Example for a particular function

Question

I am looking for an example of a nonnegative function (preferably continuous) that approaches $0$ near $0$, and for any positive number $B$, it is not monotonic on $(0,B)$ and $f(x/2) > f(x)$ for some $x\in(0,B)$. I have found that the function with the fixed discontinuity $f(x)=0$ if $x=0$ and $x\sin(1/x)$ otherwise satisfies the limit condition, it is also not monotonic on any neighborhood of $0$ but I don’t think it is possible to satisfy the last condition. My intuition why it fails is because although it is increasing and decreasing indefinitely around zero, halving has an effect too “big” on how the function approaches zero and the new point $x/2$ will be close enough to zero to disallow the possibility that it is more than $f(x)$. I would appreciate an explanation and confirmation of whether my thoughts make sense. Thanks in advance!

Answer 1

What about defining $f(\frac{1}{2^{n}}) = \frac{1}{n}$ if $n$ is even and $f(\frac{1}{2^n}) = 0$ if $n$ is odd. Then you can extend $f$ to a continuous function on $(0,1)$ by interpolating linearly.

By construction, for every $B>0$, $f(x/2)=x/2 > 0 = f(x)$ for some $x\in(0,B)$ (in fact for infinitely many $x\in (0,B)$), and $f(x)\to 0$ as $x\to 0^+$.

The reason to expect that such a function exists is because although $f(x)\to 0$ as $x\to 0$, the relative values of any two points $f(x/2)$ and $f(x)$ in a small interval containing $0$ are unconstrained by this limit condition. For example, by the limit definition, for each $\epsilon>0$, there exists $\delta>0$ so $f(x/2)$ and $f(x)$ must both be less than $\epsilon$ for $x<\delta$. However, I can make $f(x/2) = \epsilon$ and $f(x) = 0$ (or some other very small positive value like $\epsilon^3$).