continuous function can/cannot be covered by a step function

This problem described below hits me when I tried to understand the $$M_1$$ topology of Skorokhod space. The original problem is to show a non constant continuous path is bounded away (w.r.t the $$M_1$$ distance) from the space of step functions with no more than $$k$$ jumps. I rephrase it to make the notation simpler, hope this will not incur too much confusion.

Let $$f(t), t\in [0,1]$$ be a continuous function. Restrict it to be non-constant and $$f(0) = 0$$.

For any step function $$\xi(t), t\in[0,1]$$ with $$\xi(0) = 0$$, its graph $$\Gamma(\xi)$$ is a subset of $$[0,1]\times \mathbb{R}$$ containing the step function line plus lines that concatenate the jump points. Make $$\Gamma(\xi)$$ thicker will give us a blow up set around $$\Gamma(\xi)$$, defined as

$$\xi_\delta = \{(t',x')\in[0,1]\times \mathbb{R}: \|(t,x)- (t',x')\|_\infty < \delta\text{ for some }(t,x)\in\Gamma(\xi)\}$$

The parameter $$\delta$$ defines how 'thick' the set is. The motivation of the set $$\xi_\delta$$ is related to the $$M_1$$ distance.

Denote $$\mathbb{D}_{\leqslant k}$$ be the set of step functions on [0,1] that with no more than $$k$$ jumps.

My questions is, given the function $$f(t)$$ mentioned at the beginning, could we find a $$\delta$$ small enough (depends on $$f$$, $$k$$, uniform on all $$\xi\in\mathbb{D}_{\leqslant k})$$ , such that the graph $$f$$ could not be covered by $$\xi_\delta$$.

I believe such $$\delta$$ exists, but since I only require $$f$$ to be continuous, there is no further smooth information, thus I suspect some fundamental topology theorem will be required to show this. Could anyone give some hint for this? Thank you

• Just to understand your quantifiers... you want to show that there exists a continuous $f: [0,1] \rightarrow \mathbb{R}$ so that for all $k$ there exists $\delta>0$ so that there is no $\xi\in\mathbb{D}_{\leqslant k}$ with $f \in \xi_\delta$? Commented May 23, 2022 at 20:21
• To clarify, with a given $f$, $k$, I want to find a $\delta$ for all $\mathbb{D}_{\leqslant k}$, such that the graph of $f$ could not be covered by $\xi_\delta$ for any $\xi\in\mathbb{D}_{\leqslant k}$ Commented May 24, 2022 at 14:57

As $$f$$ is non-constant, $$f(t_0) \neq 0$$ for some $$t_0$$.
By mean value theorem, we can find $$2k + 1$$ points $$x_0, \ldots, x_k$$ s.t. $$f(x_i) = \frac{f(t_0)}{2k} \cdot i$$.
Take $$\delta = \min\left(\frac{f(t_0)}{2k}, \min_i |x_i - x_{i + 1}|\right) / 3$$. Then no two points of form $$(x_i, f(x_i))$$ can be covered by the same blowed segment of $$\xi_\delta$$: they can't be covered by horizontal segment because there values differ by more than $$2\delta$$, and they can't be covered by the same vertical segment because they are more then $$2\delta$$ apart of each other.
So to cover all this points we need at least $$2k+1$$ horizontal and vertical segments. As number of vertical segments is one less than number of jumps, this means we need at least $$k + 1$$ jumps (as $$k$$ jumps give us at most $$2k - 1$$ segments).