Calculate supreme of $((T+T^*)f,f)$ Let $0&lt;\theta &lt;\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 $n=0,\pm 1,\pm 2,\cdots$.
1) The set of which you are computing supremum of, is a subset of the real numbers. 

2) Therefore, the supremum is the maximum of the closure.

3) Every point in the spectrum of $T+T^*$ is in that subset.

4) Since $T+T^*$ is normal, then its spectral radius is $\|T+T^*\|$.

5) $((T+T^*)f,f)\leq \|T+T^*\|$ for $\|f\|=1$.

It follows that the supremum that you need is $\|T+T^*\|$.