What does "finite but unbounded" mean?

The term "finite" in this case is stressing the fact that it is "defined", "well defined" or "exists as a real number".

To better understand why this is important, consider the following naive attempt to give an example of a function with unbounded derivative on a closed interval:

$$f(x) = \sqrt{x}$$

The derivative, which is $$\dfrac{1}{2\sqrt{x}}$$ is clearly unbounded in $(0,1]$ but it is not even defined in zero.

The goal of the book is to disallow such examples. It technically could say "defined and unbounded", or even simply "unbounded", but to avoid any ambiguity, it is usual to say "finite and unbounded" here. Some might argue that the derivative is "defined as $\infty$" in zero (as opposed to "utterly undefined" when for example the function is not even continuous at the point), and to avoid any ambiguity the usual wording is "finite" (i.e. it is defined as a real number).

This is actually why that example in your book is interesting. By ruling out the naive examples such as $\sqrt{x}$, one might wonder if there could be any function with unbounded derivative in a closed interval.

I remember perfectly that when my teacher asked this in class, I thought that if such example would exist, surely it would involve a function that gets steeper and steeper, and that it would "end up being infinite in a closed interval" and I answered that it was impossible.

But guess what, it's possible, as the example in your book shows.


The theorem indicates that the derivative of f(x) is finite at $x=0$ and it is unbounded on $[-1,1]$

It is obvious that a finite set is bounded, therefore the author's intention is to show a function which has a finite derivative at $x=0$ but unbounded derivative on the interval [0,1].


The sequence $\{1,2,...\}$ is finite in the sense each term is finite, but the sequence is not bounded in the sense there is no fixed real number $m$ such that all the terms are bouded by this number. Similarly, $f(x)=\frac 1 x$ is finite in $(0,1)$ but it is not bounded. In the above example take $x =\frac 1 {\sqrt {2n\pi}}$ to see that $f'$ is not bounded.

Tags:

Real Analysis

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