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Thread: Lp spaces set inclusion

  1. #1
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    Lp spaces set inclusion

    I am stuck on trying to show for that a fixed measure space (X,\cal{S},\mu) with 1 \leq r \leq p \leq s < \infty we have
     L_{p} \subset L_{r} + L_{s} .

    Cheers!
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  2. #2
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    I have attempted the question. I am not sure if the proof is valid though. Any comments will be appreciated!
    Here it goes:


    Let f\in L_{p} which means we have

    \int \vert\hspace{0.5mm}f\vert^{p}\hspace{1mm}d\mu<\inf  ty.

    Write f:=g+h where the functions g and h are measurable functions.

    Since the measuable function \vert\hspace{0.5mm}g+h\vert is integrable, it is almost everywhere finite.

    Then WLOG we have

    \vert\hspace{0.5mm}g\vert \leq \vert\hspace{0.5mm}g+h\vert<\infty\hspace{3mm}\tex  t{almost everywhere}

    from which it follows that

     \Vert g\Vert_{r} := \Bigg( \int \vert g\vert^{r}\hspace{1mm}d\mu\Bigg)^{\dfrac{1}{r}}<\i  nfty_{.}

    Similarly for h, we obtain

    \Vert h\Vert_{s} := \Bigg( \int \vert h\vert^{s}\hspace{1mm}d\mu\Bigg)^{\dfrac{1}{s}}<\i  nfty_{.}
    Last edited by willy0625; May 15th 2010 at 10:23 PM.
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  3. #3
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    Quote Originally Posted by willy0625 View Post
    I have attempted the question. I am not sure if the proof is valid though. Any comments will be appreciated!
    Here it goes:


    Let f\in L_{p} which means we have

    \int \vert\hspace{0.5mm}f\vert^{p}\hspace{1mm}d\mu<\inf  ty.

    Write f:=g+h where the functions gand h are measurable functions.

    Since the measuable function \vert\hspace{0.5mm}g+h\vert is integrable, it is almost everywhere finite.

    Then WLOG we have

    \vert\hspace{0.5mm}g\vert \leq \vert\hspace{0.5mm}g+h\vert<\infty\hspace{3mm}\tex  t{almost everywhere}

    from which it follows that

     \Vert g\Vert_{r} := \Bigg( \int \vert g\vert^{r}\hspace{1mm}d\mu\Bigg)^{\dfrac{1}{r}}<\i  nfty_{.}

    Similarly for h, we obtain

    \Vert h\Vert_{s} := \Bigg( \int \vert h\vert^{s}\hspace{1mm}d\mu\Bigg)^{\dfrac{1}{s}}<\i  nfty_{.}
    I don't like your argument, you can't say |g|\leq |g+h| (what if h is negative), and the part where you took integrals is fishy at best.

    Some cases are simple enough. For example if \mu (X)<\infty and 1\leq p \leq q \leq \infty then L^q \subset L^p (this is not difficult to prove, just use Hölder's inequality).Intuitively, if on the contrary, X had no finite measure subset the inclusions would be reversed (I don't yet have a proof for this), and the general case would (?!) follow by decomposing your space in this (or some similar) fashion.
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  4. #4
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    Quote Originally Posted by Jose27 View Post
    Some cases are simple enough. For example if \mu (X)<\infty and 1\leq p \leq q \leq \infty then L^q \subset L^p (this is not difficult to prove, just use Hölder's inequality).Intuitively, if on the contrary, X had no finite measure subset the inclusions would be reversed (I don't yet have a proof for this), and the general case would (?!) follow by decomposing your space in this (or some similar) fashion.
    Following up on that idea, let f\in L^p(X), and let Y = \{x\in X:|f(x)|\geqslant1\}. Let g = f\big|_Y (that is, g equals f on Y and g is zero outside Y) and similarly h = f\big|_{X\setminus Y}. Clearly f=g+h. Then \mu(Y)<\infty and so (as Jose27 points out) g\in L^r(X). Also, it's easy to see that h\in L^s(X), because \int\!\!|h|^s = \int\!\!|h|^p|h|^{s-p}\leqslant\int\!\!|h|^p<\infty. Thus f\in L^r(X) + L^s(X), as required.
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