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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 $E$.



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.

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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
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:

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

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Great solution! Thanks a lot – Mathmo Nov 27 '12 at 17:54

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