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We have three courses called Mathematical Analysis: MA 1, MA2 and MA3. MA1 is mostly about real-valued real-variable functions, defining limits, continuity, uniform continuity, differentiability and integrals. MA2 goes into higher dimensions, starting with multivariate limits and continuity, then dealing with function successions and serieses and their pointwise, absolute (in the case of serieses), uniform and total convergences, then tackles multivariate differential and integral calculus, and ODEs, then a few more things. It has been reported to me that our MA2 teacher asked in an oral exam:

What is the MA1 concept matching Lipschitz continuity?

Now, Lipschitz continuity is a concept I have seen, briefly, in MA1, as an appendix to continuity. Since the person who reported this question to me suggested uniform continuity as the answer, I guessed it may be they are equivalent in 1 dimension and cease to be in more dimensions. Then I bumped into this question proving the opposite, i.e. providing an example of an MA1 UC but not LC function. So I'm asking: what answer would you give to that question? I have absolutely no idea what to answer. Has anyone got any?

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MickG
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    For differentiable functions, isn't Lipschitz continuity the same as having a bounded derivative? Could that be it? I guess MA1 deals mostly with differentiable functions. – bof Sep 18 '14 at 08:47
  • Since it is an oral exam we're talking about, you can ask the teacher. I mean, if you're asked this question by the person in front of you, you can tell them exactly what you posted here - explain that as far as you know LC is a MA1 concept, and ask for more elaboration. – Amitai Yuval Sep 18 '14 at 09:03

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It seems the only way to get a certain answer is to ask the teacher directly.

In any case, for 1-variable functions LC is equivalent to bounded derivative, or at least bof suggests that might be.

In any case, LC is also an MA1 concept, albeit rarely used there, so this question doesn't make too much sense.

MickG
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