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$(\Omega, \mathcal{F}, u)$ is a measure space, and $L^p(\Omega, \mathcal{F}, u)$ is its $L^p$ space. Define $N_p(f) = \int_{\Omega} |f|^p\, d\mu$. $\forall f, g \in L^p(\Omega, \mathcal{F}, u)$,

  1. when $0, $N_p(f+g) \leq N_p(f)+N_p(g) $ is true according to Wikipedia. This can help to show that $L^p(\Omega, \mathcal{F}, u)$ is a vector space. I was wondering how to prove the inequality is true?
  2. when $p \geq 1$, is the inequality $N_p(f+g) \leq N_p(f)+N_p(g) $ still true?

    If not,

    (1) can the inequality be modified to be true? Note I am not asking about the triangle inequality of $L^p$ norm.

    (2) how can one show that $L^p(\Omega, \mathcal{F}, u)$ is a vector space?

Thanks and regards!

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    For 2, consider $p=2$ with positive $f$ and $g$.2011-05-11

1 Answers 1

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  1. This holds because $(a+b)^p\leq a^p+b^p$ when $0, $a\geq 0$, and $b\geq 0$, as seen for example here.

  2. Not if $p>1$. For $p\geq 1$ you have $(a+b)^p\leq 2^{p-1}(a^p+b^p)$ when $a\geq0$ and $b\geq 0$. This can be proved by rearranging the inequality $\displaystyle \left(\frac{a+b}{2}\right)^p\leq\frac{1}{2}(a^p+b^p)$, which follows from convexity of the function $x\mapsto x^p$. You need the factor of $2^{p-1}$, as can be seen by taking $a=b=1$. A less sharp version was asked about here.

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    (1) Yes, because $|f+g|^p\leq 2^{p-1}(|f|^p+|g|^p)$ holds pointwise. (2) No. Notice that 2^{p-1}<1 in that case, and consider what happens when $g=0$. (3) No. Again consider what happens when $g=0$.2011-05-11