Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm trying to find the adjoint operator of vector spaced defined by $V=\mathbb{P}_1[x]$, and $f:V\rightarrow V$

Now I have an inner product space defined by $\langle p,q \rangle$=$\frac{1}{2}\int_{-1}^{1}{p(t)}{q(t)}dt$ I also have that $f(p)=p(0)+p(1)t$ From this, I ought to be able to find and explicit form of $f^{*}$

So far, I have that $p(0)=a+b(0)$ and $p(1)=c+d(1)$ $\\$ So $\frac{1}{2}\int_{-1}^{1}{f(p)(x)}{q(t)}dt \rightarrow\frac{1}{2}\int_{-1}^{1}{(a+ct+dt)(x)}{q(t)}dt\rightarrow \frac{1}{2}\int_{-1}^{1}{(ax+ctx+dtx)}{q(t)}dt$

So I'm wondering if I'm missing an inner integral so that I could integrate to get an explicit form for $f^{*}$, thank you.

share|improve this question

1 Answer 1

up vote 1 down vote accepted

Note that the polynomials $1$ and $t$ are orthogonal, and that $f(p)=p+p(0)t$. Then $$ \langle f^*p,q\rangle=\langle p + p(0)t,q\rangle=\langle p,fq\rangle=\langle p,q+q(0)t\rangle= \langle p,q\rangle + \langle p,q(0)t\rangle=\langle p,q\rangle + \langle p(0) + (p(1)-p(0))t,q(0)t\rangle=\langle p,q\rangle + ((p(1)-p(0))q(0)/3 =\langle p+\frac{p(1)-p(0)}3,q\rangle. $$ As we can do this for any $q$, we get that $f^*p=p+\frac{p(1)-p(0)}3$

share|improve this answer
    
Thank you, but I don't quite understand where the 3 is coming from. –  Edgar Aroutiounian Nov 13 '12 at 6:12
    
$\langle t,t\rangle =\frac12\int_{-1}^1t^2\,dt=\frac13$. So when you write your linear polynomial as $a+bt$, this is a linear combination of $1$ and $t$, which are orthogonal but not orthonormal; $1$ and $\sqrt 3\, t$ are. –  Martin Argerami Nov 13 '12 at 10:56

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.