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?

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    $\begingroup$ Your argument is correct. The two definitions are equivalent and there is reason to prefer one over the other. $\endgroup$ – Kavi Rama Murthy Aug 5 '18 at 12:03
  • $\begingroup$ Thanks . And what is that reason if I may ask? $\endgroup$ – user370967 Aug 5 '18 at 12:05

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.

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  • $\begingroup$ Thanks for your detailed answer! I was wondering if my approach would work too? $\endgroup$ – user370967 Aug 5 '18 at 14:36
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    $\begingroup$ Yes, of course, your approch works. I only wanted to round off the picture. $\endgroup$ – Paul Frost Aug 5 '18 at 14:44
  • $\begingroup$ By the way, where did you find the second definition? $\endgroup$ – Paul Frost Aug 5 '18 at 14:45
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    $\begingroup$ Thank you! I personally prefer this definition, but one hardly ever finds it in the literature. $\endgroup$ – Paul Frost Aug 5 '18 at 14:50
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    $\begingroup$ I share your point of view. Perhaps you are interested in the discussion following this question: 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. $\endgroup$ – Paul Frost Aug 5 '18 at 15:17

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