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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.

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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 $f^*p=p+\frac{p(1)-p(0)}3$

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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

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