For continuously differentiable $f,$ 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 $(0,1)^2?$

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.

• 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? – Display name Apr 21 at 6:24
• @Displayname I think the definition given in my post is correct. Can you elaborate on where should I add $1/x$? – Idonknow Apr 21 at 6:35
• 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. – Display name Apr 21 at 6:39
• Vanish at the boundary might be a less confusing terminology? – copper.hat Apr 21 at 6:44
• @copper.hat you are right. I should change to your suggestion. – 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.

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