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Let $f$ be defined (and real-valued) on $[a,b]$. For any $x\in [a,b]$ form a quotient $$\phi(t)=\dfrac{f(t)-f(x)}{t-x} \quad (a<t<b, t\neq x),$$ and define $$f'(x)=\lim \limits_{t\to x}\phi(t),$$ provided this limit exists in accordance with Defintion 4.1.

I have one question. Why Rudin considers $t\in (a,b)$? What would be if $t\in [a,b]?$

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    $\begingroup$ If you define this on the closed interval you would have to consider right and left hand limits. $\endgroup$
    – John Douma
    Oct 8, 2015 at 18:56
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    $\begingroup$ Presumably there was once a version (on paper or in Rudin's head) where $f$ was defined on the open interval $(a,b)$. There is absolutely no reason to disallow $t = a$ for $x\neq a$ or $t = b$ for $x\neq b$. $\endgroup$ Oct 8, 2015 at 18:58

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In short, this is an example of Rudin's sublime succinctness. His definition includes the usual two-sided definition when $x \in (a,b)$ and also the one-sided definition $x = a$ or $x = b$. To get the one-sided definition, notice that he requires $t \in (a,b)$.

Rudin was perhaps over-succinct in that he failed to mention that when $x = a$ or $x = b$, you have to interpret the limit as being one-sided. (Does Thm 4.1 include this?)

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    $\begingroup$ But when $x = a$, the condition $t \neq x$ excludes $t = a$. What reason could there be to exclude $t = b$ for $x = a$? $\endgroup$ Oct 8, 2015 at 19:06
  • $\begingroup$ No reason to exclude it, but no reason to include it, either! You're interested in the limit as $t\to x$. $\endgroup$
    – Austin A.
    Oct 8, 2015 at 19:07
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To answer your question directly, it doesn't matter.

Since we need the limit $t \to x$, we only really need to worry about $t$ close to (but different from) $x$. If $x$ is closer to $a$ instead of $b$ (but different from $a$), then we can worry about $$|x - t| < 1/2 |x - a| \le 1/2 | x - b |.$$ This means $t \neq a$ and $t \neq b$.

You can think of something similar for $x = a$ or $x = b$.

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the limit is defined at $a$ if there is a neighbourhood for $a$ that the function defined for all $x$ in that neighbourhood, for the limit $f'(x)=\lim \limits_{t\to x}\phi(t)$ to exist on $a$ and $b$ we need that $\phi(t)$ defined for all $x$ in some neighbourhood of $a$,but it is impossible if $a$ and $b$ are the endpoints of $f(x)$ domain,so if you want to consider the limit at $a$ and $b$ you need to add that as one-sided limit.

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