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Let $V$ be the vector space of all polynomial functions $p$ from $\mathbb{R}$ to $\mathbb{R}$ which have degree two or less.

Define three linear functionals on $V$ by $$f_1(p)=\int_0^1p(x)dx,\quad f_2(p)=\int_0^2p(x)dx , \quad f_3(p)=\int_0^{-1}p(x)dx.$$

Show that $\{f_1,f_2,f_3\}$ is a basis for $V^{\ast}$, the dual space of $V$.


Very little so far...

I imagine the easiest way to approach this is to exhibit the set in $V$ to which $\{f_1,f_2,f_3\}$ is dual, and then to show it is a basis for $V$. Not sure how one would go about this however.

Any assistance would be appreciated. Regards.

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The dual space $V^*$ is $3$-dimensional, so we only have to show that $f_1,f_2,f_3$ are linearly independent. Write $af_1+bf_2+cf_3=0$, and for particular $p$ we get equations which will give $a=b=c=0$. –  Davide Giraudo Dec 13 '11 at 10:00
See the following link (exercise 4) for a solution using your proposed method (I imagine, though, following Davide's advice will be simpler). math.sunysb.edu/~scott/mat310.spr05/hw/solution7.pdf –  David Mitra Dec 13 '11 at 10:12
@DavidMitra: Many thanks! Exactly what I was looking for. –  Mathmo Dec 13 '11 at 10:14
@DavideGiraudo: Far simpler method - great advice, thanks. –  Mathmo Dec 13 '11 at 10:14

1 Answer 1

up vote 5 down vote accepted

Each $f_i$, $1\leq i\leq 3$ is linear by linearity of integration. Since $V^*$ is $3$-dimensional, we will only show that the set $\{f_1,f_2,f_3\}$ is linearly independent.

Suppose that $af_1+bf_2+cf_3=0$ for some $a,b,c\in\mathbb R$. Taking $p(x)=1$, we get $a+2b-c=0$, taking $p(x)=2x$, we have $a+4b+c=0$, and picking $p(x)=3x^2$ we get $a+8b-c=0$. Considering first and third equations, we get $2b=8b$ so $b=0$ and first and second equations give $a-c=0$ and $a+c=0$. Finally, we got $a=b=c=0$ and we conclude that $\{f_1,f_2,f_3\}$ is a basis for $V^*$.

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