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Consider $f_n(x)=n^2xe^{-nx^2}$ on $[1,\infty)$. The exercise asks to prove that $f_n$ converges uniformly.

However it seems not to converge uniformly. Since if it were then it would converge to $0$ (because it converges pointwise to $0$). On the other hand $$f_n'(x)=n^2e^{-nx^2} (1-2nx^2)$$

then we can see $x=\frac{1}{\sqrt{2n}}$ is a global maximum point on $[1,\infty)$ but $$f_n \left( \frac{1}{\sqrt{2n}} \right)= \frac{ n^\frac{3}{2} }{\sqrt{2e}}$$ which doesn't go to $0$ as $n \rightarrow \infty$.

Am I doing something wrong? Thank you.

user16015
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1 Answers1

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In fact, $f_n' < 0$ on $[1,\infty)$ and the global maximum of $f_n$ occurs at $x=1$. (The critical point is outside of the interval in question.)

mrf
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  • Thank you. But if we compare two values $f_n(1)=n^2 e^{-n}$ is smaller than $f_n(\frac{1}{\sqrt{2n}})$ for large $n$ right? I am not getting this... – user16015 Sep 18 '15 at 20:27
  • Yes, but $1/\sqrt{2n} < 1$, so the value there is irrelevant. – mrf Sep 18 '15 at 20:29
  • Ok it seems I got it, we can't take $n$ too large since $f$ is defined on $[1,\infty)$... – user16015 Sep 18 '15 at 20:29