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    Say I have a function which is uniformly continuous on [0,a] and uniformly continuous on [a, infinity). How might I show that it is uniformly continuous on [0,infinity)?


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    Couldn't you just show that limit as x tends to a from above and below is a then it follows from what you've been given? (not sure about this though)
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    A good start with these questions in to write out exactly what uniform continuity means, in each seperate case. Then write out what it is you need to prove.
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    I've written out the things for each case, but I'm having trouble defining delta when x is in one interval and y is in the other. Any advice? Name:  ImageUploadedByStudent Room1422472258.098116.jpg
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    Bump, I know the delta isn't correct in my last post because it doesn't necessarily ensure that x is within delta1 of a and y is within delta2 of a.


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    (Original post by alexmufc1995)
    I've written out the things for each case, but I'm having trouble defining delta when x is in one interval and y is in the other. Any advice? Name:  ImageUploadedByStudent Room1422472258.098116.jpg
Views: 65
Size:  131.1 KB
    As a general comment, it's a good idea to think through what you want to say so you don't end up using the same letter for two different things (at the least, use suffixes). Seeing things like "Let \epsilon = \epsilon / 2 makes me cringe".

    Next, when you have a situation where you have something like "if x \in S then we can find \delta_1 such that <blah>, but if x \notin S then we can find \delta_2 such that <blah>", then if you want <blah> to always be true, you are nearly always going to want to pick \delta =  A\, \min(\delta_1, \delta-2) for some constant A (typically A = 1, or A = 1/2, or something like that).

    Finally, to deal with the case where x is in one interval and y is in the other, you should be thinking of considering a 3rd point - namely the point that is common to both intervals. You should then be able to finish using a triangle inequaltiy.
 
 
 
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