    This is not true. $\displaystyle f(x)=\frac{1}{\sqrt{x}}$ on [0,1] provides a counterexample. Perhaps you have been asked to prove the converse.
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  3. #3
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    Quote Originally Posted by Enrique2 View Post
    This is not true. $\displaystyle f(x)=\frac{1}{\sqrt{x}}$ on [0,1] provides a counterexample.
    In fact that is not a counterexample. $\displaystyle f(x)=\frac{1}{\sqrt{x}}$ is not Riemann integrable on $\displaystyle [0,1]$.
    Note that the function is not bounded on $\displaystyle [0,1]$.

    This happens to be a standard theorem/problem in the study Riemann integrals.
    Here is the standard hint: $\displaystyle f(x)^2 - f(y)^2 = \left( {f(x) + f(y)} \right)\left( {f(x) - f(y)} \right)$.
    Use that to show that $\displaystyle U\left[ {f^2 ;P} \right] - L\left[ {f^2 ;P} \right] \leqslant 2 B\left( {U\left[ {f;P} \right] - L\left[ {f;P} \right]} \right)$ where $\displaystyle B$ is the bound of $\displaystyle f$ on $\displaystyle [a,b]$
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    Quote Originally Posted by Plato View Post
    In fact that is not a counterexample. $\displaystyle f(x)=\frac{1}{\sqrt{x}}$ is not Riemann integrable on $\displaystyle [0,1]$.
    Note that the function is not bounded on $\displaystyle [0,1]$.

    This happens to be a standard theorem/problem in the study Riemann integrals.
    Here is the standard hint: $\displaystyle f(x)^2 - f(y)^2 = \left( {f(x) + f(y)} \right)\left( {f(x) - f(y)} \right)$.
    Use that to show that $\displaystyle U\left[ {f^2 ;P} \right] - L\left[ {f^2 ;P} \right] \leqslant 2 B\left( {U\left[ {f;P} \right] - L\left[ {f;P} \right]} \right)$ where $\displaystyle B$ is the bound of $\displaystyle f$ on $\displaystyle [a,b]$
    OK, sorry. Sure that the question is about Riemann integrability. I related the question to $\displaystyle L^p$ spaces.
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    Since it is bounded, can I just assume that there exists a partition $\displaystyle P$ of $\displaystyle [a,b]$ such that

    $\displaystyle U_{f}(P) - L_{f}(P) < \frac{\epsilon}{2B}?$
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  6. #6
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    Quote Originally Posted by Anonymous1 View Post
    Since it is bounded, can I just assume that there exists a partition $\displaystyle P$ of $\displaystyle [a,b]$ such that

    $\displaystyle U_{f}(P) - L_{f}(P) < \frac{\epsilon}{2B}?$
    That is the idea because you know that $\displaystyle f$ is integrable.
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