Functional problem

**wicked** · February 25th 2010, 05:39 AM

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

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

**Opalg** · February 25th 2010, 05:56 AM

Originally Posted by wicked

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

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

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

**wicked** · February 25th 2010, 08:46 AM

Originally Posted by Opalg

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

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

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

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

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

Thread: Functional problem

LinkBack

Thread Tools

Display

Functional problem

Similar Math Help Forum Discussions

Problem in functional analysis

Linear functional problem

problem in functional analysis

solution to a functional analysis problem

Functional

Search Tags