- #1
boneill3
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Homework Statement
We consider P2 the vector space of all real polynomials of degree at most 2.
Show that
[itex]
<f,g> = \int_{-1}^{1}f(x)g(x)dx
[/itex]
defines an inner product space
Homework Equations
I'm Using one of the Axioms of Inner product spaces IP1. which states that.
[itex]
<u,u>[/itex] [itex] \geq 0 [/itex] with equality when and only when [itex]u=0_{v}[/itex]
The Attempt at a Solution
For IP1
let [itex]a,b,c,d,e,f,x \in R[/itex]
then
[itex]
<f,g> = \int_{-1}^{1}f(x)g(x)dx
[/itex]
[itex]
<f,g> = \int_{-1}^{1}(a+bx+cx^2)(d+ex+fdx^2)dx
[/itex]
[itex]
<f,g> = \left[ \frac{cd+fx^3}{3}+\frac{bd+ex^2}{2}+adx\right]_{-1}^{1}
[/itex]
[itex]
<f,g> = 2ad+\frac{2cd+f}{3}\
[/itex]
[itex]
<f,g> \geq 0
[/itex]
To show that equality to 0 implies the zero polynomial can I use?
let [itex]a,b,c,d,e,f \in R \neq 0[/itex]
and
[itex]x\in R [/itex]
Consider
[itex]
<f,f> = \int_{-1}^{1}f(x)f(x)dx = \left[ \frac{cd+fx^3}{3}+\frac{bd+ex^2}{2}+adx\right]_{-1}^{1}
[/itex]
if we assume
[itex]
<f,f> = 0
[/itex]
then this implies
[itex]
\left[ \frac{cd+fx^3}{3}+\frac{bd+ex^2}{2}+adx\right]_{-1}^{1}
= 0
[/itex]
but a,b,c,d,e,f [itex]\neq 0[/itex]
therefore
[itex]
\left[ \frac{cd+fx^3}{3}+\frac{bd+ex^2}{2}+adx\right]_{-1}^{1}
\neq 0
[/itex]
contradicting our assumption
hence a,b,c,d,e,f [itex]= 0[/itex] the zero polynomial
So we have
[itex]
<u,u>[/itex] [itex] \geq 0 [/itex]
and
[itex]
<u,u>[/itex] [itex] = 0 \rightarrow f = 0_{v}[/itex]
So IP1 holds
Is that enough to prove IP1 ?