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I was reading BMO spaces (John-Nirenberg) on wikipidia http://en.wikipedia.org/wiki/Bounded_mean_oscillation. There they define BMO norm as $$sup_{Q}\frac{1}{Q}\int_Q |u(y) - u_Q|dy$$ where $u_Q$ is the average of $u(y)$ over $Q$ and the supremum is taken over all cubes of arbitrary diameter.

Questions: 1. I think what can be the definition of BMO spaces on a torus $T^n = S^1 \times ...\times S^1$? We cannot have cubes of arbitrary diameter. Maybe we can look at the supremum over cubes whose diameter is smaller than or equal to that of torus?

2 What will happen if we take the supremum over cubes $Q$ whose diameter is less than or equal $r$, say, r being very small? Will that give the same norm? Thanks please.

Edit: Question 3. Please also advise a little about non-quotient spaces?

Edit:Joonas Ilmavirta answers Q1 and 3 below. Someone please look at Question 2.

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  • $\begingroup$ [deleted a previous comment which was based on a misreading of the actual question] $\endgroup$ – Yemon Choi Jul 21 '14 at 21:23
  • $\begingroup$ Regarding Q2: it's been a long time since I did the relevant calculation, but I believe that (for n=2 say) if you take a square and subdivide into 4, then the mean oscillation over the big square is bounded above by the max of the mean oscillations on each subsquare. (The BMO condition is interested in behaviour on all sufficiently small scales, averaging on a coarser scale can only make things better) $\endgroup$ – Yemon Choi Jul 21 '14 at 22:34
  • $\begingroup$ @YemonChoi So basically you say we are getting equivalent norms? I was suspecting that too, but cannot carry out details..Thank you. $\endgroup$ – Kenpachi Jul 21 '14 at 22:38
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A function $T^n\to\mathbb R$ can be thought of as a periodic function $\mathbb R^n\to\mathbb R$. (Periodic in each variable, that is.) This way you can use the BMO norm of $\mathbb R^n$ on $T^n$ as well, and suddenly the size of the cube is no problem.

More explicitly, identify $T^n=\mathbb R^n/\mathbb Z^n$ and let $q:\mathbb R^n\to T^n$ be the quotient map. Then simply set $$ ||u||_{BMO(T^n)} = ||q^*u||_{BMO(\mathbb R^n)}, $$ where $q^*u(x)=u(q(x))$ is the pullback of $u$ over $q$.

This may not be the only possible choice, but at least it seems natural.

Edit (to answer the edit in the question): For a more general space $X$, we could define the BMO norm of $u:X\to\mathbb R$ as $$ \sup_{x\in X,r>0}|B(x,r)|^{-1}\int_{B(x,r)}|u-u_{B(x,r)}|. $$ To make sense of this, you only need $X$ to have a metric (or some kind of balls) and a measure. If $X$ is bounded, then it just happens that $B(x,r)=X$ for $r$ large enough, which is no issue.

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  • $\begingroup$ Thanks. Yes, this is natural. But I would also like to see something for spaces which are not quotients, maybe the sphere $S^n$? $\endgroup$ – Kenpachi Jul 21 '14 at 21:29
  • $\begingroup$ Added a more general remark, including all Riemannian manifolds. $\endgroup$ – Joonas Ilmavirta Jul 21 '14 at 21:44
  • $\begingroup$ Many many thanks Joonas Ilmavirta...Could you advise on Q 2 please? $\endgroup$ – Kenpachi Jul 21 '14 at 21:53

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