If $f(x)=f(\delta x), \delta>0$ a.e then $f$ is constant? Let $f\in L^{\infty}(0, \infty).$
For $\delta>0,$ $f(x)=f(\delta x)$ all most every where(a.e) on $(0, \infty).$
My Question: Can we expect $f$ is constant function on $(0, \infty)?$ If yes, how?
 A: There is a standard argument for problems like this:
If you have a (compactly supported, continuous) function $g \in L^1((0,\infty))$, consider the convolution (calculated w.r.t. the Haar measure on $(0,\infty)$)
$$
(f \ast g)(x)= \int_0^\infty f(x/y) g(y) \,dy/y.  
$$
The convolution is defined for every $x$ because of $f \in L^\infty$. 
Now an easy calculation using the change of variables formula yields 
$$
(f \ast g)(\delta x) = (f \ast g)(x)
$$
for all $\delta , x >0$. 
This will allow you to conclude that $f \ast g $ is constant. 
Now choosing $g = g_\varepsilon$ as a suitable approximation of unity, you get $f \ast g_\epsilon \to f$ pointwise a.e. (at every Lebesgue point of $f$). This implies that $f$ is constant a.e.   
EDIT: If you know that every (almost everywhere) translation invariant function over the reals is (almost everywhere) constant, you could also consider $x \mapsto f(\exp(x))$. 
This also allows you to calculate the convolution over the reals instead of over $(0,\infty)$. 
