Hey Brilliant users,

I haven't been on here in a while, I've been really busy with school lately. I have, however, found time to write a few original proofs in the past few weeks. At some point I'll put the better/more interesting ones up here. However, I'm having a somewhat annoying problem with one of my proofs. I probably won't post the proof on here for many reasons, one being the proof's actual length and the other being the originality of it. My problem is, the entirety of the proof rests on the fact that the function I'm considering is infinitely differentiable. Assuming this is true, the remainder of the proof is complex but manageable. For some reason, I'm having trouble proving that this function is infinitely differentiable. I know that it is, it wouldn't make sense for it not to be, but I'm having trouble applying full rigor to the problem. I don't want a direct answer, as I'd rather prove it myself, but I would like some advice if anyone can offer it. In short, my question is: How do you prove that a function is infinitely differentiable?

## Comments

Sort by:

TopNewestThe problem is to prove that the function given by y=0,x≤ 0 and y=\(e^{-1/x}\) , x>0 is C∞

When x is less than 0, there's no problem as obviously the function is infinitely differentiable. Similarly, when x is greater than zero the function is infinitely differentiable, by the properties of the exponential function. As I see it, the difficulty arises in what happens right at 0. For the function to be differentiable there, the function needs to be continuous at 0, and for that to happen both the right hand and left hand limits need to equal 0. So for the function to be infinitely differentiable, one would need to show that in the limit the function \(e^{-1/x}\) , and all its derivatives, go to zero as x goes to 0.

Am I on the right track? Thanks for any advice. – Anuj Shikarkhane · 2 years, 3 months ago

Log in to reply

– Ethan Robinett · 2 years, 3 months ago

Well my understanding is that showing that the above function is continuous at 0 does not show that the function is differentiable there (differentiability implies continuity, but the opposite is not necessarily true). Also, like you said, that function would be differentiable everywhere except x=0. My issue is how would you rigorously prove that derivatives exist everywhere and for all orders of differentiation except at x=0? Like you said "When x is less than 0, there's no problem as obviously the function is infinitely differentiable." How would you go about proving that statement?Log in to reply