While studying analysis, an interesting question came up:
Prove that if a function
$f: [a,b]\to \mathbb{R} $
Is continuous, where $[a,b]\subseteq \mathbb{R}$, then $f$ is uniformly continuous.
To prove this, I thought of defining a function
$\Delta_f: [a,b]\times \mathbb{R}^+ \to \mathbb{R} $
That sends $(c,\varepsilon)$ to the supremum of $\delta$'s (if it exists, otherwise some predetermined maximum) such that
$|x-c| <\delta \implies |f(x)-f(c)|< \varepsilon$
Uniform continuity is then equivalent to saying that for every $\varepsilon$ there is some $a>0$ such that $\Delta_f(c,\varepsilon)\geq a$ for every $c$. I spent some time trying to prove that the function $\Delta_f(c,\varepsilon)$ is itself continuous (in its first argument at least), but the proof got quite complicated, but then found a counterexample:
\begin{align} f(x) = \begin{cases} x & \text{if $x<1$} \\ 1 & \text{if $1\leq x<2$} \\ \end{cases} \end{align}
Then $\Delta_f(x,1/2)$ is discontinuous (a jump discontinuity) at $x=1/2$
However, I found it strange that there is this very natural function associated with any continuous function, which I have not heard about, so I set out to try and understand this function a bit better, and it seems that it is an approximation to the reciprocal of the derivative of the function, with the accuracy of the approximation being determined by how small you choose $\varepsilon$. more precisely, I suspect that for some differentiable function that has no turning point the following is true
\begin{align} \lim_{\varepsilon\to 0}\frac{\Delta_f(x,\varepsilon)}{\varepsilon} = \frac{1}{f'(x)} \end{align}
This seems superficially related to the derivative, but it is actually quite difference, since you are limiting the "$y$ values" of the function, and seeing how that affects the $x$ values. I tried proving this relation and finding a counterexample, but to no avail. Any insight or a reference to an existing, similar concept will be appreciated. It seems to be almost related to the modulus of continuity, but with the $\varepsilon$ and $\delta$ switched.
Your claim is correct, but first let's reformulate your claim as follows.
Proof sketch:
By the continuity of $f$, it follows $\Delta_f(c,\varepsilon)>0$ for any $\varepsilon>0$ and thus the ratio in $L(c)$ is well-defined. Now we discuss two cases:
If there exists $\varepsilon^*>0$ such that $f$ is constant on $(c-\varepsilon^*,c+\varepsilon^*)\subset(a,b)$, then $\Delta_f(c,\varepsilon)\geq\varepsilon^*$ for any $\varepsilon\in(0,\varepsilon^*)$, this shows $L(c)=0=f'(c)$.
If the above does not hold, then by the continuity of $f$ there exists $\varepsilon^*>0$ such that for all $\varepsilon\in(0,\varepsilon^*)$ it follows $f(c)-\varepsilon$ or $f(c)+\varepsilon$ is in the range of $f$. Fix such a $\varepsilon$, let
$$\underline x=c-\Delta_f(c,\varepsilon)\ \ \text{and}\ \ \bar x=c+\Delta_f(c,\varepsilon).$$
Due to $\Delta_f$ being a supremum and $f$ being continuous, we have $|f(\underline x)-f(c)|=\varepsilon$ or $|f(\bar x)-f(c)|=\varepsilon$. Suppose it is the latter, then $$\frac\varepsilon{\Delta_f(c,\varepsilon)}=\frac{|f(\bar x)-f(c)|}{|\bar x-c|}.$$ As $\varepsilon\to0$, we must have $\bar x\downarrow c$ since $f$ is continuous, this means $$L(c)=\lim_{\varepsilon\downarrow0}\frac\varepsilon{\,\Delta_f(c,\varepsilon)\,}=\lim_{\bar x\downarrow c}\frac{|f(\bar x)-f(c)|}{|\bar x-c|}=f'_+(c),$$ where $f'_+(c)$ is the right-hand derivative at $x=c$. Since $f$ is differentiable, the left-hand and right-hand limit have to agree and therefore $L(c)=f'_+(c)=f'(c)$. The case for $\underline x$ is the same with $f'_+$ replaced by $f'_-$.