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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). – 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
up vote 6 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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I solve it by Direct Solution: Suppose in VectorSpace polynomial is $p(x)=a+bx+cx^2$ Then by integration: $$f_1(p)=\int_0^1p(x)dx=a+\frac{b}{2}+\frac{c}{3}\ \ ,\ \ f_2(p)=\int_0^2p(x)dx=2a+2b+\frac{8}{3}c \, $$ $$, \,f_3(p)=\int_0^{-1}p(x)dx=-a+\frac{b}{2}-\frac{c}{3}\ \ \ \ \ (1)$$ Then use theorem 15 in Part3.5 (3-14 formula)in Hoffman,Kunze book: \, $$\alpha=\sum_{i=1}^n f_i(\alpha)\alpha_i \ \ (2) $$and $\alpha=p(x)$ and in this problem $n=3$ and { $\alpha_1,\alpha_2,\alpha_3$ } are polynomials and basis of $$\sum_{i=1}^3f_i(p).\alpha_i=P(x)\ {{\stackrel{(1)}{\Longrightarrow}}}\alpha_1(a+\frac{b}{2}+\frac{c}{3})+\alpha_2(2a+2b+\frac{8}{3}c)+\alpha_3(-a+\frac{b}{2}-\frac{c}{3})=a+bx+cx^2$$ $$\Rightarrow(\alpha_1+2\alpha_2-\alpha_3)a+(\frac{\alpha_1}{2}+2\alpha_2+\alpha_3)b+(\frac{\alpha_1}{3}+\frac{8}{3}\alpha_2-\frac{1}{3}\alpha_3)c=a+bx+cx^2$$ $\Rightarrow\left\{ \begin{array}{c} \alpha_1+2\alpha_2-\alpha_3=1 \\ \frac{\alpha_1}{2}+2\alpha_2+\alpha_3=x \\ \frac{\alpha_1}{3}+\frac{8}{3}\alpha_2-\frac{1}{3}\alpha_3=x^2 \end{array} \right.$ $\Rightarrow\left\{ \begin{array}{c} \alpha_1=p_1(x)=-\frac{3}{2}x^2+x+1 \\ \alpha_2=p_2(x)=\frac{x^2}{2}-\frac{1}{6} \\ \alpha_3=p_3(x)=-\frac{x^2}{2}+x-\frac{1}{3} \end{array} \right.$ Then by $f_i(p_j)=\delta_{ij}$ The result is obtained.

Note: 1- This solution obtain basis of V and $V^*$ Direct and don't use Kronecker delta because in theorem15 that is occult. 2- In this solution I don't consider remainder of polynoimals and consequently isn't like DavidMitra pdf. 3- This solution work for all problem of Dual basis.

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