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
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).