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# converges uniformly and converge pointwise

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1. a) Show converges pointwise to
b) Show does not converge uniformly to
2. (Original post by vanessa_tram)

a) Show converges pointwise to
b) Show does not converge uniformly to
What have you tried?
3. (Original post by Zacken)
What have you tried?
this is how I did it ( note epsilon is "e"
a) given , take . Then for n>= N we have

take x = n then we obtain

b) suppose converges uniformly to . Then there exist N such that for all and for all n>= N we have

i stopped ther because i realised i was doing it very similar way to (a) , I dont really understand the difference between converges uniformly and converges pointwise
4. (Original post by vanessa_tram)

i stopped ther because i realised i was doing it very similar way to (a) , I dont really understand the difference between converges uniformly and converges pointwise
I'll have a look at your proof in a sec, but for now - I found that a really good example of the difference pointwise and uniform convergence and how to prove each one in turn is located here. :-)
5. (Original post by Zacken)
I'll have a look at your proof in a sec, but for now - I found that a really good example of the difference pointwise and uniform convergence and how to prove each one in turn is located here. :-)
thank you
6. (Original post by vanessa_tram)
thank you
Let me know if it makes sense!
7. (Original post by Zacken)
Let me know if it makes sense!
if like what it said from the link, then what i have done for (a) should be for (b) and change to to get contradiction. then to show pointwise we have to choose x, i dont know what x to choose.
8. (Original post by vanessa_tram)
if like what it said from the link, then what i have done for (a) should be for (b) and change to to get contradiction. then to show pointwise we have to choose x, i dont know what x to choose.
To show pointwise, work backwards from the fact that to find a suitable as a function of and . Then re-write your proof in the forward direction.
9. (Original post by Zacken)
To show pointwise, work backwards from the fact that to find a suitable as a function of and . Then re-write your proof in the forward direction.
yes, i always work backward to find my N, but as we said in term of x, then u think my first line for (a) was right? from there say
so for that "< e" we choose so
10. (Original post by vanessa_tram)
yes, i always work backward to find my N, but as we said in term of x, then u think my first line for (a) was right? from there say
so for that "< e" we choose so
That seems to work, that gets us .

Then again, I'm just a puny A-Level student and prone to errors.
11. (Original post by Zacken)
That seems to work, that gets us .

Then again, I'm just a puny A-Level student and prone to errors.
omg, r u serious, u r really good, ur knowledge is far better than a level students. i though u were doing Phd
12. (Original post by vanessa_tram)
omg, r u serious, u r really good, ur knowledge is far better than a level students. i though u were doing Phd
Haha, thanks. But FWIW, I think you're correct with - if I'm wrong, somebody will jump in and correct me soon enough, hopefully!
13. (Original post by Zacken)
Haha, thanks. But FWIW, I think you're correct with - if I'm wrong, somebody will jump in and correct me soon enough, hopefully!
ok, i will come to my lecturer so he can check them again, but thank you so much mate.
14. (Original post by vanessa_tram)
ok, i will come to my lecturer so he can check them again, but thank you so much mate.
No problem!
15. (Original post by vanessa_tram)
ok, i will come to my lecturer so he can check them again, but thank you so much mate.
Yes, this is all correct. The point about uniform convergence is that the value of depends on . So, the convergence would be uniform if you restrict attention to any compact subset of , as the value of would itself be bounded. But if you are dealing e.g. with the whole of , then it would not. To finish off the proof of non-uniform convergence, derive a contradiction: "suppose there were an independent of ..."
16. (Original post by Gregorius)
Yes, this is all correct. The point about uniform convergence is that the value of depends on .
Wait, what? I thought that was pointwise convergence?
17. (Original post by Zacken)
Wait, what? I thought that was pointwise convergence?
Ah, poor use of language on my behalf. If I said "When we turn to the question of uniform convergence, the issue is the fact that the value of N we have calculated for point-wise convergence depends upon x", would that be clearer?
18. (Original post by Gregorius)
Ah, poor use of language on my behalf. If I said "When we turn to the question of uniform convergence, the issue is the fact that the value of N we have calculated for point-wise convergence depends upon x", would that be clearer?
Ah, I see. So proving pointwise doesn't help us one lick for proving uniform. But the converse is true?
19. (Original post by Zacken)
Ah, I see. So proving pointwise doesn't help us one lick for proving uniform. But the converse is true?
Yup. Uniform implies pointwise.
20. (Original post by Gregorius)
Yup. Uniform implies pointwise.
Thanks!

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