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 $\gamma$ .

We may assume that $T=0$ by subtracting $Tx$ from our function. Also assume $a=0$ .
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 $k$ .
Passing to a subsequence, make sure that $v_k/|v_k|$ tends to a unit vector $v$ , and also that $|v_{k+1}|\le 2^{-k}|v_k|$ .
Connect the points $v_k$ by line segments. Parametrize this piecewise-linear curve by arc length.
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 $k\to \infty$ .
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 $0$ .
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.

1 thought on “How much multivariable calculus can be done along curves?”

Asaf Shachar says:

2016-04-29 at 07:35

Very nice post. Regarding your counter-example (differentiability along paths does not imply standard differentiability), where did you use the fact that $g$ preserves the antipodes? (Perhaps in the proof that $f$ is Lipschitz, since I am not sure how to show this). Thanks.

How much multivariable calculus can be done along curves?

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