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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!

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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
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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

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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 $\beta\gt2\alpha$.

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