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Let me recall some notations: The mean oscillation of a locally integrable function $u$ (i.e. a function belonging to $ L^1_{\textrm{loc}}(\mathbb{R}^n))$ over a cube Q in $\mathbb R^n$ (which has sides parallel to axis) is defined as the following integral: $\frac{1}{|Q|}\int_{Q}|u(y)-u_Q|\,\mathrm{d}y$,

where $|Q|$ is the volume of $Q$, i.e. its Lebesgue measure, $u_Q$ is the average value of $u$ on the cube $Q$, i.e. $u_Q=\frac{1}{|Q|}\int_{Q} u(y)\,\mathrm{d}y$.

A BMO function is any function u belonging to $L^1_{\textrm{loc}}(\mathbb{R}^n)$ whose mean oscillation has a finite supremume over the set of all cubes $Q$ contained in $\mathbb R^n$.

I could find many examples for functions in BMO. But I could not find a function $u:\mathbb R^n\to\mathbb R$ which is not constant, so that $u$ in BMO and $u(tx)=u(x)$ for almost everywhere $t\in[0;1]$, and for every $x\in \mathbb R^n$.

I also want to find such function $u$ in BMO so that $u(tx)=u(t)$ for almost every $t\neq0$, and every $x$

So my question is that: does exits such function $u$, and could you give me any example.

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You have to assume $u(tx)=u(x)$ for all $t\in\mathbb R_{>0}$ otherwise take $t=0$. –  Davide Giraudo Mar 18 '12 at 12:59
    
Thank you, I have edited my question –  Hai Minh Mar 18 '12 at 13:56
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2 Answers 2

up vote 3 down vote accepted

There are many such functions.

Consider: $$f(x) = [x_i \geqslant 0 \text{ for all $i = 1, \ldots, n$}].$$

Or (thanks robjohn) for all $i = 1, \ldots n$ the functions $$g_i(x) = \frac{x_i}{|x_i|}.$$

For another example for your new function. Say you are working in $\mathbf R^2$. Then define $f(x, y) = 1$ everywhere except on the line $x = y$. There we let $f = 0$.

So, $f(tx, ty)$ is $1$ as long as $tx \neq ty$ or $x \neq y$. It is $0$ if $tx = ty$. That is if $x = y$. So, your function is not constant.

Of course, this is only a measure $0$ set where the function is different, however, you can swipe the line from $x = y$ to $x = -y$ and get both in positive measure.

To make this explicit consider the set in $\mathbf R^2$, $A := \{(x, y) \mid |x| \leq |y|\}$, that is this region: The set $A$

Now consider the function $f(x, y) := [(x, y) \in A]$. This notation for the indicator function is the Iverson Bracket.

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Yes I also edit my question in which I need to find function $u$, so that $u(tx)=u(x)$ for any $t\neq0$. It's easy to see that your second example is not satisfied. By the way, could you explain me your first example? –  Hai Minh Mar 19 '12 at 0:29
    
The original question stated $t \in [0,1]$, then it is satisfied. The first example is the indicator function where all $x_i$ are positive, then it is $1$ otherwise $0$. Otherwise, you can put a sign flip somewhere. –  Jonas Teuwen Mar 19 '12 at 9:15
    
Could you help me to find examples for second question? –  Hai Minh Mar 21 '12 at 11:12
    
@HDHung Why would you want that? The definition of homogeneity usually only requires $t > 0$. –  Jonas Teuwen Mar 21 '12 at 18:00
    
@ Janas Teuwen: I know the definition of homogeneity usually like that. But, though, these new type of functions (if they exists) shall help me in extending some new results on BMO. I think such functions exist, but I could not find any non-trivial example. –  Hai Minh Mar 21 '12 at 22:49
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Let $S_{n-1}=\{x\in\mathbb{R}^n:|x|=1\}$ be the unit sphere in $\mathbb{R}^n$ and $\phi\colon S_{n-1}\to\mathbb{R}$ a bounded even function, that is $$ \sup_{\omega\in S_{n-1}}|\phi(\omega)|=\|\phi\|_\infty<\infty,\quad\phi(-\omega)=\phi(\omega)\quad\forall\omega\in S_{n-1}. $$ Define $$ u(x)=\begin{cases} \phi(x/|x|) & x\ne0,\\ 0 & x=0. \end{cases} $$ Then $$ u\in L^\infty(\mathbb{R}^n)\implies u\in BMO(\mathbb{R}^n)\text{ and }u(t\,x)=u(x)\quad\forall t\ne0,\forall x\in\mathbb{R}^n. $$

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Very nice, thanks both of you Julian Aguirre and Jonas Teuwen –  Hai Minh Mar 27 '12 at 15:41
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