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.
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.