Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $\beta>2\alpha\ge0$. What must be the relation between $\alpha$ and $\beta$ so that $$f_n(x)={2n^\alpha x\over 1+n^\beta x^2}$$ is uniform convergent over the interval $[0,1]$. Thanks in advance!

share|cite|improve this question
Hint: The pointwise limit is the zero function. What is the maximum value of $f_n$ over $[0,1]$? – David Mitra Feb 7 '13 at 17:47
I tried that way but could not get to somewhere. @DavidMitra I should check whether maximum of the functions has limit zero right? – ciceksiz kakarot Feb 7 '13 at 17:49
Yes, that's correct. It's not to hard to do using a derivative analysis. I obtained that the maximum value of $f_n$ is attained when $x^2=n^{-\beta}$. – David Mitra Feb 7 '13 at 17:56
I found that too, but after I work the rest, I get something like $2\alpha \le 5\beta$, and I think this is not a sufficient relation. – ciceksiz kakarot Feb 7 '13 at 18:44
Hmm.. $${2n^\alpha n^{-\beta/2}\over 1+n^\beta n^{-\beta}}={2n^{\alpha-\beta/2}\over 2} = n^{2\alpha-\beta\over 2}.$$ So it seems no further conditions on $\alpha$ and $\beta$ are required to insure uniform convergence on $[0,1]$. – David Mitra Feb 7 '13 at 18:51
up vote 1 down vote accepted

Hint: Apply the inequality $z^2+1\geqslant 2|z|$ to $z=n^{\beta/2}x$, to deduce that $f_n\to0$ uniformly on $\mathbb R$ as soon as $\beta\gt2\alpha$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.