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Is there a function which is infinitely-differentiable everywhere but continuous nowhere?

EDIT: The graphic which has since been deleted was taken from my own notes, and I can clarify what definition of "differentiable" was being used there (presumably, the meaning the OP had in mind was the same).

A function is differentiable at $x$ iff (i) all directional derivatives at $x$ exist, and (ii) the map that sends a tangent vector to the corresponding directional derivative is continuous and linear. (Of course, in finite dimensions, the condition "continuous" is superfluous.) This is strictly stronger than gâteaux differentiable and strictly weaker than fréchet differentiable. (FWIW, that this definition is nonstandard is made clear immediately following the definition, where it is contrasted with both gâteaux differentiability and fréchet differentiability.)

If you're curious, the motivation for this terminology is as follows.

  1. Using all three terms allows me to be more precise, and as "fréchet differentiable" and "gâteaux differentiable" already have names, by necessity the above condition is referred to as simply "differentiable" (which I found more palatable than just making up a new term).

  2. This definition is easier (and, IMHO, more natural) than fréchet differentiable, and furthermore, almost everything that is true for fréchet differentiable functions is true of functions which are differentiable in this sense (the fact in question here being the biggest exception I am aware of).

  3. While this is (slightly) more difficult than the definition of gâteaux differentiable, I found that not having the derivative be a one-form 'broke' things to an unacceptable degree (for example, as I only defined the derivative of tensor fields, if the derivative itself were not a tensor field, then strictly speaking the second derivative would have been left undefined).

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    $\begingroup$ What do you mean when you say "infinitely-differentiable"? $\endgroup$ – Noah Schweber Aug 8 '16 at 3:57
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    $\begingroup$ @Michael The fact that the function is on $\mathbb{R}^2$ is absolutely crucial because the answer is false in $\mathbb{R}$ (and I assumed that was what your question was about given your attempt). This is why it's important that you include all the relevant information in your post when you make it, rather than 10 comments later. $\endgroup$ – T. Bongers Aug 8 '16 at 4:08
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    $\begingroup$ It didn't matter whether the function is on $\mathbb{R}^2$ or not. The function in the picture is not differentiable at the origin. So there is no contradiction that it is not continuous at the origin. Infinitely differentiable everywhere except at one single point is not strong enough to force the function to be continuous at the specific point. $\endgroup$ – achille hui Aug 8 '16 at 4:13
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    $\begingroup$ @MichaelHardy I know what "infinitely differentiable" means; I was asking the OP because - based on their question - I wasn't sure they were using the term correctly. $\endgroup$ – Noah Schweber Aug 8 '16 at 4:15
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    $\begingroup$ It looks like the OP is talking about infinitely differentiable in a given direction (i.e. I am talking about a directional derivative) -- this seems like it is just confusion between the existence of directional derivatives, Gateaux differentiability, and Frechet differentiability. $\endgroup$ – Chill2Macht Aug 8 '16 at 4:18
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Your function is infinitely differentiable everywhere but $(0,0)$. As you say, it is not even continuous there. These are not contradictory. It is not differentiable at $(0,0)$ either. It is differentiable in every specific direction from $(0,0)$, but for every $\epsilon \gt 0$ you can't find a $\delta \gt 0$ such that blah-blah for all points within $\delta$ of the origin. It is crucial whether you choose the $\delta$ before I choose the direction or the other way around. The definition I know of differentiable requires that the $\delta$ be chosen first, and your proof of non-continuity is an even stronger proof of non-differentiabiliy. It requires that the derivative be well defined based on all points within $\delta$ of the point you are taking the derivative at. If you have a different definition of differentiable, you should state it.

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