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Let $ {P_{2}}([0,1]) $ be the Hilbert space consisting of all polynomials of degree at most $ 2 $ (including the zero polynomial on $ [0,1] $) equipped with the inner product $ \displaystyle \langle f,g \rangle \stackrel{\text{def}}{=} \int_{0}^{1} f(x) g(x) ~ d{x} $. Define a linear functional on $ {P_{2}}([0,1]) $ by $ \phi(p) \stackrel{\text{def}}{=} p(1) $ for all $ p \in {P_{2}}([0,1]) $.


  1. How can I show that $ \phi $ is linear, and how can I find $ \| \phi \|$? I guess I don’t quite understand what $ \phi(p) = p(1) $ means. As for the norm $ \| \cdot \| $ , can I use the assertion of the Riesz Representation Theorem that the operator norm of the linear functional is equal to the vector norm of the vector that induces it?

  2. Find the polynomial $ q \in {P_{2}}([0,1]) $ such that $ \phi(p) = \langle p,q \rangle $ for all $ p \in {P_{2}}([0,1]) $. I know that for this part, a unique such polynomial $ q $ exists by the Riesz Representation Theorem and that I have to get an orthonormal basis for $ {P_{2}}([0,1]) $. But I’m stuck from there. Appreciate any help, dear colleagues.

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An evaluation is very easily seen to be linear. –  1015 Mar 18 '13 at 21:52

2 Answers 2

up vote 1 down vote accepted

Just a hint:

$\phi(p +aq) = p(1) + aq(1) = \phi(p) + a\phi(q)$ therefore $\phi$ is linear.

$||\phi|| = \sup_{p : \int_0^1p^2 = 1}\phi(p) = \sup_{p : \int_0^1p^2 = 1}p(1)$.

$p = ax^2 + bx +c$, $\int_0^1p^2 = 1 \Longleftrightarrow \frac{a^2}{5} + \frac{ab}{2} + \frac{2ac+b^2}{3} + bc + c^2 = 1$.

Now you have to maximize $p(1) = a +b + c$ bearing in mind the previous condition.

Also: $p = ax^2 + bx +c$, $q = dx^2 + ex +f$. $\int_0^1pq = a(\frac{d}{5} + \frac{e}{4} + \frac{f}{3}) + b(\frac{d}{4} + \frac{e}{3} + \frac{f}{2}) + c(\frac{d}{3} + \frac{e}{2} + f)$. You are looking for $q$ such that the previous integral equals $p(1) = a + b +c$, so you look for $d,e,f$ such that $\frac{d}{5} + \frac{e}{4} + \frac{f}{3} = 1$, $\frac{d}{4} + \frac{e}{3} + \frac{f}{2} = 1$ and $\frac{d}{3} + \frac{e}{2} + f = 1$.

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Why are you squaring p? –  User69127 Mar 18 '13 at 23:46
    
because the norm of $p$ is the square root of the scalar product of $p$ with itself. mathworld.wolfram.com/OperatorNorm.html So I'm taking the $\sup$ over the polynomials of unitary norm. –  user01123581321345589144... Mar 18 '13 at 23:53

Hints:

1) Try a few examples to get the hang of the action of $\phi$, e.g. let $p(x)=1+x+x^2$; then, $\phi(p)=p(1)=1+1+1^2=3$. Try a few more! To show this is a linear operator, we need to show that $\phi(\alpha p+\beta q)=\alpha\phi(p)+\beta\phi(q)$. Equivalently, we could show both $\phi(p+q)=\phi(p)+\phi(q)$ and $\phi(\alpha p)=\alpha\phi(p)$ (which might be a little easier on the algebra eyes). As far as finding the norm, you certainly could use the norm of the Riesz representative, and since you have to find it for part 2), you may as well. Though it shouldn't be too hard to go "from the definition".

2) Every function in $P_2$ can be written as $ax^2+bx+c$. So, let $p=ax^2+bx+c$, and $q=\alpha x^2+\beta x+\gamma$. Explicitly compute $\langle p,q\rangle$ (it's a bit messy, but it's simple calculus). You want $\langle p,q\rangle=p(1)=a+b+c$. Match up terms to find $\alpha,\beta$ and $\gamma.$

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So then the norm will be $||\phi||=||q||= \sqrt{<q,q,>}$ right? –  User69127 Sep 12 '13 at 21:32

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