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.

Problem

Let $V$ be the vector space of real polynomials in one variable $X$ with degree at most $1$. Define the inner product $\langle f,g\rangle:=\int^{1}_{-1}f(t)g(t) \, dt$ on this space and the map $E:V\rightarrow V$ by $E(f)(X)=f(X+1)-f(X)$.

Find the adjoint map $E^{\star}$ of $E$.

Progress

EDIT:

We require $E^{\star}$ such that $\langle E(f),g\rangle=\langle f,E^{\star}(g)\rangle$.

We note that $\langle E(f),g\rangle =\int^1_{-1}[f(t+1)-f(t)]g(t)dt$.

$f,g \in V \implies f=a_f+b_ft$ and $g=a_g+b_gt$.

Thus, $\int^1_{-1}[f(t+1)-f(t)]g(t) \, dt=2b_fa_g$ I think.

So, $\langle f,E^{\star}(g)\rangle=2b_fa_g$, but how do we find $E^{\star}$ from this?

Any help would be very appreciated. Thanks.

share|improve this question
    
You're right, I'll edit in the correction. –  Mathmo Nov 27 '12 at 16:08
    
I changed $<f,g>$ to $\langle f,g\rangle$. That is standard notation. –  Michael Hardy Nov 27 '12 at 18:58
add comment

1 Answer 1

up vote 1 down vote accepted

Here's another approach using the basis $\{1, X\}$ of $V$. Check that $\langle f, 1 \rangle = \langle 1, f \rangle = 2a_f$ and $\langle f, X \rangle = \langle X, f \rangle = \tfrac{2}{3}b_f$ for all $f \in V$. Using this we can decompose $E^{\ast}(g)$ in this basis:

$$ \begin{eqnarray} E^{\ast}(g) & = & \tfrac{1}{2}\langle 1, E^{\ast}(g) \rangle + \tfrac{3}{2} \langle X, E^{\ast}(g) \rangle \, X\\ & = & \tfrac{1}{2} \langle E(1), g \rangle + \tfrac{3}{2} \langle E(X), g \rangle \, X \\ & = & \tfrac{1}{2} \langle 0, g \rangle + \tfrac{3}{2} \langle 1, g \rangle \, X\\ & = & 3 a_g \, X \end{eqnarray} $$

share|improve this answer
    
Great solution! Thanks a lot –  Mathmo Nov 27 '12 at 17:54
add comment

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.