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Uniform convergence
Posted: Fri Oct 21, 2011 4:45 am
by nayel
Determine whether the sequence of functions $\displaystyle f_n(x)=e^{-x^2}\sin\frac xn$ converges uniformly on $\mathbb R$.
Re: Uniform convergence
Posted: Fri Oct 21, 2011 10:05 am
by tanvirab
How about this :
for any $\epsilon$ we can choose $x_0$ = $inf\{x : e^{-x^2} < \epsilon \}$
now we can choose $n$ large enough so that $sin(\frac{x}{n}) < \epsilon$ for all $|x| \leq x_0$
Re: Uniform convergence
Posted: Fri Oct 21, 2011 5:56 pm
by nayel
Does it not need to hold for all $x\in\mathbb R$?
Re: Uniform convergence
Posted: Fri Oct 21, 2011 6:03 pm
by tanvirab
It does (right?)
For $|x| \leq x_0$ the $sin$ part is smaller than $\epsilon$ and for $|x|>x_0$ the $exp$ part is smaller than $\epsilon$. So the product is always smaller than $\epsilon$. (?)
Re: Uniform convergence
Posted: Fri Oct 21, 2011 8:33 pm
by nayel
That looks so easy compared to my one page long solution!
Re: Uniform convergence
Posted: Sun Oct 23, 2011 4:51 am
by tanvirab
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