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Math Help - does 1/x in Lp

  1. #1
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    Question does 1/x in Lp

    I need to check whether or not \frac{1}{x} \in {L}^p

    \int_{0}^{1} \frac{1}{|x|^p} ~dx = \frac{1}{-p + 1} x^{-p + 1}|_{0}^1

    because this integral divergent, so \frac{1}{x} \notin {L}^p

    is it correct??

    thanks for any comment and suggestion
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  2. #2
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    \vert x \vert ^p is integrable in (0,1) iff 1+p>0 iff p>-1. So yeah, for any 1\leq p we have that \frac{1}{\vert x \vert } \notin L^p
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  3. #3
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    Question

    Quote Originally Posted by Jose27 View Post
    \vert x \vert ^p is integrable in (0,1) iff 1+p>0 iff p>-1. So yeah, for any 1\leq p we have that \frac{1}{\vert x \vert } \notin L^p
    sure??

    but i need to check \frac{1}{|x|^p}, not \vert x \vert ^p

    \int_{0}^{1} \frac{1}{|x|^p} ~dx = \frac{1}{-p + 1} x^{-p + 1}|_{0}^1 convergent for 0 < p < 1

    so the right answer is \frac{1}{x} \in {L}^p for 0 < p < 1, and \frac{1}{x} \notin {L}^p for 1 \leq p \leq \infty

    is it correct?need verification

    thanks for any comment and suggestion
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  4. #4
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    Quote Originally Posted by dedust View Post
    sure??

    but i need to check \frac{1}{|x|^p}, not \vert x \vert ^p

    \int_{0}^{1} \frac{1}{|x|^p} ~dx = \frac{1}{-p + 1} x^{-p + 1}|_{0}^1 convergent for 0 < p < 1

    so the right answer is \frac{1}{x} \in {L}^p for 0 < p < 1, and \frac{1}{x} \notin {L}^p for 1 \leq p \leq \infty

    is it correct?need verification

    thanks for any comment and suggestion
    Notice that my statement is equivalent to yours (maybe it's confusing since I named all the exponents p). As for your last statement, how do you define L^p for 0<p<1?
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  5. #5
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    Smile

    Quote Originally Posted by Jose27 View Post
    Notice that my statement is equivalent to yours (maybe it's confusing since I named all the exponents p). As for your last statement, how do you define L^p for 0<p<1?
    ohh,..i see,..

    thanks
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