Functional problem

• Feb 25th 2010, 05:39 AM
wicked
Functional problem
Let $\displaystyle V=R_n[x]$ and $\displaystyle \alpha_i(p)=p^i(0)$ be a functional over $\displaystyle V$( $\displaystyle p^i$ is the number i derivative of p)
Show that $\displaystyle \{\alpha_i\}_\forall\i$ is a basis of a dual space $\displaystyle V^*$

What I first should do is prove that $\displaystyle \alpha_i \forall i$ are linearly independent.
How should I do it?
Any ideas?
• Feb 25th 2010, 05:56 AM
Opalg
Quote:

Originally Posted by wicked
Let $\displaystyle V=R_n[x]$ and $\displaystyle \alpha_i(p)=p^i(0)$ be a functional over $\displaystyle V$( $\displaystyle p^i$ is the number i derivative of p)
Show that $\displaystyle \{\alpha_i\}_\forall\i$ is a basis of a dual space $\displaystyle V^*$

What I first should do is prove that $\displaystyle \alpha_i \forall i$ are linearly independent.
How should I do it?
Any ideas?

Hint: notice that $\displaystyle \alpha_i(x^k)$ is $\displaystyle k!$ if i=k, and 0 if i≠k.
• Feb 25th 2010, 08:46 AM
wicked
Quote:

Originally Posted by Opalg
Hint: notice that $\displaystyle \alpha_i(x^k)$ is $\displaystyle k!$ if i=k, and 0 if i≠k.

$\displaystyle \alpha_0,....,\alpha_n$are independent iff
$\displaystyle \sum\lambda_i\alpha_i=0$

So if $\displaystyle p=a_0+a_1x+....+a_nx^n$
$\displaystyle \alpha_0(p)=(a_1+2a_2+...+na_nx^(n-1))(0)=a_0$
.
.
.
$\displaystyle \alpha_n(p)=(0+0+...+n!a_nx^0)(0)=n!a_n$
right?
Then $\displaystyle \forall a_1....a_n$
$\displaystyle \lambda_0\alpha_0+....+\lambda_n\alpha_n=0$

and therefore it's obvious that $\displaystyle \lambda_0=\lambda_1=.....=\lambda_n=0$

And therefore $\displaystyle dimV^*=dimV=n+1$ => $\displaystyle \{\alpha_i\}$ is basis.
This is more or less is the answer right?