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converges uniformly and converge pointwise

fn(x)=x2+1/nf_n(x) = x^2 + 1/n
a) Show (fn2) (f_n^2) converges pointwise to x4 x^4
b) Show (fn2) (f_n^2) does not converge uniformly to x4 x^4
Reply 1
Original post by vanessa_tram
fn(x)=x2+1/nf_n(x) = x^2 + 1/n
a) Show (fn2) (f_n^2) converges pointwise to x4 x^4
b) Show (fn2) (f_n^2) does not converge uniformly to x4 x^4


What have you tried? :smile:
Original post by Zacken
What have you tried? :smile:

this is how I did it ( note epsilon is "e":wink:
a) given e>0e >0 , take N<(e/2) N< (e/2) . Then for n>= N we have
fn2(x)x4=x4+2x21n+1n2x4=2x21n+1n2>=2x21n |f_n^2 (x) - x^4| = |x^4 + 2x^2 \frac {1}{n} + \frac{1}{n^2} - x^4 | = |2x^2 \frac {1}{n} + \frac{1}{n^2}| >= 2x^2 \frac {1}{n}
take x = n then we obtain
fn2(x)x4>=2n21n=2n>=2N<e |f_n^2 (x) - x^4| >= 2n^2 \frac{1}{n} = 2n >= 2N < e

b) suppose fn2 f_n^2 converges uniformly to x4 x^4 . Then there exist N such that for all xR x \in R and for all n>= N we have
fn2x4=2x21n+1n2 |f_n^2 - x^4| = |2x^2 \frac{1}{n} + \frac{1}{n^2}

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
(edited 7 years ago)
Reply 3
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. :-)
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 :smile:
Reply 5
Original post by vanessa_tram
thank you :smile:


Let me know if it makes sense! :smile:
Original post by Zacken
Let me know if it makes sense! :smile:


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.
Reply 7
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 fn2x4<ϵ|f^2_n - x^4| < \epsilon to find a suitable NN as a function of xx and ϵ\epsilon. Then re-write your proof in the forward direction.
Original post by Zacken
To show pointwise, work backwards from the fact that fn2x4<ϵ|f^2_n - x^4| < \epsilon to find a suitable NN as a function of xx and ϵ\epsilon. 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 2x21n<=2x21N 2x^2 \frac{1}{n} <= 2x^2 \frac{1}{N}
so for that "< e" we choose 1N<e2x2 \frac{1}{N} < \frac{e}{2x^2} so N>2x2e N > \frac {2x^2}{e}
Reply 9
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 2x21n<=2x21N 2x^2 \frac{1}{n} <= 2x^2 \frac{1}{N}
so for that "< e" we choose 1N<e2x2 \frac{1}{N} < \frac{e}{2x^2} so N>2x2e N > \frac {2x^2}{e}


That seems to work, that gets us fn2x4ϵ+ϵ24x2<ϵ|f_n^2 - x^4| \leq \epsilon + \frac{\epsilon^2}{4x^2} < \epsilon.

Then again, I'm just a puny A-Level student and prone to errors. :colondollar:
Original post by Zacken
That seems to work, that gets us fn2x4ϵ+ϵ24x2<ϵ|f_n^2 - x^4| \leq \epsilon + \frac{\epsilon^2}{4x^2} < \epsilon.

Then again, I'm just a puny A-Level student and prone to errors. :colondollar:


omg, r u serious, u r really good, ur knowledge is far better than a level students. i though u were doing Phd :smile:
Reply 11
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 :smile:


Haha, thanks. But FWIW, I think you're correct with N>2x2ϵN > \frac{2x^2}{\epsilon} - if I'm wrong, somebody will jump in and correct me soon enough, hopefully! :smile:
Original post by Zacken
Haha, thanks. But FWIW, I think you're correct with N>2x2ϵN > \frac{2x^2}{\epsilon} - if I'm wrong, somebody will jump in and correct me soon enough, hopefully! :smile:


ok, i will come to my lecturer so he can check them again, but thank you so much mate.
Reply 13
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! :smile:
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 NN depends on xx. So, the convergence would be uniform if you restrict attention to any compact subset of R\mathbb{R}, as the value of xx would itself be bounded. But if you are dealing e.g. with the whole of R\mathbb{R}, then it would not. To finish off the proof of non-uniform convergence, derive a contradiction: "suppose there were an NN independent of xx..."
Reply 15
Original post by Gregorius
Yes, this is all correct. The point about uniform convergence is that the value of NN depends on xx.


Wait, what? I thought that was pointwise convergence?
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?
Reply 17
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?
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.
Reply 19
Original post by Gregorius
Yup. Uniform implies pointwise.


Thanks!

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