Problem on a rank 1 perturbation of an self-adjoint operator

Question

With my teacher we have been trying to solve for a while a problem that consists of two parts, and each one in three sections. The first part, which has to do with the problems that I was able to solve, I will detail in a summarized way:

Considering that we are in a Hilbert space

Let $H$ be a selfadjoint operator which is applied to Hilbert space. Consider the operator

$$H_\lambda = H + \lambda V $$

where $V = \lvert \delta_0\rangle\langle\delta_0 \rvert$.

We defined operator position momentum associated to $H$ as

$$\langle \lvert x \rvert^p \rangle := \sum_{x \in Z^d} \lvert x \rvert^p \langle \delta_x, e^{-itH} \delta_0 \rangle$$

Part one:

1. Write it using Duhamel expansion
2. Find the relation between $\mu(P,\lambda)$ and $\mu(P,0)$ if $\mu(P,\lambda) := \| \lvert x \rvert^{P/2} e^{-itH_\lambda} \delta_0 \|^2$
3. Prove: $\| \lvert x \rvert^{P/2} e^{-itH_\lambda} \delta_0 \|^2 \leq 2 \| \lvert x \rvert^{P/2} \delta (t) \|^2 + 2 \lvert \lambda \rvert^2 \int_0^t \lvert \langle \delta_0, e^{-i(t-s)H_\lambda} \delta_0 \rangle \rvert^2 ds$

My answer: 1)

$$\frac{d}{dt} (e^{-itH_\lambda} e^{itH}) = \frac{d}{dt}(e^{-itH_\lambda}) e^{itH} + e^{-itH_\lambda} \frac{d}{dt} (e^{itH})$$

I used the commutativity properties and

$$=-ie^{-itH_\lambda} (H_\lambda -H) e^{itH} $$

Now, integrating both sides, and multiplying by $e^{-itH}$

$$e^{-itH_\lambda}=e^{-itH} -i\lambda \int_0^t e^{-isH_\lambda} V e^{-i(t-s)H} ds$$

Next, items 2 and 3 were detached from the previous development and were satisfactorily achieved, here the problems begin, in part 2.

Part two:

a) Find the upper bound of the following expression

$\int_0^T |\langle\delta_0, e^{-i(t-s)H} \delta_0\rangle|^2 ds$

(is just the beginning of this part, then I have to get the time periods from this expression and finally set up a coefficient consisting of the lower bound of the logarithm of that whole expression, divided by the logarithm of $T$)

What I tried, based on the fact that the $\delta_0$ is independent of time, is the Cauchy Schwarz inequality

$$\int_0^T \lvert \langle \delta_0, e^{-i(t-s)H} \delta_0 \rangle \rvert^2 ds \leq \int_0^T \lvert \delta_0 \rvert^4 ds \int_0^T \lvert e^{-i(t-s)H} \rvert^2 ds \\ \int_0^T \lvert \langle \delta_0, e^{-i(t-s)H} \delta_0 \rangle \rvert^2 ds \leq T \lvert \delta_0 \rvert^4 \int_0^T \lvert e^{-i(t-s)H} \rvert^2 ds $$

I don't know if it's right, much less if I can solve the expression that I have in the integral or work with it in some way. If there are no more ways, when incorporating it in the other expressions I have something difficult to observe, that's why I think I'm developing it wrong, but I need advice, help, some theory that could be useful to me.

I am grateful in advance.

P.S my teacher is on vacation, so I'm consulting you in the meantime.

Answer 1

We had \begin{gather*} |\langle \delta_0 , e^{-i(t-s)H} \delta_0 \rangle |^2 \end{gather*} $\delta_0$ corresponds to an operator in its canonical base, therefore, the following relation exists : $\delta_0^2=1$.
The operator $e^{-itH}$ it's unitary, implies that its norm is equal to $1$. \ Now we applied Cauchy-Schwarz for this expression and \begin{equation*} \begin{split} |\langle \delta_0, e^{-i(t-s)H} \delta_0 \rangle |^2 & \leq |\delta_0|^2 |{e^{-i(t-s)H}| \delta_0}^2 \\ & \leq 1 \cdot |{e^{-i(t-s)H}| \delta_0}^2 \\ & \leq 1 \end{split} \end{equation*} Finally: \begin{equation*} \int_0^T |\langle \delta_0 , e^{-i(t-s)H} \delta_0 \rangle |^2 ds \leq T \end{equation*}