Motivation for a new definition of the derivative using the concept of average velocity

Question

As everyone knows, for a function $ f: \mathbb{R} \to \mathbb{R} $ and a point $ a \in \mathbb{R} $, we say that the derivative of $ f $ at $ a $ equals $ L $ if and only if $$ \lim_{x \to a} \frac{f(x) - f(a)}{x - a} = L. $$

Now, most textbooks on particle mechanics offer the following recipe for defining the instantaneous velocity of a particle traveling along a straight line.

• Let $ f: \mathbb{R} \to \mathbb{R} $ denote the displacement function of a particle traveling along the $ x $-axis, with respect to time. The goal is to define the instantaneous velocity of the particle at time $ t_{0} $.

• Pick a sequence of non-degenerate bounded closed intervals $ ([a_{n},b_{n}])_{n \in \mathbb{N}} $ such that $ t_{0} \in (a_{n},b_{n}) $ for all $ n \in \mathbb{N} $ and $ \displaystyle \lim_{n \to \infty} (b_{n} - a_{n}) = 0 $.

• Compute the sequence of average velocities $ (v_{\text{ave},n})_{n \in \mathbb{N}} $ of the particle over these closed intervals, i.e., $$ (v_{\text{ave},n})_{n \in \mathbb{N}} \stackrel{\text{def}}{=} \left( \frac{f(a_{n}) - f(b_{n})}{a_{n} - b_{n}} \right)_{n \in \mathbb{N}}. $$

• Define the instantaneous velocity of the particle at time $ t_{0} $ as $ \displaystyle \lim_{n \to \infty} v_{\text{ave},n} $, if it exists.

This recipe is somehow proposing a new definition of the derivative of a function. Let me describe it in precise mathematical terms.

Consider $ f: \mathbb{R} \to \mathbb{R} $. Let $ a \in \mathbb{R} $, and let $ \Lambda $ be the set of non-degenerate bounded closed intervals $ I $ such that $ a \in I^{\circ} $. Define a partial ordering $ \preceq $ on $ \Lambda $ by $ I \preceq J \iff I \supseteq J $ for all $ I,J \in \Lambda $. One can easily verify that $ (\Lambda,\preceq) $ is a directed set. Next, define a net $ \lambda: \Lambda \to \mathbb{R} $ by $$ \forall I \in \Lambda: \quad \lambda(I) \stackrel{\text{def}}{=} \frac{f({\frak{l}}(I)) - f({\frak{r}}(I))}{{\frak{l}}(I) - {\frak{r}}(I)}, $$ where $ {\frak{l}}(I) $ and $ {\frak{r}}(I) $ denote the left- and right-endpoints of $ I $ respectively. Then define $$ f'(a) \stackrel{\text{def}}{=} \lim_{I \in \Lambda} \lambda(I), $$ if it exists.

Question: How can we show that this new definition of the derivative agrees with the usual one?

Answer 1

In the following it is assumed that $a<0<b$ whenever the letters $a$ and $b$ appear. Then one has the following

Proposition. Let $f:\ U\to{\mathbb R}$ be defined in a neighborhood of $0$. If the limit $$\lim_{b-a\to0}{f(b)-f(a)\over b-a}=:p$$ exists then the one-sided limits $\lim_{x\to0-} f(x)$ and $\lim_{x\to0+} f(x)$ exist and are equal. If $f(0)$ equals this common limit then $f'(0)=p$.

Proof. Subtracting a linear function from $f$ we may assume $p=0$. Let an $\epsilon>0$ be given. Then by assumption there is a $\delta\in\ \bigl]0,{1\over4}\bigr]$ such that for $$-\delta < a < 0 < b <\delta$$ one has $$|f(b)-f(a)|<\epsilon(b-a)<{\epsilon\over2}\ .\tag{1}$$ It follows that for arbitrary $x$, $x'\in\ ]0,\delta[\ $ and $a:=-{\delta\over2}$ we are sure that $$|f(x)-f(x')\leq |f(x)-f(a)|+|f(a)-f(x')|<\epsilon\ .$$ This implies by Cauchy's criterion the existence of the $\lim_{x\to0+} f(x)$. Similarly for $\lim_{x\to0-} f(x)$, and then the equality of the two limits is obvious.

For the last statement we now assume $f(0)=\lim_{x\to0} f(x)$. Letting $a\to0-$ in $(1)$ we see that $$|f(b)-f(0)|\leq\epsilon b\qquad(0<b<\delta)\ ,$$ and as $\epsilon>0$ was arbitrary this implies $$\lim_{b\to0+}{f(b)-f(0)\over b}=0\ .$$ Arguing similarly about the lefthand limit we are done.$\qquad\square$

The converse of this proposition has been dealt with by coffeemath.

Answer 2

The two versions can be compared using the equation (where $h,k>0$) $$\frac{f(x+h)-f(x-k)}{h+k}=\frac{h}{h+k}\frac{f(x+h)-f(x)}{h}+\frac{k}{h+k}\frac{f(x)-f(x-k)}{k}.$$

If $f$ is differentiable in the usual sense, then the two difference quotients on the right agree and have common value $f'(x)$ as $h,k \to 0.$ Then in the limit the right side is $f'(x)$ in the usual definition, since the right is an affine combination.

On the other hand, using the "new" derivative definition, one is allowed to let either of $h$ or $k$ approach zero first, and then have the other approach zero. Assuming $f$ is continuous, if we first let $h\to 0$ we end up with the usual left derivative, and if we first let $k$ go to zero we end up with the usual right derivative, so that these match and $f$ is differentiable in the usual sense with value as in the limit of the "new" definition.

EDIT: Haskell Curry has pointed out that one cannot let one of $h,k$ approach zero first, and then the other. In other words, during the approach of each to zero, both of $h,k$ must remain positive. Given this, the "new" definition will say that the function $f(x)=0$ for $x\ne 0$ and $f(0)=1$ has the derivative $0$ at $x=0$, but this function $f$ is not differentiable in the usual sense at $x=0$ (it's not even continuous there).