Hi,

problem:If $\displaystyle (\alpha_0,\alpha_1,\alpha_2)$ is an arbitrary sequence of complex numbers, and if $\displaystyle x$ is an element of $\displaystyle \mathbb{P}$, $\displaystyle x(t)=\sum^n_{i=0}\xi_it^i$,

write $\displaystyle y(x)=\sum^n_{i=0}\xi_i\alpha_i$. Prove that $\displaystyle y$ is an element of $\displaystyle \mathbb{P}'$ and that ever element of $\displaystyle \mathbb{P}'$ can be obtained in this manner by a suitable choice of the $\displaystyle \alpha$'s

$\displaystyle \mathbb{P}'$ is the collection of all linear functionals on $\displaystyle \mathbb{P}$

attempt:To prove that $\displaystyle y$ is an element of $\displaystyle \mathbb{P}'$ I show that $\displaystyle y$ is a linear functional:

$\displaystyle

\begin{aligned}

y(x+z)=&y\left(\sum^n_{i=0}(\xi_i+\eta_i)t^i\right )\\

=&\sum^n_{i=0}(\xi_i+\eta_i)\alpha_i\\

=&\sum^n_{i=0}\alpha_i\xi_i + \sum^n_{i=0}\alpha_i\eta_i\\

=&y(x)+y(z)

\end{aligned}

$

$\displaystyle

\begin{aligned}

y(\beta x)=&\sum^n_{i=0}\beta\xi_i\alpha_i\\

=&\beta\sum^n_{i=0}\xi_i\alpha_i\\

=&\beta y(x)

\end{aligned}

$

Prove that every element can be obtained in this manner by a suitable choice of the $\displaystyle \alpha$'s.

I am not sure how to prove this, but perhaps I can say that if the set of $\displaystyle \alpha$'s are a basis for $\displaystyle \mathbb{P}'$, then every element of $\displaystyle \mathbb{P}'$ can be written as $\displaystyle \sum^n_{i=0}\xi_i\alpha_i$.

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Thanks!