How much multivariable calculus can be done along curves?

Working with functions of two (or more) real variables is significantly harder than with functions of one variable. It is tempting to reduce the complexity by considering the restrictions of a multivariate function to lines passing through a point of interest. But standard counterexamples of Calculus III, such as \displaystyle f(x,y)=\frac{xy^2}{x^2+y^4}, f(0,0)=0, show that lines are not enough: this function f is not continuous at (0,0), even though its restriction to every line is continuous. It takes a parabola, such as x=y^2, to detect the discontinuity.

Things look brighter if we do allow parabolas and other curves into consideration.

Continuity: f is continuous at a\in\mathbb R^n if and only if f\circ \gamma is continuous at 0 for every map \gamma\colon \mathbb R\to \mathbb R^n such that \gamma(0)=a and \gamma is continuous at 0.

Proof: If f is not continuous, we can find a sequence a_n\to a such that f(a_n)\not\to f(a), and run \gamma through these points, for example in a piecewise linear way.

Having been successful at the level of continuity, we can hope for a similar differentiability result:

Differentiability, take 1: f is differentiable at a\in\mathbb R^n if and only if f\circ \gamma is differentiable at 0 for every map \gamma\colon \mathbb R\to \mathbb R^n such that \gamma(0)=a and \gamma'(0) exists.

Alas, this is false. Take a continuous function g\colon S^{n-1}\to \mathbb R which preserves antipodes (i.e., g(-x)=-g(x)) and extend it to \mathbb R^n via f(tx)=tg(x). Consider \gamma as above, with a\in \mathbb R^n being the origin. If \gamma'(0)=0 when (f\circ \gamma)'(0)=0 because f is Lipschitz. If \gamma'(0)\ne 0, we can rescale the parameter so that \gamma'(0) is a unit vector. It is easy to see that \displaystyle \frac{f(\gamma(t))}{t}= \frac{f(\gamma(t))}{|\gamma(t)|\mathrm{sign}\,t} \frac{|\gamma(t)|}{|t|}\to g(\gamma'(0)), hence f\circ \gamma is differentiable at 0. However, f is not differentiable at a unless g happens to be the restriction of a linear map.

I can’t think of a way to detect the nonlinearity of directional derivative by probing f with curves. Apparently, it has to be imposed artificially.

Differentiability, take 2: f is differentiable at a\in\mathbb R^n if and only if there exists a linear map T such that (f\circ \gamma)'(0)=T\gamma'(0) for every map \gamma\colon \mathbb R\to \mathbb R^n such that \gamma(0)=a and \gamma'(0) exists.

Note that the only viable candidate for T is given by partial derivatives, and those are computed along lines. Thus, we are able determine the first-order differentiability of f using only the tools of single-variable calculus.

Proof goes along the same lines as for continuity, with extra care taken in forming \gamma.

  1. We may assume that T=0 by subtracting Tx from our function. Also assume a=0.
  2. Suppose f is not differentiable at 0. Pick a sequence v_k\to 0 such that |f(v_k)|\ge \epsilon |v_k| for all k.
  3. Passing to a subsequence, make sure that v_k/|v_k| tends to a unit vector v, and also that |v_{k+1}|\le 2^{-k}|v_k|.
  4. Connect the points v_k by line segments. Parametrize this piecewise-linear curve by arc length.
  5. The distance from v_{k+1} and v_k is bounded by |v_{k+1}|+|v_k|\le (1+2^{-k})|v_k|, the triangle inequality. Hence, the total length between 0 and v_k does not exceed \sum_{m\ge k}(1+2^{-m})|v_m| \le (1+c_k)|v_k|, where c_k\to 0 as k\to \infty.
  6. By 3, 4, and 5 the constructed curve \gamma has a one-sided derivative when it reaches 0. Shift the parameter so that \gamma(0)=0. Extend \gamma linearly to get two-sided derivative at 0.
  7. By assumption, |f(\gamma (t))|/|t|\to 0 as t\to 0. This contradicts 2 and 5.

Can one go further and detect the second order differentiability by probing f with paths? But the second derivative is not a pointwise asymptotic condition: it requires the first derivative to exist in a neighborhood. The pointwise second derivative might be possible to detect, but I’m not sure… and it’s getting late.

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