# Is there an alternative definition of uniform differentiability that implies differentiability, instead of requiring it?

Let $I \subseteq \mathbb{R}$ be an interval and let $f: I \to \mathbb{R}$.

Let's start looking at the definitions of continuity, uniform continuity and differentiability.

Yes, I really like sentences in the language of set theory.

Definition 1: $f$ is continuous at $t \in I$ if

$$\forall \, \varepsilon > 0 \quad \exists \, \delta > 0 \quad \forall \, x \in (t-\delta, t+\delta) \cap I \quad |f(x)-f(t)| < \varepsilon$$

Definition 2: $f$ is continuous on $I$ if

$$\forall \, t \in I \quad \forall \, \varepsilon > 0 \quad \exists \, \delta > 0 \quad \forall \, x \in (t-\delta, t+\delta) \cap I \quad |f(x)-f(t)| < \varepsilon$$

Definition 3: $f$ is uniformly continuous on $I$ if

$$\forall \, \varepsilon > 0 \quad \exists \, \delta > 0 \quad \forall \, t \in I \quad \forall \, x \in (t-\delta, t+\delta) \cap I \quad |f(x)-f(t)| < \varepsilon$$

or, put more nicely,

$$\forall \, \varepsilon > 0 \quad \exists \, \delta > 0 \quad \forall \, x \in I \quad \forall \, y \in I \quad \bigg[ \quad |x-y|<\delta \implies |f(x)-f(t)| < \varepsilon \quad \bigg]$$

Definition 4: $f$ is differentiable at $t \in I$ if

$$\exists \, v \in \mathbb{R} \quad \forall \, \varepsilon > 0 \quad \exists \, \delta > 0 \quad \forall \, x \in (t-\delta, t+\delta) \cap I \setminus \!\{t\} \quad \left|\,\dfrac{f(x)-f(t)}{x-t} - v\,\right| < \varepsilon$$

Definition 5: $f$ is differentiable on $I$ if

$$\forall \, t \in I \quad \exists \, v \in \mathbb{R} \quad \forall \, \varepsilon > 0 \quad \exists \, \delta > 0 \quad \forall \, x \in (t-\delta, t+\delta) \cap I \setminus \!\{t\} \quad \left|\,\dfrac{f(x)-f(t)}{x-t} - v\,\right| < \varepsilon$$

Now, note that in the definition of uniform continuity, we did not require the function to be continuous; instead, our definition implies it (quite trivially!).

Now let's proceed to the most common definition1 of uniform differentiability:

Definition 6: $f$ is uniformly differentiable on $I$ if

$$f \,\,\text{is differentiable on} \,\,I \text{,} \qquad \text{and}$$

$$\forall \, \varepsilon > 0 \quad \exists \, \delta > 0 \quad \forall \, t \in I \quad \forall \, x \in (t-\delta, t+\delta) \cap I \setminus \!\{t\} \quad \left|\,\dfrac{f(x)-f(t)}{x-t} - f'(t)\,\right| < \varepsilon$$

or, equivalently,

$$f \,\,\text{is differentiable on} \,\,I \text{,} \qquad \text{and}$$

$\quad \forall \, \varepsilon > 0 \quad \exists \, \delta > 0$ $$\quad \forall \, t \in I \quad \forall \, h \quad \bigg[ \enspace \big(\,\, 0 < |h| < \delta \,\,\big) \land \big(\,\, t+h \in I \,\,\big) \implies \left|\,\dfrac{f(t+h)-f(t)}{h} - f'(t)\,\right| < \varepsilon \enspace \bigg]$$

or, as I prefer,

$$f \,\,\text{is differentiable on} \,\,I \text{,} \qquad \text{and}$$

$$\forall \, \varepsilon > 0 \quad \exists \, \delta > 0 \quad \forall \, x \in I \quad \forall \, y \in I \enspace \bigg[ \enspace 0 < |x-y| < \delta \implies \left|\,\dfrac{f(x)-f(y)}{x-y} - f'(x)\,\right| < \varepsilon \enspace \bigg]$$

The given definition of uniform differentiability requires differentiability! (And as it stands it is indeed required because we refer to $f'(t)$ inside the definition)

Is there a way to "fix" this? In other words:

## Is there an alternative definition of uniform differentiability, that does not require differentiability, and follows the same set-theory, $\varepsilon$-$\delta$ theme of the definitions 1 to 5?

The problems I encountered in my attempt:

I tried to alter the given Definition 6, to include the quantifier $\exists \, v \in \mathbb{R}$, and then use $v$ in place of $f'(t)$, but I couldn't make this work, because there seems to be a unbreakable loop:

• I need the "$\forall \, t$" somewhere
• $v$ clearly depends on $t$, so the "$\exists \, v$" would have to be placed to the right of "$\forall \, t$"
• $\delta$ cannot depend on $t$, so the "$\exists \, \delta$" would have to be placed to the left of "$\forall \, t$"
• But $v$ cannod depend on $\delta$ (and therefore on $\varepsilon$), since $v$ is some sort of limit, which means the same $v$ must work for all epsilons! Then "$\exists \, v$" would have to be placed to the left of "$\exists \, \delta$", which is impossible given the above considerations!

After thinking about this, it seemed to me that I would need two different epsilons and two different deltas inside the definition, but that looks too ugly, and too close to simply replacing the sentence "$f \,\,\text{is differentiable on} \,\,I$" by its set-theory equivalent, which is clearly not what I want here. Then I got stuck.

1 the three forms presented in Definition 6 were manipulated and "converted to the language of set theory" by me; but are grounded on many other posts on Math.SE such as here, here, here and here.

• What is the \{t} supposed to mean in definitions 4,5,and 6? – R. Kap Aug 25 '16 at 2:16

## 1 Answer

Why not just replace $f'(x)$ with some other function?

$$\exists g: I \rightarrow \Bbb R, \quad \forall \epsilon > 0 \quad \exists \delta >0 \quad, \quad \forall x,y \in I \\ 0<|x-y|<\delta \implies \left| \frac{f(x)-f(y)}{x-y} - g(y) \right|< \epsilon$$

• Brilliant, this works! Thank you! – Pedro A Sep 7 '16 at 18:43