# Thread: Prove that y is an element of the dual space of P.

1. ## Prove that y is an element of the dual space of P.

Hi,

problem:

If $(\alpha_0,\alpha_1,\alpha_2)$ is an arbitrary sequence of complex numbers, and if $x$ is an element of $\mathbb{P}$, $x(t)=\sum^n_{i=0}\xi_it^i$,
write $y(x)=\sum^n_{i=0}\xi_i\alpha_i$. Prove that $y$ is an element of $\mathbb{P}'$ and that ever element of $\mathbb{P}'$ can be obtained in this manner by a suitable choice of the $\alpha$'s

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

attempt:

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


\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}


\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 $\alpha$'s.

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

2. The problem with that is that the " $\alpha$s" are sequences of complex numbers, not members of P' and so cannot be a basis.

But it is easy to construct a basis:

A basis for P itself is $\{1, t, t^2, \cdot\cdot\cdot, t^n\}$.

Let $L_i$ be the linear functional that assigns the value 1 to $t^i$ and 0 to all other basis members.

Then any member of P' can be written as $L= a_0L_0+ a_1L_1+ a_2L_2+ \cdot\cdot\cdot+ a_nL_n$ and $L(\xi_0+ \xi_1t+ \cdot\cdot\cdot+ \xi_nt^n)= a_0\xi_0+ a_1\xi_1+ \cdot\cdot\cdot+ a_n\xi_n$