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 $(0,1)^2?$

Question

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 $(0,1).$

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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.

Answer 1

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.