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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 $k\in\mathbb R, \,\,\,\,\,\,x\in[0,1]$

Then

  • \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.

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    I.S.: In fact $f\sim g$ means there exists $n_0$ and $h$ such that $f(n)=g(n)h(n)$ for every $n\geqslant n_0$ and $h(n)\to1$ when $n\to\infty$. For example $f(n)=(1+(-1)^n)$ and $g(n)=(1+(-1)^n)n/(n+1)$ are such that $f(n)\sim g(n)$.2011-10-19

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