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Let $H=l^2(N\cup \{0\})$.

a. Show that if $\{\alpha_n\}\in l^2$, then the power series $\sum_{n=0}^\infty \alpha_nz^n$ has radius of convergence $\geq 1$.

b. If $|\lambda|< 1$ and $L:H\to C $ is defined by $L(\{\alpha_n\}) = \sum_{n=0}^\infty \alpha_n \lambda^n$, find the vector $h_0\in H$ such $L(h)=(h,h_0)$ for every $h\in H$

c. What is the norm of L?

For this exercise I do not have any idea about part a. Because I know that if $z=1$ and put $\{\alpha_n\} = \{\frac{1}{n}\}$ then clearly $\sum_{n=0}^\infty \alpha_nz^n =\infty$. so, the power series is not convergent in 1.

For part b and c, I put $h_0=\{\lambda^n\}_{n=0}^\infty$ then by this definition $\|L\|\neq \|h_0\|$. Please help me. Thanks.

saeed
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1 Answers1

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a) Radius convergence only tells you the behavior of the power series inside the radius, but says nothing about it on the boundary. Notice for part $b$, $|\lambda|<1$, not $\leq$.

For your example of $a_n=1/n$, for $z<1$, this still converges by the Dirichlet test. This can be found http://en.wikipedia.org/wiki/Dirichlet's_test . The first part follows by Cauchy-Schwarz inequality. Please check you can do this.

b) I think your $h_0$ is wrong, shouldn't it be the complex conjugate of $\lambda^n$? (is the field you are working complex or real?)

c) Please also write down your calculation for norm of $L$ and norm of $h_0$. As you noticed, they should be the same.

Lost1
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  • Thanks for your attention. I agree with you about radius convergence; but in exercise get it $\geq 1$ which means that it can be > 1+$\epsilon$. – saeed Dec 29 '13 at 16:06
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    @saeed It says, the radius of convergence can be 1. By Cauchy Schwarz, it means, the series is convergent for all $|z|<1$. This means the radius of convergence $\geq 1$.

    How does $\geq 1$ mean $>1+\epsilon$?!???!?!?

    BTW, you need to accept some of the answers to your previous questions, else many won't be interested even to look at your questions...

     – Lost1 Dec 29 '13 at 16:10
  • About $h_0$ why do you say that it should be the complex conjugate of λn? I showed that $||L||\leq ||h_0||$ and $||h_0||=\sum|\lambda|^{2n}=\frac{1}{1-|\lambda|^{2}}>1$. so I cannot show that $||h_0||\leq||L||$ – saeed Dec 29 '13 at 16:25
  • @saeed how is the inner product defined on C? $(x.y) = x_1y^_1+x_2y^_1+...$ where * is conjugation. surely you know this. How else can $(x,x) = ||x||^2$? – Lost1 Dec 29 '13 at 16:29