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Mar 25, 2010 at 18:15 comment added Dyke Acland No, I'm quite happy with an example where the partial derivatives exist at some non-continuous point.
Mar 25, 2010 at 18:13 comment added Georges Elencwajg Bjorn is right, of course: the example was only meant to show that a function can be quite pathological and yet have directional derivatives in all directions at the origin. The construction is the source of a few amusing exercises: e.g. the extended function is continuous at the origin iff the function on the circle is bounded. On the other hand even if you start with a $C^{\infty}$ function on the circle (seen as submanifold of $\mathbb R^2$) the extended function on the plane will NOT be differentiable at the origin in general ( think bump function on the circle).
Mar 25, 2010 at 16:36 comment added Bjorn Poonen In George's "wilder example", further conditions on f will be needed to make the partial derivatives exist everywhere (is that what the proposer wanted?)
Mar 25, 2010 at 15:58 comment added Georges Elencwajg However you define f at the origin, it will be discontinuous since its limit along the coordinate axes is zero, whereas it is 1 along the diagonal of $\mathbb R^2$. You get wilder example by starting with any antisymmetric function on the unit circle (say non-measurable) and extending it linearly on all (vector)lines of the plane i.e. defining $f(ra)=rf(a)$ for $r \in \mathbb R$ and $a$ on the circle.
Mar 25, 2010 at 14:53 vote accept Dyke Acland
Mar 25, 2010 at 14:25 comment added Mark Meckes Define it to be 0 at (0,0) and it's discontinuous there, although the partial derivatives exist.
Mar 25, 2010 at 14:23 comment added Dyke Acland But surely that's continuous on its domain of definition.
Mar 25, 2010 at 13:45 history answered Thomas Kragh CC BY-SA 2.5