# 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 $\displaystyle (\alpha_0,\alpha_1,\alpha_2)$ 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!

2. The problem with that is that the "$\displaystyle \alpha$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 $\displaystyle \{1, t, t^2, \cdot\cdot\cdot, t^n\}$.

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

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

3. Helpful as always,thank you.