Adjoint Operator to the Derivative 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
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:

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

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