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The oscilation $o(f,x_{0})$ of a function $f:\mathbb{R}\rightarrow \mathbb{R}$ at a point $x_{0}$ is the infimum of the oscilations of the function $f(x)$ at each neighbourhood of the point $x_{0}$, where the oscilation at a set $U$, denoted by $o(f,U)$, is the difference $\sup_{x\in U}f(x)-\inf_{x\in U}f(x)$.

Suppose that $f(x)$ has a oscilation $o(f,x)=1$ at each $x\in \mathbb{R}$ with $x\neq 0$. Which are the possible value of the oscilation of $f$ at $x=0$?.

Attempt of proof: I have considered the following neighbourhoods:

  • $B(\frac{1}{2^{n}},\frac{1}{2^{n-1}})$, the ball with center at $\frac{1}{2^{n}}$ and radius $\frac{1}{2^{n-1}}$.
  • $B(0,\frac{1}{2^{n}})$, the ball with center at $0$ and radius $\frac{1}{2^{n}}$.

Then we have that $B(\frac{1}{2^{n+1}},\frac{1}{2^{n}})\subset B(\frac{1}{2^{n}},\frac{1}{2^{n-1}})$, and this implies that $1 \leq o(f,B(\frac{1}{2^{n+1}},\frac{1}{2^{n}}))\leq o(f,B(\frac{1}{2^{n}},\frac{1}{2^{n-1}}))$. Due to this inequality we have that the limit $lim_{n\to \infty} o(f,B(\frac{1}{2^{n}},\frac{1}{2^{n-1}})))$ exists, and shall call it $\alpha$. Observe that $1 \leq \alpha$.

On the other $B(0,\frac{1}{2^{n}})\subset B(\frac{1}{2^{n}},\frac{1}{2^{n-1}})$, so $0\leq o(f,B(0,\frac{1}{2^{n}}))\leq o(f,B(\frac{1}{2^{n}},\frac{1}{2^{n-1}}))$. Taking limit as $n\to \infty$ we have

$0\leq o(f,0) \leq \alpha$

Remark. This is an estimation of the range of values that $o(f,0)$ can take, but I am not sure that this solve the problem.

DIEGO R.
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  • This argument is not going in the right direction. It is always true that $o (f,0) \geqslant 0$. That does not mean it is in an interval $[0,\alpha]$ where $\alpha < 1$. The term $o(f,o(f,B(\frac{1}{2^{n}},\frac{1}{2^{n-1}})))$ has no meaning. – RRL Apr 13 '19 at 05:04
  • @RRL. There was a mistake. It has been edited. – DIEGO R. Apr 13 '19 at 05:43
  • Ok, thanks. But you have not solved the problem. We must have the oscillation at 0 at least $1$. I showed this in two different ways. – RRL Apr 13 '19 at 05:46
  • Yes that is a concrete answer. Thanks. – DIEGO R. Apr 13 '19 at 05:47

1 Answers1

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The function $x \mapsto o(f,x)$ is upper semi-continuous, proved here, so $\{x:o(f,x) < \alpha\}$ is open for any $\alpha$. Thus it is impossible that $o(f,0) < 1$ since the oscillation equals $1$ in every punctured neighborhood of $x =0$. The set $\{x:o(f,x) <1\} = \{0\}$ is not open.

We can also argue directly using the fact that if $A \subset B$ then $o(f,A) \leqslant o(f,B)$. Take any interval $I_\delta= (-\delta,\delta)$. For any $x \neq 0$ in $I_\delta$, take $\delta_1$ such that $I_1 =(x- \delta_1,x+\delta_1) \subset (-\delta,\delta)$. We then have

$$1 = o(f,x) \leqslant \sup_{u,v \in I_1}|f(u) - f(v)| \leqslant \sup_{u,v \in I_\delta}|f(u) - f(v)| $$

Thus,

$$o(f,0) = \lim_{\delta \to 0} \sup_{u,v \in I_\delta}|f(u) - f(v)|\geqslant 1$$

Can we have $o(f,0) > 1$? Yes -- for example take $f(0) =2 $ and $f$ taking values of $1$ and $0$ at all other rational points and irrational points in $[-1,1]$, respectively.

RRL
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