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I just wish to check that I have got these right. Please just indicate whether the ans are right -- please don't show steps (I wish to figure those out myself).

Given $\phi_n (x)=n^k(1-x)x^n$ where $k\in\mathbb R, \,\,\,\,\,\,x\in[0,1]$


  • $\phi_n'(x)$ converge pointwise to $0$ for $k<0$ only

$\phi_n'(x)=n^{k+1}x^{n-1}[1-(1-{1\over n})x]$

Clearly $\phi_n'(x)$ converge pointwise to $0$ $\forall x\in[0,1)$

For $x=1$, $\phi_n'(x)=-n^k$ so we need $k<0$

  • $\phi_n'(x)$ converge uniformly to $0$ for $k<0$ only

$c_n:=\sup_{x\in[0,1]} |\phi_n(x)-0|=\sup_{x\in[0,1]} n^kx^{n-1}[n-(n+1)x]$

Then $c_n$ is continuous on a closed bounded interval hence attains its sup

$\phi_n'(x_{stationary point})=n^k({n-1\over n+1})^{n-1}$ obtained by differentiation then substitution. Noting that this is $\phi_n'(0)=0$ and $\phi_n'(1)=-n^k$ we have that $c_n=n^k({n-1\over n+1})^{n-1}$. It then follows that we need $k<0$.

  • $\int_0^1\phi_n(x)\,\,dx$ converge to $0$ for $k<2$ only

$\int_0^1\phi_n(x)\,\,dx={n^k\over(n+1)(n+2)}\to{n^k\over n^2}$

Hence we need $k<2$

Thank you.

share|cite|improve this question
"No steps/solutions/proofs please" - well I've never seen that before on – anon Sep 18 '11 at 20:37
Maybe it would be more interesting if you write what you have done, and then we will see if it's correct. For example, if an answer you gave is wrong, we will be able to tell you why. – Davide Giraudo Sep 18 '11 at 20:39
Edited to include steps. :-) – I. S. Sep 18 '11 at 21:06
I agree for the first question. For the second one, we have $c_n=n^k$, since $\frac{n-1}{n+1}\leq 1$ (but it doesn't change the conclusion). For the last question, maybe you should write $\sim$ instead of $\to$ (but the conclusion is good). – Davide Giraudo Sep 18 '11 at 21:51
Thanks, Davide. :-) Btw, what does ~ mean? – I. S. Sep 18 '11 at 21:58

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