Let $0<\theta <\pi $, and continuous linear operator $T:L^2(0,1)\to L^2(0,1)$ $Tf(x):=\int _0^x e^{i\theta} f(t)dt$
Then, what is $\sup_{\{||f||\leq 1\}}\int_0 ^1 (T+T^*)f(t) \bar{f(t)}dt$ ?
My idea :$T+T^*$ is symmetry, so $\int (T+T^*)f(t)\bar{f(t)}dt$ is real value. I calculated $\int_0 ^1 (T+T^*)f(t) \bar{f(t)}dt=\int \int \cos{\theta}f(t)\bar{f(x)}+i\sin{\theta} \bar{f(x)}(\chi _{[0,x]}(t)-\chi_{[0,t]}(x))dtdx$ by Fubini's theorem, but I don't know how to find sup. I guess supremum is $\sqrt{2}\cos{\theta}$
The adjoint of $Tf = e^{i\theta}\int_{0}^{x}f(t)dt$ is $T^*f = e^{-i\theta}\int_{x}^{1}f(t)dt$. Both $T$ and $T^*$ are compact operators and, therefore $T+T^*$ is selfadjoint and compact. Therefore, $$ \sup_{\|f\|=1}\langle (T+T^*)f,f\rangle $$ is the largest eigenvalue of $T+T^*$. The eigenvalue problem for $T+T^*$ is
$$ e^{i\theta}\int_{0}^{x}f(t)dt+e^{-i\theta}\int_{x}^{1}f(t)dt=\lambda f $$ Any such $f$ must satisfy $\lambda f(0)=e^{-i\theta}\int_{0}^{1}f(t)dt$ and $\lambda f(1)=e^{i\theta}\int_{0}^{1}f(t)dt$, or
$$ e^{i\theta}f(0)-e^{-i\theta}f(1)=0. $$ $$ f(1)=e^{2i\theta}f(0). $$ And, $f$ must satisfy the ODE $$ e^{i\theta}f(x)-e^{-i\theta}f(x)=\lambda f'(x) $$
$$ 2i\sin\theta f(x)=\lambda f'(x) $$
$$ \frac{f'(x)}{f(x)}=\frac{2i\sin\theta}{\lambda} $$
$$ f(x) = Ce^{2i\sin(\theta) x/\lambda} $$ The endpoint condition is satisfied for $C\ne 0$ iff
$$ e^{2i\sin(\theta)/\lambda}=e^{2i\theta} $$ $$ \sin(\theta)/\lambda=\theta+\pi n $$ $$ \lambda = \frac{\sin(\theta)}{\theta+\pi n}. $$ So the supremum is the largest value of $\lambda$ given above for $n=0,\pm 1,\pm 2,\cdots$.