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 , , show that lines are not enough: this function is not continuous at , even though its restriction to every line is continuous. It takes a parabola, such as , to detect the discontinuity.
Things look brighter if we do allow parabolas and other curves into consideration.
Continuity: is continuous at if and only if is continuous at for every map such that and is continuous at .
Proof: If is not continuous, we can find a sequence such that , and run 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: is differentiable at if and only if is differentiable at for every map such that and exists.
Alas, this is false. Take a continuous function which preserves antipodes (i.e., ) and extend it to via . Consider as above, with being the origin. If when because is Lipschitz. If , we can rescale the parameter so that is a unit vector. It is easy to see that , hence is differentiable at . However, is not differentiable at unless 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 with curves. Apparently, it has to be imposed artificially.
Differentiability, take 2: is differentiable at if and only if there exists a linear map such that for every map such that and exists.
Note that the only viable candidate for is given by partial derivatives, and those are computed along lines. Thus, we are able determine the first-order differentiability of using only the tools of single-variable calculus.
Proof goes along the same lines as for continuity, with extra care taken in forming .
- We may assume that by subtracting from our function. Also assume .
- Suppose is not differentiable at . Pick a sequence such that for all .
- Passing to a subsequence, make sure that tends to a unit vector , and also that .
- Connect the points by line segments. Parametrize this piecewise-linear curve by arc length.
- The distance from and is bounded by , the triangle inequality. Hence, the total length between and does not exceed , where as .
- By 3, 4, and 5 the constructed curve has a one-sided derivative when it reaches 0. Shift the parameter so that . Extend linearly to get two-sided derivative at .
- By assumption, as . This contradicts 2 and 5.
Can one go further and detect the second order differentiability by probing 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.