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Assume that $f:[a,b]\to\mathbb R$ is a continuous function. For $x_0\in (a,b)$ and $\alpha,\beta>0$, we define the (asymetric) difference quotient $$(\Delta_{\alpha,\beta,h}f)(x_0):=\frac{f(x_0+\alpha h)-f(x_0-\beta h)}{(\alpha+\beta)h},\qquad h\neq 0.$$ Given $\alpha,\beta>0$ with $\alpha\neq \beta$, I want to find an example of $f$ such that $$\lim_{h\to0}(\Delta_{\alpha,\beta,h}f)(x_0)\ \ \text{exists }\ \ \not\!\!\!\implies\ f'(x_0)\ \ \ \text{exists}.$$

This is a problem that I came up with while I'm preparing my recitation class as a teaching assistant in Analysis (I) course.

If $f'(x_0)$ exists, we can prove that $\lim_{h\to0}(\Delta_{\alpha,\beta,h}f)(x_0)=f'(x_0)$ by Taylor's theorem: It follows from $f(x_0+\alpha h)=f(x_0)+\alpha f'(x_0)h+o(h)$ and $f(x_0-\beta h)=f(x_0)-\beta f'(x_0)h+o(h)$ that $(\Delta_{\alpha,\beta,h}f)(x_0)=f'(x_0)+o(1)$ as $h\to0$, hence $\lim_{h\to0}(\Delta_{\alpha,\beta,h}f)(x_0)=f'(x_0)$.

Now, a natural question appears: What about the reverse problem? If $\alpha=\beta$, then for $f(x)=|x|$ and $x_0=0$ we have $(\Delta_{\alpha,\beta,h}f)(0)=0$ for all $h\neq0$, but $f'(0)$ doesn't exist.

However, for $\alpha\neq \beta$, I failed to find an example. I found that if $\alpha\neq \beta$, and if $f_-'(x_0)$, $f_+'(x_0)$ both exist, then we must have $$\lim_{h\to0}(\Delta_{\alpha,\beta,h}f)(x_0)\ \ \text{exists }\ \ \implies\ f'(x_0)\ \ \ \text{exists}.$$ So, for any example in which $\alpha\neq \beta$ and $\lim_{h\to0}(\Delta_{\alpha,\beta,h}f)(x_0)\ \ \text{exists }\ \ \not\!\!\!\implies\ f'(x_0)\ \ \ \text{exists}$, there must be one of $f_-'(x_0)$ and $f_+'(x_0)$ that doesn't exist. (Note that $f$ should be continuous in this post.)

Any help would be appreciated!

Feng
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  • How about $f(x) = 2x$ when $x \ge 0$ and $f(x) = -x$ when $x < 0$. It is continuous and works for any $\beta = 2\alpha$? – Gregory Nov 04 '22 at 18:33
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    @Gregory The right and left sided limits are not equal. – Mark Viola Nov 04 '22 at 19:43
  • Was thinking about the counterexample $f(x) = |\beta x|$ for positive $x$ and $f(x) = | \alpha x| $ for negative $x$. This would have worked for $h \to 0^{+}$, but for negative $h$ we get a different limit. Starting from this idea of comparing negative and positive $x$, you can show that if the weird limit exists, then $lim_{h \to 0^+} \frac{ f(x_0 - (\alpha/\beta) h) - f(x_0 - (\beta/\alpha) h) }{ h} $ exists. This is not analogous to the above limits, since both the arguments are at the left of $x_0$. One could search for a "multiplicative fractal" counterexample of factor $(\alpha/\beta)^2$ – Andrea Marino Nov 04 '22 at 22:35
  • I mean something like this: https://math.stackexchange.com/questions/1837259/functions-invariant-under-scaling But even simpler and constructed by hand, since we don't need such generality for parameters, nor smoothness. I was trying to define the function piecewise on intervals of the form $[\pm 2^{n}, \pm 2^{n+1}]$ for $ n \in \mathbb{Z}$, where $\beta = 2, \alpha = 1$. However, this seems to mess up continuity in zero. – Andrea Marino Nov 04 '22 at 23:01
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    Relevant: If $f$ is differentiable at $k$, find: $\lim_{h \to 0} \frac{f(k + ph) - f(k - ph)}{h}$ Also, this google search for "pseudo-symmetric derivative" and Parametric differentiation by Evans/Humke (1981). – Dave L. Renfro Nov 05 '22 at 13:10
  • @DaveL.Renfro Thanks for your comment! It seems that the predecessors have considered more general “derivatives” than mine. This is interesting! – Feng Nov 05 '22 at 14:07

2 Answers2

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I have to apologize... Thinking for more time leads to the conclusion that we cannot find any desired example. Let me explain.

Proposition. Let $f$ be a function defined near $x_0\in\mathbb R$ and $\alpha, \beta\in\mathbb R$ be such that $|\alpha|\neq |\beta|$. Assume that $f$ is continuous at $x_0$, and $$\lim_{h\to0}\frac{f(x_0+\alpha h)-f(x_0+\beta h)}{(\alpha-\beta)h}=A\in\mathbb R.$$ Then $f$ is differentiable at $x_0$ and $f'(x_0)=A$.

Proof. Without loss of generality, we assume that $x_0=0$, $|\alpha|>|\beta|$ and $A=0$. Let $k=\frac\beta\alpha$, then $|k|<1$ and we have $$\lim_{h\to0}\frac{f(h)-f(kh)}{(1-k)h}=0.$$ We are going to prove that $f'(0)=0$. For any $\epsilon>0$, there exists $\delta>0$ such that $$\left|\frac{f(h)-f(kh)}{(1-k)h}\right|<\epsilon,\qquad 0<|h|<\delta.$$ Now, for $0<|h|<\delta$, we have $0<|k^nh|<\delta$ for all $n\in\mathbb N_{>0}$, hence $$\left|\frac{f\left(k^{n-1}h\right)-f\left(k^{n}h\right)}{h}\right|<(1-k)|k|^{n-1}\epsilon,\qquad \forall n\in\mathbb N_{>0},$$ and thus \begin{align*} \left|\frac{f\left(h\right)-f\left(k^{n}h\right)}{h}\right|&\leq\sum_{j=1}^n\left|\frac{f\left(k^{j-1}h\right)-f\left(k^{j}h\right)}{h}\right|\\ &<\sum_{j=1}^n(1-k)|k|^{j-1}\epsilon\\ &<\frac{1-k}{1-|k|}\epsilon, \qquad \forall n\in\mathbb N_{>0}. \end{align*} Letting $n\to\infty$, the continuity of $f$ at $x=0$ gives that $$\left|\frac{f\left(h\right)-f\left(0\right)}{h}\right|\leq\frac{1-k}{1-|k|}\epsilon,\qquad 0<|h|<\delta.$$ Therefore, $f$ is differentiable at $x=0$ and $f'(0)=0$. The proof is complete.

Feng
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I think this is too long for a comment, but I'm not sure how complete this is. Hopefully it is helpful.

Consider \begin{align} \lim_{h \to 0^{\pm}} \frac{f(x_0+\alpha h)-f(x_0-\beta h)}{(\alpha+\beta)h} & = \lim_{h \to 0^{\pm}} \frac{f(x_0+\alpha h) - f(x_0) + f(x_0)-f(x_0-\beta h)}{(\alpha+\beta)h} \\ & = \frac{\alpha}{\alpha + \beta} f'(x_0^{\pm}) + \frac{\beta}{\alpha + \beta} f'(x_0^{\mp}). \end{align} Now since $\lim_h\to0 \Delta_{\alpha,\beta,h}$ exists, we must have equality when taking the limit from either direction. This leads to $$ (\alpha-\beta) f'(x_0^+) = (\alpha - \beta)f'(x_0^-).$$ By hypothesis if $\alpha \neq \beta$, we may cancel and have shown that $f'(x_0)$ exists.

Gregory
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  • Thanks for your comments. I’ve noticed this. As I wrote in my post, for any desired example, there must be one of the right derivative and the left derivative that doesn’t exist. Thanks for writing the details of this fact! – Feng Nov 04 '22 at 13:45
  • What I am trying to say is that if $\alpha \neq \beta$, then no such example can exist. Because than the required equality can't be satisfied – Gregory Nov 04 '22 at 13:50
  • Only if $\alpha = \beta$ can you then freely choose $f'(x_0^+)$ and $f'(x_0^-)$ – Gregory Nov 04 '22 at 13:51
  • You proved that if $f'(x_0^+)$ and $f'(x_0^-)$ both exist, then no such example can exist. We may probably find an example where $f'(x_0^+)$ and $f'(x_0^-)$ don’t exist. – Feng Nov 04 '22 at 13:57
  • I'm trying to say that I don't think there is. – Gregory Nov 04 '22 at 14:08
  • Your continuity of the function implies that these left- right- handed derivatives exist. Can you describe an example where $f'(x_0^+)$ does not exist for a continuous function? – Gregory Nov 04 '22 at 14:28
  • For example, $f(x)=x\sin\frac1x$. – Feng Nov 04 '22 at 14:31
  • These limits don't exist at 0 do they, from either direction... and hence the $\Delta$ function won't exist. – Gregory Nov 04 '22 at 14:39
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    You split the limit of the sum into the sum of limits, neither of which is assumed to exist. – Mark Viola Nov 04 '22 at 14:44