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?
Showing that a function below is in VMO
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0Vanishing mean oscillation, see here: http://en.wikipedia.org/wiki/Bounded_mean_oscillation – 2012-04-17
1 Answers
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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0I used the substitution $as=r$. – 2012-04-23