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Is an almost everywhere (ae) Homogeneous function $f$of degree $0$ equals to a constant for almost every $x \in (0,\infty)$ given that $ f $ is measurable?

Let $f : \mathbb R \to \mathbb R$.

If $f(ax)=f(x)$ ae for any $a>0$

Then $f(x)=c$ for almost every $x \in (0,\infty)$, where $c$ is a constant.

Is the above true?

I know it is true if $ f $ is locally integrable see here

I encountered this problem while studying bounded linear operators $ T:L^2 \to L^2$

ibnAbu
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    I think that this is a classical functional equation in disguise. If $f\colon \mathbb R_{>0}\to \mathbb R$, letting $ g(y):=f(e^y)$ yields a solution to the functional equation $g(y_1+y_2)=g(y_1)$. – Giuseppe Negro May 08 '19 at 17:54
  • @GiuseppeNegrog yeah it is. Any idea regarding the functional equation ? – ibnAbu May 08 '19 at 18:42
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    From what I wrote in the previous comment, it seems obvious to me that $g$ must be constant. Please check carefully, I ignored the "almost everywhere" part which is the trick one. – Giuseppe Negro May 09 '19 at 08:03

2 Answers2

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Lemma $1$ $\quad A\subset\mathbb{R}$ is measurable and $m(A)>0$. then $m\left(\mathbb{R}-\bigcup_{q\in\mathbb{Q}}q\cdot A\right)=0$.

${ Proof}$ It's sufficient to show $m\left([1/n,n]-\bigcup_{q\in\mathbb{Q}}qA\right)=0$, where $n\in\mathbb{Z}-\{0\}$.

$\forall \alpha<1$, $\exists$ an interval $I$ s.t. $m(I\cap A)>\alpha\cdot m(I)$. Obviously, $[1/n,n]\subset\bigcup_{q\in\mathbb{Q}}qI$. By arranging the intervals properly, we can find finitely many $\{q_k\}_{k=1}^N\subset\mathbb{Q}$ such that the corresponding $\{I_k\}_{k=1}^N:=\{q_kI\}_{k=1}^N$ satisfies

$$[1/n,n]\subset\bigcup_{k=1}^N I_k,\quad \sum_{k=1}^n m(I_k)\leq3n.$$

Then we have \begin{eqnarray*} [1/n,n]-\bigcup_{q\in\mathbb{Q}}q_kA&\subset&[1/n,n]-\bigcup_{k=1}^Nq_kA\\ &\subset&\left\{[1/n,n]-\bigcup_{k=1}^NI_k\right\}\bigcup\left\{\bigcup_{k=1}^N[(q_kA)^c\cap I_k]\right\}\\ &\subset&\left\{\bigcup_{k=1}^N[(q_kA)^c\cap I_k]\right\} \end{eqnarray*}

Thus, $$m\left\{[1/n,n]-\bigcup_{q\in\mathbb{Q}}\{qA\}\right\}\leq m\left\{\bigcup_{k=1}^N[(q_kA)^c\cap I_k]\right\}\leq (1-\alpha)3n.$$

Let $\alpha\to 1$, so we proved the lemma, and directly have lemma 2.

Lemma $2$ $\quad r\in\mathbb{R},$ $m\{f\leq(\geq)r\}>0\Rightarrow f\leq(\geq)r\ a.e.$

$ Proof:$ $f\overset{a.e.}{\leq} r$ on $q\cdot\{f\leq r\}$ because of homogeneity, where $q\in\mathbb{Q}$. Unit of all $q\cdot\{f\leq r\}$ with respect to $q$ covers $\mathbb{R}$ almost everywhere due to lemma 1, so we proved lemma 2.

Let $R_1:=\{r\in\mathbb{R}:m\{f\leq r\}>0\},\ R_2:=\{r\in\mathbb{R}:m\{f\geq r\}>0\}$. $f\overset{a.e.}{\equiv}\inf R_1=\sup R_2$.

Seems we don't need any extra condition here, just measurability.

Selene
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  • I don't understand. What 's lemma 2 about ? – ibnAbu May 08 '19 at 18:44
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    @ibnAbu Use this lemma to conclude $f$ is bounded. So $f$ is locally integrable naturally. – Selene May 08 '19 at 18:59
  • @ibnAbu The proposition is still true when you allow $f$ to take $+\infty$ and $-\infty$. – Selene May 08 '19 at 20:46
  • Lemma 1 seems to prove the proposition but i'm finding it difficult to understand the proof of the Lemma 1.Can give reference where the proof can be found ? For example it is not always true that you can always have I interval $ I $ you can always have $ m (I \bigcap A) > \alpha m (I) $ however $ I $ can be a countable union of intervals covering $ A $ – ibnAbu May 09 '19 at 13:08
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    @ibnAbu I proved this myself. Not sure if there's any reference. That conclusion involving $I$ is true. Take ${I_k}k$ s.t. $\sum m(I_k\cap E)\geq m(E)>\alpha\sum m(I_k)$, so there must be one $I{k_0}$ that we want. – Selene May 09 '19 at 13:29
  • The proof of Lemma 1 is a simple consequence of the fact that the it is true for an interval and $ m (q \cdot A) = qm (A) $ – ibnAbu May 09 '19 at 15:21
  • I think the lemma is true by replacing $ m$ with outer measure $\mu^*$ without the condition $A $ is measurable – ibnAbu May 09 '19 at 15:33
  • @ibnAbu If do so, you may get into trouble estimating $\mu^*(I\cap A^c)$. but you can try if you want. – Selene May 09 '19 at 19:34
  • I tried to simplify your proof – ibnAbu May 24 '19 at 17:47
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The proof follows from lemma 2 given by XIADO : Here is a simplified proof

Proof of Lemma 1:

It suffice to prove it for some measurable $ B \subset A $ and $ B\subseteq I=[a ,b]$ and $ m (B) > 0,|a|,|b| > 0$:

There some measurable $ S$ with $ m (S) > 0$ and by Lebesgue density theorem there some $ x \in S \subseteq I $ and some $ 0<r<\frac{m(I)}{2},1>\epsilon > 0$ such that $ I_r=(x-r, x+r)$, $ m (S \bigcap I_r) > (1-\epsilon)m (I_r)$

It also obvious that for $ S_q =q\cdot S $, $ I_{r, q}=q\cdot I_r $,$ m(S_q)=qm (S) , m(I_{r, q})=qm (I_r) $, and $ (S_q\bigcap I_{r, q}) >q (1-\epsilon)m (I_r)$ ,where $q \in \mathbb {Q} $

Now we can take subsequence {$ I_{r, q} $}that is pairwise disjoint such that $\bigcup_q I_{r, q} \supseteq I $ and $max${$|q|$} $\le n^2,n^2=\frac {|b|}{|a|}$

Define $ B = S \bigcap I_r $ , $ B_q=q\cdot B $, It follows $ m(\bigcup_q B_{q}) >(1-\epsilon)m (I)$

$ m(I-\bigcup_q B_{q}) = m (I) - m ( I\bigcap\bigcup_q B_{q}) $

$ m(I-\bigcup_q B_{q}) \le \epsilon m (I)+ 2rn^2 $

$A_q=q\cdot A$

$ I- \bigcup_q A_q \subseteq I-\bigcup_q B_{q} $

Therefore $ m(I-\bigcup_q A_q) \le \epsilon m (I)+2rn^2 $

Letting $r,\epsilon \to 0$ the result follows

Proof of Lemma 2

define $A=${$x \in \mathbb{R} : f(x) < r$}

$m(q\cdot A-A)=0$ since $f(qx)=f(x)$ ae

so if $m(A) >0$ ,since $ m(I-\bigcup_q A_q)=0$

It follows $f(x) < r$ ae, hence it is bounded almost everywhere and so it is locally integrable

Using the result of see here shows $ f (x) $ is constant ae on $(0, \infty)$

ibnAbu
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