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Assume there has the power series $\sum_{n=1}^{\infty}a_nx^n$,the convergent radius of it $r>0$,prove:

If $a_1\neq 0$,and in a neighborhood of zero point,$|\sum_{n=1}^{\infty}a_nx^n|\geq|a_1||x|-2x^2$,then

$|a_2|\leq2$


It is a problem in my exercise book, I don't know how to start.

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To begin with, can you show that $g(x)=\sum\limits_{n\geqslant2}a_nx^{n-2}$ is such that $|g(x)|\leqslant2$ for $|x|$ small enough? –  Did Jul 25 '12 at 6:59
    
@did Since $\sum_{n=1}^{\infty}a_nx^n\to0$ when $x\to0$,I think it is right. –  Joe Jul 25 '12 at 7:02
1  
This is unrelated, please read again the result I propose to prove (in particular, $g(x)$ does not depend on $a_1$). And note that this result is wrong in general hence you must use a hypothesis you did not use so far. –  Did Jul 25 '12 at 7:04
    
Try figuring it out with $a_n = 0$ for $n>2$ first. Remember you are working in $\mathbb{R}$, and you can choose the sign of $x$ to suit your needs. –  copper.hat Jul 25 '12 at 7:35
    
89085731: Any progress? –  Did Jul 26 '12 at 12:01

1 Answer 1

up vote 3 down vote accepted

If $a_2=0$, there is nothing to prove, hence from now on we assume that $a_2\ne0$.

Assume first that $a_n=0$ for every $n\geqslant3$ (as @copper.hat suggested). Then the hypothesis is that $|a_1+a_2x|\geqslant|a_1|-2|x|$ for every $|x|$ small enough, say $|x|\leqslant x_0$. Choose $x$ such that $x\ne0$, $|x|\leqslant x_0$, $a_1a_2x\leqslant0$, and $|a_2|\,|x|\leqslant |a_1|$. Then $|a_1|-|a_2|\,|x|=|a_1+a_2x|\geqslant|a_1|-2|x|$ and $x\ne0$, hence $|a_2|\leqslant2$.

In the general case, consider $A(x)=a_2+\sum\limits_{n=3}^{+\infty}a_nx^{n-2}$. Since the series $\sum\limits_na_nx^n$ has a positive radius of convergence, the function $A$ is continuous at $0$ and $A(0)=a_2\ne0$, hence $A(x)\ne0$, $|A(x)|\leqslant2|a_2|$, and $A(x)$ has the sign of $a_2$, for every $|x|$ small enough, say $|x|\leqslant x_0$. One can, and we will, assume without loss of generality that, furthermore, $|a_1+A(x)x|\geqslant|a_1|-2|x|$ for every $|x|\leqslant x_0$. Choose $x$ such that $x\ne0$, $|x|\leqslant x_0$, $a_1A(x)x\leqslant0$ and $2|a_2|\,|x|\leqslant |a_1|$. Then $|a_1|-|A(x)|\,|x|=|a_1+A(x)x|\geqslant|a_1|-2|x|$ and $x\ne0$, hence $|A(x)|\leqslant2$. One can choose $x$ with all these properties as close to $0$ as desired, hence, by the continuity of $A$ at $0$, $|a_2|\leqslant2$.

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