Timeline for Good example of a non-continuous function all of whose partial derivatives exist
Current License: CC BY-SA 2.5
8 events
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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 |