Possible proof of Gabriel's Theorem?
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From: John Gabbriel on 8 Mar 2005 14:52 Jason wrote: > You are an imposter! As promised I will be reporting you! > > What a coward you are! Too afraid to post anything under your real > name? > > Silly boy. How hard is it to comprehend the fact that there could be more than one John Gabriel in this world? Why, there are at least 10 in the state of Texas alone.
From: examachine on 8 Mar 2005 15:35 David C. Ullrich wrote: > Also, I _know_ that in general interchanging limits is > the hard part when you're proving almost anything in > analysis - if you just assume that that works you're > usually sweeping the entire proof under the rug. Hmm. This was a quite nice statement, I think. -- Eray
From: Jason on 8 Mar 2005 16:24 Q: If f' exists everywhere need f' be continuous? A: No. I cannot think of *one* example where this is true. Again you are stating something irrelevant and skirting the main issue at hand.
From: Jason on 8 Mar 2005 16:48 > Silly boy. How hard is it to comprehend the fact that there could be > more than one John Gabriel in this world? Why, there are at least 10 in > the state of Texas alone. There are a lot of fools in this world. How hard is it for you to comprehend that you are one?
From: Randy Poe on 8 Mar 2005 17:00
Jason wrote: > Q: If f' exists everywhere need f' be continuous? > A: No. > > I cannot think of *one* example where this is true. Again you are > stating something irrelevant and skirting the main issue at hand. Surely you're not saying you can't think of a function which exists everywhere but is not continuous. If the left and right limits of f(x) as x->x0 are not the same, f is not continuous. Consider: 1. Does a step function exist everywhere? 2. Is the step function continuous? 3. Can you think of a function whose derivative is a step function? - Randy |