Prove or disprove that sequence of functions is uniform convergent

Question

I would like to see if $$f_n(x) = n^{-x}x^n\cos (nx)$$ converges uniformly on $[0,1]$

The sequence converges pointwise to $f(x)=0$.

By using the crieria the $f_n$ converges if and only if $$M_n = sup_{x\in[0,1]} |f_n(x) - f(x)|$$ converges to $0$, I obtain an equation for $x$ that is not solvable analytically.

So I though of using the fact that $$f_n(x) <= n^{-x} x^n \forall x\in [0,1]$$.

By obtaining the $M_n$ for this new sequence, I get that the supremum is obtained by $x_0 = \frac{n}{\log(n)}$.

What is then left to show is that $M_n = n^\frac{-(n)}{\log(n)} \left(\frac{n}{\log (n)}\right)^n $ tends to $0$ as $n$ tends to infinity. This however, I am not able to prove (and I am not even sure this is the case).

Can someone help/give me a hint with this? Help for the starting question or/and the subsequent question which I ended up with are most welcome.

Thank you!

Fix $x \in [0,1] $ then $$f_n(x) = n^{-x}x^n\cos (nx) <= n^{-x}x^n$$ which tends to 0 as $x^n$ tends to 0 and $n^{-x}$ tends to 0. So $f_n(x)$ tends to 0 as n approaches infinity. — elaRosca, Apr 28 '13 at 20:59
Use the following systematic technique. Do not forget to up vote the answer if you find it useful. — Mhenni Benghorbal, Apr 29 '13 at 02:58
And don't forget to vote it down if you find it impossible to apply in this case :-) — TonyK, Apr 29 '13 at 05:51

score 3 · Accepted Answer · answered Apr 28 '13 at 21:23

3

Hint:

$|f_n(x)| = |n^{-x} x^n \cos (nx)| \le n^{-x} x^n$.

Both the factors on the right-hand side are $\le 1$ in absolute value; one of them tends to zero uniformly on $[0,\frac{1}{2}]$, and the other on $[\frac{1}{2},1]$.

answered Apr 28 '13 at 21:23

TonyK

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Prove or disprove that sequence of functions is uniform convergent

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