Let $V \in K (L^2 [0,1])$ be the Volters operator $Vf(s) = \int_0^s f(t) dt$, and let $A = e^{i \theta} V + e^{-i \theta} V^*$ with $0 < \theta < \pi$. Find $\underset{{{\|f\|}_{2}}=1}{\max} <Af,f>$.
I am 'in general' not very familiar with the subject, however I found some solution of how to find the the solution for the Eigen value people $Vf = \lambda f$ where V is Volterra operator.
In the posted problem I have a linear combination of Volterra operator and its Self-adjoint operator. I have two possible methods to solve the problem; The first one is:
If $V: L^2[0, 1]\to L^2 [0,1]$ then its adjoint operator $V^*: L^2[0, 1]\to L^2 [0,1]$ is given by $$V^*f(t)=\int_t^1 f(s)ds.$$
As an application of Arzela-Ascoli's theorem that $V$, as well as its adjoint $V^*$, is a compact operator and that it maps to $C[0, 1]$ (it is Hölder continuous). On the other hand, for a fixed $\theta$, it is easily shown that $A: L^2[0, 1]\to L^[0,1]$ is self-adjoint,
$$A^* = (e^{i \theta} V + e^{- i \theta} V^*)^* = (e^{i \theta} V)^* + (e^{- i \theta} V^*)^* = e^{- i \theta} V^* +e^{i \theta} V =A$$
Therefore by the spectral theorem $A$ is diagonalizable.
Now, the eigenvalue equation $Af=\lambda f$ reads as $$e^{i\theta}\int_0^t f(s)ds+e^{-i\theta}\int_t^1 f(s)ds=\lambda f(t).$$ From the properties of $V$ listed above, $f$ is continuous and $Af$ is also continuous. Then $f\in C^1$ and by differentiating the eigenvalue equation $f$ must satisfy $$e^{i\theta}f(t)-e^{-i\theta}f(t)=\lambda f'(t).$$
The solution is
$$f(t) = C e^\frac{e^{i \theta } + e^{- i \theta} t}{\lambda}.$$
But when we find the initial data, we have
$$f(0) = \frac{e^{-i \theta}}{\lambda} \int_0^1 f(s) ds$$ and $$f(1) = \frac{e^{i \theta}}{\lambda} \int_0^1 f(s) ds$$
I stoped here, I could not reach any further.
The next approach that I have tried:
$$A A^* f = \lambda f.$$ $$(e^{i \theta} V + e^{-i \theta} V^*) (e^{-i \theta} V^* + e^{i \theta} V) f = \lambda f.$$
$$(V^*V + e^{-2i \theta } (V^*)^2 + e^{2 I \theta} V^2 + VV^*) f = \lambda f$$
I started to doubt here because so many integrations are coming... Could you please show me which one is promising method and help me to do this problem. Is there any easer method to find the Maximus of the mentioned inner product?