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 $\{a_k\}$.
An Inequality involving integrations (Hölder-like).
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real-analysis
inequality
1 Answers
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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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0You are welcome. – 2012-06-17