How to prove $\left\Vert\nabla \frac{f}{||\nabla f||+\epsilon}\right\Vert\leq {\rm constant}$?

Question

How to prove $\left\Vert\nabla \frac{f}{||\nabla f||+\epsilon}\right\Vert\leq {\rm constant}$?

The $f:\mathbb{R}^n\rightarrow\mathbb{R}$ is a continuous function. The $\epsilon$ is a small positive constant. The $||\cdot||$ stand for norm operation. You can choose any norm (2, Frobenius, etc.) for convenience.

Note: if suppose $f$ is piecewise linear function, it is easy to prove (see Theorem 6 in link). But what I concern is how to prove it without piecewise linear assumption.

Answer 1

From the reference you give, I understand that you are asking for the following result.

Question. Does there exist $C, \epsilon> 0$ such that, for every $f \in C^2(\mathbb{R}^n;\mathbb{R})$, there holds $$\forall x \in \mathbb{R}^n, \quad \left\Vert\left(\nabla \frac{f}{||\nabla f||+\epsilon}\right)(x)\right\Vert \leq C.$$

Outside of the particular case of affine functions (which is considered in the reference you mention), such a result is false even with $n=1$, because the considered quantity contains a second-order derivative which has no reason to be bounded by the first-order one.

Let us consider $f \in C^\infty(\mathbb{R};\mathbb{R})$ and $x^* \in \mathbb{R}$ such that $f(x^*) > 0$ and $f'(x^*) > 0$. Hence, we want to prove that $|g'(x^*)| \leq C$ where $$g(x) := \frac{f(x)}{f'(x) + \epsilon}.$$ Thus $$ g'(x^*) = \frac{f'(x^*)}{f'(x^*) + \epsilon} - \frac{f(x^*) f''(x^*)}{(f'(x^*)+\epsilon)^2}. $$ While the first term is indeed bounded (by $1$), the second one can be arbitrarily large. Consider $$f_n(x) := 1 + (x-x^*) - n \frac{(x-x^*)^2}{2}.$$ Then $f_n(x^*) = 1$, $f_n'(x^*) = 1$ and $f_n''(x^*) = n$. Thus $$ g'(x^*) = \frac{1}{1+\epsilon} + \frac{1 \cdot n}{(1+\epsilon)^2}. $$ Hence, taking $n$ arbitrarily large prevents the existence of such a constant $C$.