Notations: We denote $C_0^1(0,1)$ the collection of all real-valued continuously differentiable function $f$ on $(0,1)$ that vanish at boundary, that is, for any $\epsilon>0,$ the set $$\{x\in (0,1): |f(x)|\geq \epsilon\}$$ is compact in $(0,1).$

Question: For any $f\in C_0^1(0,1)$ and any $\epsilon>0,$ is it true that the set $$\{(x_0,x_1)\in (0,1)^2: |f(x_0)| + |f'(x_1)| \geq \epsilon\}$$ not compact in $(0,1)^2?$

Intuitively the statement seems correct as $f$ and $f'$ may not 'vanish at infinity' at different points. However, I do not know how to formulate this vigorously.

  • $\begingroup$ Vanishing at infinity is not a tight restriction on real-valued functions. I suspect that the functions could do whatever they wanted (within reason of differentiability) in the interval $(0,1).$ Did you by any chance mistype the definition of vanishing at infinity? Maybe there should there be a $1/x$ somewhere? $\endgroup$ – Display name Apr 21 at 6:24
  • $\begingroup$ @Displayname I think the definition given in my post is correct. Can you elaborate on where should I add $1/x$? $\endgroup$ – Idonknow Apr 21 at 6:35
  • $\begingroup$ The functions are defined on $(0,1)$ yet we ask about behavior at infinity. To solve this problem, $(0,1)$ can be mapped to $(1,\infty)$ by inversion. $\endgroup$ – Display name Apr 21 at 6:39
  • $\begingroup$ Vanish at the boundary might be a less confusing terminology? $\endgroup$ – copper.hat Apr 21 at 6:44
  • $\begingroup$ @copper.hat you are right. I should change to your suggestion. $\endgroup$ – Idonknow Apr 21 at 6:46

Try $f(x) = (x \sin {1 \over x} ) ((1-x) \sin {1 \over 1-x} ) $.

Note that $f$ is smooth on $(0,1)$, $\lim_{x \downarrow 0} f(x) = \lim_{x \uparrow 1} f(x) =0$.

The derivative is straightforward, if messy, to compute and we see that $\limsup_{x \downarrow 0} f'(x) = \limsup_{x \uparrow 1} f'(x) = \infty$ and $\liminf_{x \downarrow 0} f'(x) = \liminf_{x \uparrow 1} f'(x) = -\infty$.

In particular, the set $\{ (x_0,x_1) | |f(x_0)|+|f'(x_1)| \ge \epsilon \}$ is contains points of the form $(x_0(n),x_1(n))$ where $(x_0(n),x_1(n)) \to (0,0)$, hence the set is not compact.

  • $\begingroup$ May I know your intuition of obtaining such function? Thanks. $\endgroup$ – Idonknow Apr 21 at 7:59
  • $\begingroup$ Can you elaborate your answer? $\endgroup$ – Idonknow Apr 21 at 15:57
  • $\begingroup$ In what way? ${}$ $\endgroup$ – copper.hat Apr 21 at 16:14
  • $\begingroup$ How can this set be unbounded? It is still a subset of $(0,1)^2$. $\endgroup$ – Nathanael Skrepek Apr 21 at 21:09
  • $\begingroup$ @NathanaelSkrepek: Thanks for catching that. $\endgroup$ – copper.hat Apr 21 at 21:26

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