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I am interested in the following inequality: \begin{equation} \int\limits_\Omega \left\vert f_1 - f_2 \right\vert^p ~ d\mu \le C_p \left[ \int\limits_\Omega \left\vert f_1 \right\vert^p ~ d\mu + \int\limits_\Omega \left\vert f_2 \right\vert^p ~ d\mu - 2\int\limits_\Omega \left\vert \frac{f_1 + f_2}{2} \right\vert^p ~ d\mu \right]^{\min(1,p/2)}. \end{equation} This is supposed to hold for functions $f_1,f_2 \in L_p$ (for an arbitrary measure space) such that $\Vert f_i \Vert_p \le 1$, $i=1,2$, and with $1 < p < \infty$. (In fact, the expressions make no sense when $p=\infty$, and for $p=1$ we immediately run into a contradiction when choosing one of the functions to vanish identically.)The constant $C_p$ should depend only on $p$.

I stumbled on this in the book Topics in almost everywhere convergence by Adriano Garsia. It is a crucial part of a proof, yet the only given hint as to why this is true is that it is obtained by expressing the RHS integrand \begin{equation} \left\vert f_1 \right\vert^p + \left\vert f_2 \right\vert^p - 2 \left\vert \frac{f_1 + f_2}{2} \right\vert^p \end{equation} as an integral involving the second derivative of $\left\vert x \right\vert^p$. I am not sure what to make of this and, having thought about it, I do not think it is that obvious. However, I don't think it is a very deep result, either.

My question is: Can anyone give me a reference where this is shown? Or can anyone point me a little more to the right direction, so that I might be able to prove it myself?

Thanks in advance, I appreciate any thoughts!

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I suggest you look for "uniform convexity of L^p spaces" or the "modulus of convexity of L^P spaces". –  Dirk Sep 8 '13 at 14:15
    
Thank you Dirk, exactly what I needed! –  Henry Wegener Sep 8 '13 at 15:24
    
So, to whom it may concern: This is basically Clarkson's inequality. The constant C_p can be chosen as 2^p for 1<p<2 and 2^(p-1) for p>2. :-) –  Henry Wegener Sep 8 '13 at 15:35
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