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Let $V \subset \Bbb R[X]$ be the Vectorspace of all Polynomials of degree $\le 3$. The inner product on $V$ is defined as follows: $$\langle f,g \rangle:=\int^1_{-1}f(t)g(t)dt$$Let $L:V \to V$ be the derivative $L(p) = p'$

I now have to determine the adjoint Operation $L^*$ so that $$ \langle L(f),g \rangle=\langle f,L^*(g) \rangle$$ Any ideas? ( i tried partial Integration but i don't get anywhere.)

Thanks in advance

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  • $\begingroup$ i saw another post with a similar problem but they came to no solution. The problem i have is that even with partial integration L* would be dependent on f but it should of course be independently defined from g and f $\endgroup$ – DeltaChief Apr 23 '16 at 18:15
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Integration by parts $\int_{-1}^1 f'g=[fg]_{-1}^1-\int_{-1}^1 fg'$ is a good idea. But you have to be in a context where the term $[fg]_{-1}^1$ is zero. In order that this condition is always fulfilled, you must restrict your working space to functions that are zero at both bounds $-1$ and $1$ (it is no longer a vector space). In this framework, we have

$$\int_{-1}^1 f'g=\int_{-1}^1 f(-g)'$$

proving that the adjoint of $L$ is its opposite

$$L^*=-L$$

Remark (a discrete analogy): If you are acustomed to the discrete equivalent of the derivation operator $L$, represented by the infinite matrix:

$$\begin{bmatrix}&\cdots&&&&&\\&-1&1&&&&\\&&-1&1&&&\\&&&-1&1&&\\&&&&-1&1&\\&&&&&\cdots&\\&&&&&&\cdots\\\end{bmatrix}$$ its transpose is as well its opposite.

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