Construct the sequence $\{a_n\}\subset\mathbb R$ such, that for every $f\in C[0,1]$ with $f(0)=0$, there exists a sequence $\{n_k\}$ for which \begin{equation} \lim_{k\to\infty}\sup_{x\in[0,1]}\left|\sum_{n=1}^{n_k}a_nx^n-f(x)\right|=0 \end{equation}
construct $\{a_n\}$, for which exists $\{n_k\}$ and $\sum_{n=1}^{n_k}a_nx^n$ converges uniformly to $f$, where $f\in C[0,1]$, $f(0)=0$
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0I presume you have looked at the Bernstein polynomials http://en.wikipedia.org/wiki/Bernstein_polynomial ? – 2012-07-30
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0Is this homework? If it is, the convention is to label it as such. – 2012-07-30
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0Do you really mean “construct”, in the sense of giving an explicit construction, or merely “show the existence”? The former looks quite hard; the latter, not quite so hard. – 2012-07-30
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0yes, I looked, but I failed to apply them. No, this is not homework. Unfortunately, I even cannot show the existence of such sequence. – 2012-07-30
1 Answers
Here is an outline: First you need a lemma:
If $f\in C[0,1]$ with $f(0)=0$ and $\epsilon>0$ and a natural number $n$ are given, there exists a polynomial $p$ with $\lvert x^np(x)-f(x)\rvert<\epsilon$ for all $x\in[0,1]$.
Then you pick a dense sequence $\{f_n\}$ in $\{f\in C[0,1]\colon f(0)=0\}$ where,moreover, every function in the sequence occurs infinitely often in the sequence. (Edited: There is no need for all this repetition, since every tail of a dense sequence is itself dense!)
Let $\{\epsilon_k\}$ be a sequence of positive numbers converging to $0$.
Now we can start the work: Pick a polynomial $p_1$ approximating $f_1$ better than $\epsilon_1$.
Let $p_1$ have degree $n_1-1$. Pick a polynomial $q_2$ so that $x^{n_1}q_2(x)$ approximates $f_2-p_1$ better than $\epsilon_2$. Let $p_2(x)=p_1(x)+x^{n_1}q_2(x)$. Note that $p_2$ approximates $f_2$ better than $\epsilon_2$.
Repeat the procedure of the previous paragraph. The coefficients of the polynomials $p_k$ as $k\to\infty$ will become the sequence $\{a_n\}$.