TITLE: Adjoint of Derivative Operator
QUESTION [2 upvotes]: I've been working on this problem, and it seems like it shouldn't be too difficult, but I can't seem to piece it together.
Let $V$ be the space of 2-periodic (period of 2) functions $f$ with the inner product 
$$ \langle f, g \rangle = \int_{1}^{2} f(t)g(t)\,dt $$ and let $T$ be the derivative operator. 
I need to find $T^*.$ I tried using the definition of an adjoint map, but I wasn't able to figure out how exactly it follows.

REPLY [3 votes]: $\langle f, D^{*}g \rangle = \langle Df, g\rangle  \iff \int_{0}^{2} f(x) (D^{*}g)(x) dx = \int_{0}^{2} (Df)(x) g(x) dx = I$.
Letting $u = g(x)$, $u' = g'(x)$, $v' = f'(x), v = f(x)$ in the integration by parts, we get:
$I = \left[g(0 +2)f(0 +2) - g(0)f(0) \right] - \int_{0}^{2} f(x)g'(x)dx = \int_{0}^{2} f(x) \left(- g'(x) \right)dx$. 
It follows that $(D^{*}g)(x) = - g'(x)$.
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