13
$\begingroup$

Are there functions whereby the left-handed and right-handed derivatives are always defined but different? What made me think about this is price elasticity of demand which for psychological reasons I think wouldn't be the same from the left and right.

  • 1
    There are continuous functions which are not differentiable at any point, such as the Weierstrass function, if that's what you mean: https://en.wikipedia.org/wiki/Weierstrass_function – 2017-01-07
  • 2
    @TheCount "Are there functions whereby the left-handed and right-handed derivatives are always defined but different?" – 2017-01-07
  • 0
    @Alephnull For a function to be "differentiable" at a point $c$, this means by definition both the left-sided limit and the right-sided limit of $\lim\limits_{x\to c} \frac{f(x)-f(c)}{x-c}$ must be equal, so for a function to not be differentiable at a point either the limits from the left and right both exist but are unequal or the limits from left or right simply don't exist. A function whose left-handed and right-handed derivatives always exist but are different is by definition a function which is not differentiable at any point, though other examples of undifferentiable functions exist. – 2017-01-07
  • 0
    That being said, TheCount's comment was indeed valid and relevant and did not warrant your reply, which was seemingly to point out the first sentence of the post again and nothing else. – 2017-01-07
  • 3
    But do the left and right limits of the difference quotient of the W-M function exist? – 2017-01-07
  • 4
    It's not difficult to construct functions where the left- and right-hand derivatives exist everywhere and fail to agree on a dense set (e.g., the rationals); I don't know the answer to your question, but offhand would guess the answer is "no". – 2017-01-07
  • 0
    @AndrewD.Hwang The function would have to have the left-handed derivative continuous and the right-handed derivative continuous. I don't know if it makes any difference otherwise which derivatives you choose but it doesn't seem to work (a left handed tangent for one point is the right-handed tangent for another close-by point) – 2017-01-07
  • 2
    @Alephnull: The construction I had in mind is outlined in my answer to [Are all continuous one one functions differentiable?](http://math.stackexchange.com/a/1234465/86418). There exists a strictly increasing function $f$ having a jump discontinuity at every rational; the integral $F$ of such a function has one-sided derivatives at each point by the fundamental theorem, but due to the jump discontinuities of $f$, the one-sided derivatives disagree at each rational number. – 2017-01-07
  • 0
    @Alephnull I don't get what I am supposed to take away from you repeating your opening sentence to me. – 2017-01-08
  • 1
    @TheCount I would assume the point is that you've made no effort to connect the W function to the OP's hypotheses. – 2017-01-08
  • 1
    @StellaBiderman I think the fact that I was spitballing is clear by the wording of my original question, and the fact that I made no effort is clear as well. It was a passing comment, not a master's thesis. – 2017-01-08
  • 0
    @TheCount the point is "what you mean" is what I said. – 2017-01-08
  • 0
    @Alephnull I suppose it's a new concept to some people that not all communication that is obvious to the transmitting party is obvious to the receiving party. Case in point, the Weierstrass function does what you want, *as near as I can tell*, but you seem to not like it. – 2017-01-08
  • 0
    @TheCount I was not commenting on the function but your comment. At any rate, things seem to be cleared up. – 2017-01-08
  • 0
    @Alephnull True, true. Fair enough. – 2017-01-08

1 Answers 1

2

In Klambauer's Real Analysis pg. 101, he proves that if $f$ is a function on the open interval $(a,b)$ then there are most a countable number of points $x$ such that both $f_l'(x)$ and $f_r'(x)$ exist (including the infinite cases) but not equal.

I'll repeat his argument. Let $$A=\{x\in(a,b): \text{both }f_l'(x)\text{ and } f_r'(x) \text{ exist, but }f_l'(x)< f_r'(x)\}$$ $$B=\{x\in(a,b): \text{both }f_l'(x)\text{ and } f_r'(x) \text{ exist, but }f_l'(x)> f_r'(x)\}$$

For each $x\in A$, chose a rational number $r_1^{(x)}$ such that $f_l'(x)r_1^{(x)}$$ $$\text{whenever } x

This process (with the Axiom of Choice) let's us construct a function $\varphi$ from $A$ into $\mathbb{Q}^3$ given by $\varphi(x)=(r_1^{(x)}, r_2^{(x)}, r_3^{(x)})$.

This function $\varphi$ is also injective. For suppose that $\varphi(x)=\varphi(y)$. This implies that $$(r_2^{(x)}, r_3^{(x)})=(r_2^{(y)}, r_3^{(y)})\,.$$ Recall that both $x$ and $y$ are within this open interval. Thus we can infer both of these inequalities $$f(y)-f(x)< r_1^{(x)}(y-x)$$ $$f(x)-f(y)< r_1^{(y)}(x-y)$$ Since we assumed that $r_1^{(x)}=r_1^{(y)}$, adding these inequalities gives $0<0$. Which is nonsense.

This shows that $A$ is countable. And by the same reasoning, $B$ is countable.

  • 1
    I don't have anything important to add, but I don't think you need the axiom of choice here. Namely, you can first enumerate the rationals with your favorite bijection between $\mathbb{N}$ and $\mathbb{Q}$, and then e.g., you let $r_1$ be the least such rational(with respect to this weird enumeration) between $f'_l(x)$ and $f'_r(x)$, etc. – 2018-01-19