1
$\begingroup$

Recently I am considering the following question: Let $[a,b]$ be a non-degenerate bounded closed interval in the set $\mathbb{R}$ of real numbers. Suppose a function $f\colon [a,b]\to \mathbb{R}$ is strictly increasing on $[a,b],$ and continuous there. If $f$ is not differentiable from left at the right end point $b,$ then is it true that $\lim_{h\to 0^-}\frac{f(b+h)-f(b)}{h}=+\infty?$

Clearly, from strict increasing of $f$ on $[a,b],$ the difference quotient $g(h):=\frac{f(b+h)-f(b)}{h}$ is nonnegative on $(a-b,0).$ But I do not know how to answer this question. It is apparent that if we change the domain $[a,b]$ into $\mathbb{N},$ the set of positive integers, the answer is negative, for each real number can not be a limit point of $\mathbb{N},$ thus, we can not pass limit to $g(h).$ But as for $[a,b],$ the situation is different. Can anyone help me? Thanks!

PS: For convenience, we can suppose that $a=0, b=1,$ and so, the previous question turns to be as following

Suppose that $f\colon [0,1]\to\mathbb{R}$ is continuous and strictly increasing on $[0,1].$ If $f$ is not differentiable at $x=1,$ then can we say that $\lim_{h\to 0^-}\frac{f(1+h)-f(1)}{h}=+\infty?$

$\endgroup$

1 Answer 1

1
$\begingroup$

The answer is no. Here is a counterexample: We let $[a,b]=[0,1]$, $f(0)=0$ and define $f(\frac{1}{3^n})=\frac{1}{3^n}$ if $n$ is even and $f(\frac{1}{3^n})=\frac{2}{3^n}$ if $n$ is odd. the values of $f$ on the other points of $(0,1)$ can be set in a way that keeps $f$ continuous and monotonic on $[0,1]$, and the limit of $\frac{f(\frac{1}{3^n})}{\frac{1}{3^n}}$ does not exist and is not $\infty$.

$\endgroup$
4
  • $\begingroup$ But I am considering the case that if the left limit of the difference quotient of $f$ at the right endpoint $1,$ other than at other point! $\endgroup$
    – azc
    Commented Dec 30, 2017 at 9:14
  • $\begingroup$ @azc Then you can take $-f(1-x)$ instead. $\endgroup$
    – idok
    Commented Dec 30, 2017 at 9:20
  • $\begingroup$ I see what you suggest. But as for the piecewise function you given above, how can we evaluate the difference quotient of $f$ at the origin $\frac{f(0)-f(h)}{-h},$ where $0<h<1$ and $h$ can not be expressed as $1/3^n$ for some positive integer? $\endgroup$
    – azc
    Commented Jan 1, 2018 at 1:10
  • $\begingroup$ @azc it doesn't matter - we found two sequences that converge to 0, and when you evaluate the quotient at the points of the different sequences you get two different partial limits. From Heine's lemma, the quotient cannot have a finite or infinite limit at 0. $\endgroup$
    – idok
    Commented Jan 1, 2018 at 4:24

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .