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(a) Does there exist a function $f$ defined on the open interval $(a,b)$ such that $f'(b^-)$ exists, and $\lim_{x\to b-}f'(x)\neq f'(b^-)$, or (b) where $f'(b^-)$ exists and $\lim_{x\to b-}f'(x)$ does not exist?

Since $f(b)$ is undefined, define $$f'(b^-)=\lim_{h\to0+}\frac{f(b-h)-f(b-2h)}h.$$ Are there any difficulties with this definition as compared to the standard definition of the one-sided derivative?

Just reading my analysis textbook and thought this would make an interesting problem.

Related: $f(x)=x^2\sin\frac1x$ has a derivative which is defined at $0$ (equal to $0$), but $\lim_{x\to 0}f'(x)$ does not exist. (c) Is it always true that if the limit exists, it is equal to $f'(0)$? Even more curiously, $\limsup_{x\to0}f'(x)+\liminf_{x\to0}f'(x)=2f'(0)$ for this function. (d) Is this always the case, when the quantity on the left side of the equality is defined?

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This post may be of interest. –  David Mitra Dec 17 '12 at 21:17
    
$f(x)=\sin(2\pi \log_2|x|)$ has $f'(0^-)=0$ but $\lim_{x\to 0^-}f'(0)$ doesn't exist, so (b) has a positive answer. –  user108903 Dec 17 '12 at 21:23

2 Answers 2

up vote 2 down vote accepted

Consider e.g. a function of the form $f(x) = g(\log_2(b-x))$ where $g$ is periodic with period $1$. This satisfies $f(b-2x) = f(b-x)$, and thus according to your definition $f'(b-) = 0$, while $f'(x) = - \dfrac{g'(\log_2(b-x))}{(b-x) \ln 2}$, so $\lim_{x \to b-} f'(x)$ will not exist unless $g$ is constant.

EDIT: On the other hand, if $f$ is differentiable on $[b-2h,b-h]$ the Mean Value Theorem says
$ \dfrac{f(b-h) - f(b-2h)}{h} = f'(\xi)$ for some $\xi \in [b-2h,b-h]$, so if $\lim_{x \to b-} f'(x)$ exists then so does $f'(b-)$ and the two are equal.

Moreover, if $f'(b)$ exists, since $$\frac{f(b-h) - f(b-2h)}{h} = \frac{f(b-h) - f(b)}{h} - 2 \frac{f(b-2h) - f(b)}{2h}$$ we must have $f'(b-) = f'(b)$.

EDIT: As in your comments, suppose $g$ and $h$ are continuous and decreasing on $[0,\epsilon]$ with $h(x) < g(x) < b$ for $x > 0$ and $g(0) = h(0) = b$, and you define $f'(b-) = \lim_{x \to 0+} \dfrac{f(g(x)) - f(h(x))}{g(x) - h(x)}$.
Define a sequence $x_n$ by $x_0 = \epsilon$ and $h(x_{n+1}) = g(x_n)$. Then $x_{n+1} < x_n$ and $\lim_{n \to \infty} x_n = 0$. Take any nonconstant continuous function $f$ on $[h(x_0), g(x_0)] = [h(x_0), h(x_1)]$ with $f(h(x_0)) = f(h(x_1))$, and then define $f$ on $[h(x_n),h(x_{n+1})] = [g(x_{n-1},g(x_n)]$ iteratively by $f(t) = f(h(g^{-1}(t)))$. We obtain a continuous function $f$ on $[h(x_0), b)$ with $f(g(x)) = f(h(x))$, and thus $f'(b-) = 0$, but $\lim_{x \to b-} f(x)$, and thus also $\lim_{x \to b-} f'(x)$, does not exist.

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I hadn't thought about using $\log_2$ to get periodicity that way. Would you consider this an artifact of my formula, or would a more clever way of letting the points approach zero of the form $\lim_{x\to0+}\frac{f(g(x))-f(h(x))}{g(x)-h(x)}$ where $g,h\to b^-$ as $x\to0^+$ prevent that? –  Mario Carneiro Dec 17 '12 at 21:33
    
I'm not sure what "that" you are hoping to prevent. –  Robert Israel Dec 17 '12 at 21:37
    
Well, in your example, you have an oscillating function, and I sample the function at two points, but you've set up the function so that the two points are always a period apart, so I can't "tell" that the function is oscillating, and hence declare it to have zero derivative. Indeed, it is clear that your function is "adversarial", in the sense that the $\log_2$ part was chosen specifically so that this behavior occurs ($\log_3$ would not have worked for this). My question is whether I could define a $g$ and $h$ such that you can't hide periodicity in this manner. –  Mario Carneiro Dec 17 '12 at 21:42
    
@MarioCarneiro $\frac{f(g(x))-f(h(x))}{g(x)-h(x)}=f'(\xi)$ with $\xi\in(h(x),g(x))$ will always hold by MVT. Then if $\lim f'$ exists, it will also equal this revised definition of $f(b^{-})$. –  Hagen von Eitzen Dec 17 '12 at 21:47
    
@HagenvonEitzen In this case, my attempt is to make it so that whenever $\lim_{x\to b}f'(x)$ doesn't exist, the revised definition of $f(b^-)$ will not be defined either. (The example in my mind right now is $g(x)=b-x$ and $h(x)=b-x^2$.) –  Mario Carneiro Dec 17 '12 at 21:52

(a) If $\lim_{x\to b} f'(x)$ and $f'(b^{-})$ exist, then the limit equals $f'(b^-)$ because $\frac{f(b-h)-f(b-2h)}h=f'(b-\theta h) $ with some $\theta\in(1,2)$.


You've already given $f(x)=x^2\sin\frac1x$ as example for (b). Indeed, we also have $f'(0^-)=0$.


For (c) the same MVT argument applies as for (a).


(d) Let $g$ be smooth periodic and $$f(x)=\begin{cases}x^2\cdot g\left(-\frac1x\right)&\text{if }x>0\\ 0&\text{if }x=0.\end{cases}$$ Then $f'(0)=0$ and $f'(x)=2x g(-1/x)+g'(1/x)$, hence $\limsup f'=\limsup g'$ and $\liminf f'=\liminf g'$. If $g$ is asymmetric, we will have $\limsup f'+\liminf f'\ne0$. Indeed, consider $g(t)=\sin t+\sin 2t$, a first approximation to a sawtooth curve, which has $g'(0)=3$ but $g'(t)>-3$ for all $t$.

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