# 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.

$$\langle Df,g\rangle=\int_0^1f'(x)g(x)dx=f(x)g(x)\Bigg|_0^1-\int_0^1f(x)g'(x)dx\\=-\int_0^1f(x)g'(x)dx=\langle f,-Dg\rangle$$ since $f(1)=f(0)$ and $g(1)=g(0)$. Hence $D^*=-D$.
• I get everything you did but I don't understand why $g(1)=g(0)$. Thanks for the answer though! Commented Mar 31, 2015 at 21:24
From integration by parts, $$\langle Df,g \rangle = \int_0^1 f'g = fg\big|_0^1 - \int_0^1fg'=-\int_0^1fg'$$
because if $f,g \in V$, then $fg|_0^1=0$.
So maybe try defining $D^*g=-g'$