Q:

adjoint map of differentiation space

"Let $V$ be the vector space of real-valued inﬁnitely diﬀerentiable
functions $f$ on R that are periodic with period 1, that is, $f(x) = f(x + 1)$
for all $x ∈$ R. Consider the inner product $<f,g> = \int_0^1 f(x)g(x) $. Let $
D : V → V$ be the diﬀerentiation map. Find the adjoint map $D^∗
: V → V $."
I am really stuck on this problem I tried using the fact that $<Df(x),g(x)>=<f(x),D^*g(x)>$. By what is defined as the inner product, this means $\int_0^1 Df(x)g(x)dx=\int_0^1 f(x)D^*g(x)dx$. But I have no clue what to do from here... Thanks in advance for your help.

A:

We integrate by parts
$$\langle Df,g\rangle=\int_0^1f'(x)g(x)dx=f(x)g(x)\Bigg|_0^1-\int_0^1f(x)g'(x)dx\\=-\int_0^1f(x)g'(x)dx=\langle f,-Dg\rangle$$
since $f(1)=f(0)$ and $g(1)=g(0)$. Hence $D^*=-D$.

