Let $f\colon [0,1]\to\mathbb C$ be a Lipschitz function. Then $f^\prime \in L_\infty[0,1]$. For a given $\varepsilon > 0$, we may approximate $f^\prime$ in the supremum norm by a simple function $h$ within $\varepsilon$.
It seems that often I can manipulate the values of $h$ (if I start with some specific situations) so that the condition $\|f^\prime - h\|_\infty < \varepsilon$ is satisfied. Is there a way concluding that one can amend $h$ further so that:
- $\|f^\prime - h\|_\infty < \varepsilon$
- $\|f - h(0) - \int_0^\cdot h \|_\infty < 2 \varepsilon $,
- $\inf_{x\in [0,1]}|h(0) + \int_0^x h| > 0$?
Please note that there is room for changing $h$, so if initially we had $0\in h^{-1}(\{0\})$, nothing stops us from changing slightly the value on that preimage, etc.
I am seeking an argument that avoids invoking the initial form of $h$ explicitly (such as this one).
First, I would like to point out that those conditions imply that $|f'(0)-h(0)|<\varepsilon$ and $|f(0)-h(0)|<2\varepsilon$, which gives $|f'(0)-f(0)|<3\varepsilon$. Therefore I assume there's a small mistake in the formulation of the problem.
Now I assume that it's clear that it's possible to find a simple function $h$ such that
Put $g_a(x):=a + \int_0^x h(t)dt $. It's a Lipschitz function, so the image $g_a([0,1])$ is compact. Therefore the condition $\inf_{x\in [0,1]}|g_a(x)| > 0$ is equivallent to $0\notin g_a([0,1])$.
We need a version of Sard's Theorem for Lipschitz functions. I think here is the proper version.
The function $g_{f(0)}$ has image of measure zero, therefore there exists $v\in B(0,\varepsilon)$ such that $v\notin g_{f(0)}([0,1])$, so $0\notin g_{f(0)-v}([0,1])$. Now observe that $$\|f-f_{f(0)-v}\|_\infty = \left\|f-f(0)+v-\int_0^\cdot h\right\|_\infty \leq \left\|f-f(0)-\int_0^\cdot h\right\|_\infty+|v|<\varepsilon+\varepsilon. $$ Therefore for $a:=f(0)-v$ we have