Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

The standard definition is that a function $f:\mathbb{R}^n\to \mathbb{R}$ is differentiable at a point $x$ if there exists a linear map $\mathrm{d}f_x: \mathbb{R}^n \to \mathbb{R}$ such that

$$f(x+h) = f(x) + \mathrm{d}f_x(h) + \epsilon \|h\|$$

where $\epsilon\to 0$ as $h\to 0$. This is stronger than the existence of all partial (or directional) derivatives, but weaker than their continuity. However, when talking about higher differentiability, one usually switches to talking about partial derivatives, asking them to be continuous in order to prove basic properties.

Suppose that instead we define $f$ to be "twice differentiable" at $x$ if in addition to $\mathrm{d}f_x$ as above there exists a quadratic form $\mathrm{d}^2f_x$ such that

$$f(x+h) = f(x) + \mathrm{d}f_x(h) + \frac{1}{2}\mathrm{d}^2f_x(h) + \epsilon \|h\|^2$$

where $\epsilon\to 0$ as $h\to 0$. This is true if $f$ has continous second-order partials (it's the multidimensional Taylor expansion, with $\mathrm{d}^2f_x$ the Hessian matrix).

  1. Does this imply that all second-order partial derivatives of $f$ exist?
  2. If so, does it imply that the mixed second-order partials are equal?
share|improve this question
    
I am not sure. I will think some about this. My definition of higher order differentiability would be the following: First $f$ needs to be differentiable in a nbhd of $p$. Then $f$ is twice differentiable at $p$ if there is a bilinear form $D^2f(p)$ such that $Df(p+v_1)(v_2) = Df(p)(v_2)+D^2f(v_1,v_2)+\epsilon |v_1||v_2|$. For some reason it is very hard to find sources where people talk about the second derivative as a bilinear form rather than just a quadratic form. –  Steven Gubkin May 6 at 21:31
    
You may also be interested in my online course with Jim Fowler here: ximera.osu.edu/course/kisonecat/m2o2c2/course/activity/week1. It is a little buggy, and a little sparse in terms of examples, but it covers this perspective on higher order derivatives in a kind of "interactive textbook" format. –  Steven Gubkin May 6 at 21:36
4  
See Chapter XIII, section 3--6 of Lang's "Real and Functional Analysis" for a very elegant treatment under which higher derivatives are defined via multilinear maps and moreover the $p$th higher derivative is genuinely the "derivative" of the $(p-1)$th. One virtue of this approach is that it permits both a formulation and proof of the higher-dimensional Taylor formula which looks and feels exactly like the 1-dimensional case (with the mess of factorials hidden away within a clean formalism); of course, one can bust out coordinates and recover the usual messier explicit version from that. –  user76758 May 6 at 21:47
    
I sort of doubt this even implies that $f$ is differentiable except at $x$. For example, let $n = 1$, let $w(t)$ be your typical nowhere-differentiable continuous (i.e. Weierstrass) function, and let $f(t) = t^2 + t^3 w(t)$. Then this is "twice differentiable" according to you, but only at $t = 0$. –  Ryan Reich May 6 at 21:54
1  
The first derivative of $f$ at $a$ is the linear function of $h$ that best approximates $f(a+h)-f(a)$. The second derivative of $f$ at $a$ is the bilinear function of $(h_1,h_2)$ that best approximates $f(a+h_1+h_2)-f(a+h_1)-f(a+h_2)+f(a)$. Etc. –  Tom Goodwillie May 7 at 12:13

3 Answers 3

up vote 8 down vote accepted

The funny thing is that I got almost the same question from one of my student last semester. The idea was to neglect both the linear and quadratic parts of the hypothetical expansion (without loss of generality, we may assume they are zero), and focus on $ \epsilon \|h\|^2$. We came to the following (obvious) one-dimensional counterexample:

$$f(h) = h^3 \sin(\frac{1}{h})$$

for $h \not= 0$, and: $$f(0) = 0$$

share|improve this answer

No, if I understand the question right, not even in one variable. Let $f(x)$ be $x^3g(x)$ where $g$ is bounded and, except at $x=0$, infinitely differentiable. If $g$ oscillates fast enough near $x=0$ then even though it is the sum of a degree two polynomial (namely zero) and an error term of the kind you are allowing, it will not have a second derivative at $0$.

share|improve this answer

Thanks Michal and Tom for the answer. Let me improve on it slightly with an example of a function which is well-approximated by a polynomial of every degree near 0 (in fact, the zero polynomial), but is still not even twice differentiable there:

$$f(x) = e^{-1/x^2} \sin(e^{1/x^2}) $$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.