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Let $ p,r \geqslant 1, \; f \in L^r (\mathbb R), g \in L^p (\mathbb R), \; 2/q = 1/p + 1/r$. Here $ 2p / q > 1 $ and $ 2r / q > 1 $ . Also $ \frac{q}{2p} + \frac{q}{2r} = 1$. I want to prove following. $$ \sum_{k=1}^n \left( \int_{a_{k-1}}^{a_k} |g|^p \right)^{\frac{q}{2p}} \left( \int_{a_{k-1}}^{a_k} |f|^r \right)^{\frac{q}{2r}} \leqslant \left( \sum_{k=1}^n \int_{a_{k-1}}^{a_k} | g| ^p \right)^{\frac{q}{2p}}\left( \sum_{k=1}^n \int_{a_{k-1}}^{a_k} | f| ^r \right)^{\frac{q}{2r}} $$ for given real sequence $\{a_k\}$.

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Let $b_k:=\left(\int_{a_{k-1}}^{a_k}|g|^p\right)^{\frac q{2p}}$ and $c_k:=\left(\int_{a_{k-1}}^{a_k}|f|^q\right)^{\frac q{2r}}$. We have by Hölder's inequality for sums, applied to the exponents $\frac{2r}q$ and $\frac{2p}r$ that, $$\sum_{k=1}^nb_kc_k\leq \left(\sum_{k=1}^nb_k^{\frac{2p}q}\right)^{\frac q{2p}} \left(\sum_{j=1}^nc_j^{\frac{2r}q}\right)^{\frac q{2r}}.$$ The LHS corresponds to the LHS in the wanted inequality, and the same for the RHS.

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Oh, I understand this, really thank you. –  Misaj Jun 17 '12 at 17:43
    
You are welcome. –  Davide Giraudo Jun 17 '12 at 17:47
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