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I have two definitions of derivatives. The first one is the one provided in Rudin:

Let $f: [a,b] \to \mathbb{R}$ be a function. Let $x \in [a,b]$.

Let $\phi: (a,b) \setminus \{x\}\to\mathbb{R}: t \mapsto \frac{f(t)- f(x)}{t-x}$

Then we define $f'(x) = \lim_{t \to x} \phi(t)$, provided that the limit exists.

Here's the second one:

Let $f: E \subseteq \mathbb{R} \to \mathbb{R}$ be a function, $x \in E$ and $x$ a limit point of $E$. Put

$\phi: E \setminus \{x\} \to \mathbb{R}: t \mapsto \frac{f(t)- f(x)}{t-x} $.

Then we define $f'(x) = \lim_{x \to t} \phi (t)$, provided that the limit exists.

I want to show that the second definition implies the first, when we put $E = [a,b]$.

So, in particular, I want to show that for every $x \in [a,b]$

$\lim_{t \to x, t \in (a,b) \setminus \{x\}} \frac{f(t)-f(x)}{t-x}$ exists $\iff$ $\lim_{t \to x, t \in [a,b] \setminus \{x\}} \frac{f(t)-f(x)}{t-x}$

and if one of the two exists, both are equal.

Clearly, $\Leftarrow$ is satisfied (immediately from the limit definition).

For the other direction, put $q:= \lim_{t \to x, t \in (a,b) \setminus \{x\}} \frac{f(t)-f(x)}{t-x}$. If $x \in (a,b)$, then it is an interior point and the limit is equal to the other one.

WLOG, assume $x = a$. Let $\epsilon > 0$. Choose $\delta > 0$ such that $|q - \frac{f(t)-f(a)}{t-a}| < \epsilon$ for all $t \in (a,b)$ with $0 < |t-x|< \delta$.

Then, if $t \in [a,b] \setminus \{a\}$ with $0 < |t-a| < \min \{\delta, |b-a|\}$, then $t \neq b, t \neq a$ and we can make the quantity smaller than $\epsilon$.

Does this seem correct? Are there any arguments against using the second definition over the first one, as it seems more general?

1 Answers1

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Let us generalize the second definition. For $x \in \mathbb{R}$ we denote by $\mathfrak{N}(x)$ the set of all neighborhoods of $x$ in $\mathbb{R}$ (a neighborhood of $x$ is a set $N \subset \mathbb{R}$ such that $(x -\varepsilon, x +\varepsilon ) \subset N$ for some $\varepsilon > 0$).

Let $x \in E \subset \mathbb{R}$. Then obviously the following are equivalent:

(1) $x$ is a limit point of $E$.

(2) There exists $N \in \mathfrak{N}(x)$ such that $x$ is a limit point of $N \cap E$.

(3) For all $N \in \mathfrak{N}(x)$, $x$ is a limit point of $N \cap E$.

Now let $f: E \to \mathbb{R}$ be a function and $x \in E$ be a limit point of $E$. For each $N \in \mathfrak{N}(x)$ define $$\phi_N: N \cap E \setminus \{x\} \to \mathbb{R}: t \mapsto \frac{f(t)- f(x)}{t-x} .$$ The map $\phi$ from the second definition is given as $\phi = \phi_{\mathbb{R}}$. If $\lim_{x \to t} \phi_N (t)$ exists, we denote it by $f'_N(x)$. In case $N = \mathbb{R}$ we simply write $f'(x)$.

The following are obvious:

(1) If $\lim_{x \to t} \phi (t)$ exists, then $\lim_{x \to t} \phi_N (t)$ exists for all $N \in \mathfrak{N}(x)$ and $f'_N(x) = f'(x)$.

(2) If $\lim_{x \to t} \phi_N (t)$ exists for some $N \in \mathfrak{N}(x)$, then $\lim_{x \to t} \phi (t)$ exists and $f'(x) = f'_N(x)$.

Now let $E = [a,b]$. To avoid confusion the function $\phi$ from the first definition will be denoted by $\Phi$.

For $x \in (a,b)$ we have $\Phi = \phi_{(a,b)}$, for $x = a$ we have $\Phi = \phi_{(a-1,b)}$ and for $x = b$ we have $\Phi = \phi_{(a,b+1)}$.

This shows that the first and the second definition are equivalent.

Paul Frost
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  • Thanks for your detailed answer! I was wondering if my approach would work too? –  Aug 05 '18 at 14:36
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    Yes, of course, your approch works. I only wanted to round off the picture. – Paul Frost Aug 05 '18 at 14:44
  • By the way, where did you find the second definition? – Paul Frost Aug 05 '18 at 14:45
  • It's a modification of Tao's definition in his analysis volume 1. –  Aug 05 '18 at 14:47
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    Thank you! I personally prefer this definition, but one hardly ever finds it in the literature. – Paul Frost Aug 05 '18 at 14:50
  • I browsed through many books. It frustrated me that only Tao's book mentioned differentiability on something other than open intervals. Guess it has to do something with the physical interpretation of tangent lines. –  Aug 05 '18 at 14:54
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    I share your point of view. Perhaps you are interested in the discussion following this question: https://math.stackexchange.com/q/2830875 Most people seem to dislike the more general definition for functions defined on an arbitrary $E$ and prefer to regard $f$ as differentiable at a point $x \in E$ if there exist an open neighborhood $U$ of $x$ in $\mathbb{R}$ and a differentiable function $\hat f : U \to \mathbb{R}$ such that $\hat f \mid_{U \cap E} = f \mid_{U \cap E}$. In my eyes this is more a theorem than a definition. – Paul Frost Aug 05 '18 at 15:17
  • In the last comment I should have written correctly "and a function $\hat{f} : U \to \mathbb{R}$ which is differentiable at $x$ such that ...". – Paul Frost Aug 05 '18 at 23:40
  • See also math.stackexchange.com/q/161910 with an even more interesting discussion. – Paul Frost Aug 05 '18 at 23:41