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I am trying to show that the function $\log||\log |x||$ is in VMO. The book that I was reading just mentioned that it was a straightforward verification. Can some help me to figure it out?

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What is VMO...? –  Aryabhata Apr 17 '12 at 22:55
    
Vanishing mean oscillation, see here: en.wikipedia.org/wiki/Bounded_mean_oscillation –  Paul Apr 17 '12 at 22:59

1 Answer 1

Since we have a positive function, we can deal with ball instead of cubes (it's allowed without this assumption, but not trivial and needed here, and will allow us to use the fact that the function, called $f$, is radial over $\mathbb R^d$), and which are centered at $0$. First, we write \begin{align}m(B(0,a),f)&=\frac 1{a^d}\int_0^ar^{d-1}\log|\log r|dr\\\ &=\int_0^1 s^{d-1}\log|\log s+\log a|ds\\\ &=\int_0^1s^{d-1}\log|\log a|\left|\frac{\log s}{\log a}+1\right|ds\\\ &=\frac{\log |\log a|}d+\int_0^1s^d\log\left|\frac{\log s}{\log a}+1\right|ds. \end{align} Now, denoting $\mathcal O(B(0,a),f)$ the oscilation over the ball $B(0,a)$ we get \begin{align} \mathcal O(B(0,a),f)&=\int_0^1\left|\int_0^1\left((\log|\log t+\log a|)t^{d-1}-m(B(0,a),f)\right)ds\right|dt\\\ &\leq 2\int_0^1s^d\log\left|\frac{\log s}{\log a}+1\right|ds. \end{align} An argument of dominated convergence gives us the result.

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Thank you very much Davide! –  Sarah Grady Apr 22 '12 at 20:27
    
Davide. Can you please explain how we go from the first equality (the one containing |log r|) to the second equality (the one containing |log s + log a|? Thanks in advance! –  Sarah Grady Apr 23 '12 at 1:12
    
I used the substitution $as=r$. –  Davide Giraudo Apr 23 '12 at 9:51
    
Thank you very much ! –  Sarah Grady Apr 23 '12 at 16:47
    
You are welcome. –  Davide Giraudo Apr 23 '12 at 16:51

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