We consider the operator $L=-i\frac{d}{dx}$ with $x \in (a,b)$. Check whether L is a self-adjoint operator for
$\mathcal{D}(L)=\{u \in C[a,b]:u(a)=u(b)\}$
My thought is the following
To check whether $L$ is self-adjoint on the domain $\mathcal{D}(L)={u \in C[a,b]:u(a)=u(b)}$, we need to verify whether it satisfies the following property: $$\langle Lu,v \rangle=\langle u,Lv \rangle$$
Let $u,v \in \mathcal{D}(L)$. Then, we have:
\begin{align*} \langle Lu, v \rangle &= \int_a^b (-i\frac{d}{dx}u(x))\overline{v(x)} dx \newline &= -i[u(x)\overline{v(x)}]_a^b + i\int_a^b u(x)\frac{d}{dx}\overline{v(x)} dx \newline &= i\int_a^b u(x)\frac{d}{dx}\overline{v(x)} dx \hspace{15mm} \text{(since $u(a)=u(b)$)}\newline &= i[u(x)\overline{v(x)}]_a^b - i\int_a^b \frac{d}{dx}u(x)\overline{v(x)} dx \newline &= i[u(x)\overline{v(x)}]_a^b + \langle u, -i\frac{d}{dx}v \rangle \newline &= \langle u, Lv \rangle \hspace{16mm} \text{(since $v(a)=v(b)$)} \end{align*}
Therefore, we have shown that $L$ is self-adjoint on the domain $\mathcal{D}(L)=\{u \in C[a,b]:u(a)=u(b)\}$
Is this way of thinking correct?