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I have the squared modulus of an integral defined over all real axis, that is,

$$P(y)=\left|\int_{-\infty}^{+\infty} \psi^{\ast}(x) \psi(x-y) f_{G}(x) ~dx \right|^2 \tag{1}$$

where $f_{G}(x)= (C/2\pi)^{1/4} e^{-C x^2/4}$ with $C \in \mathbb{R}$; then, $f_{G}(x)$ is a normalized function since $\int_{-\infty}^{+\infty}\left| f_{G}(x)\right|^{2} dx =1$. Besides $\psi(x)$ is a complex valued function which is also normalized: $\int_{-\infty}^{+\infty}\left| \psi(x)\right|^{2} dx =1$ and $\psi(x-y)$ is $\psi(x)$ displaced by a $y$-parameter. Then, my question is: there is some hint to find the maximum value for the expression of Eq. (1)?

I am trying to use the inequality:

$$\left|\int_{a}^{b} f(t) dt\right|^{2} \leq \int_{a}^{b} \left|f(t) \right|^{2} dt\tag{A}$$

In the case of Eq. (1), the closed interval $\left[a, b \right]$ is all integration space. Then, if it is valid to use Eq. (A), we can use Eq. (1) to write

$$\text{max}\left[ \left|\int_{-\infty}^{+\infty} \psi^{\ast}(x) \psi(x-y) f_{G}(x) ~dx \right|^2\right]=\text{max}\left[ \int_{-\infty}^{+\infty} \left|\psi(x) \psi(x-y) f_{G}(x) ~\right|^2 dx \right] \tag{B}$$

where we use $\left|\psi^{\ast}(x) \psi(x-y) f_{G}(x) ~\right|^2=\left|\psi(x) \psi(x-y) f_{G}(x) ~\right|^2$; then, with Eqs. (A) and (B), we can write the maximum of Eq. (1) as

$$\text{max}\left[ P(y)\right]=\text{max}\left[ \int_{-\infty}^{+\infty} \left|\psi(x) \psi(x-y) f_{G}(x) ~\right|^2 dx \right] \tag{2}.$$

Then, it is correct my procedure?, There is a way to write a more simplified version of Eq. (2)

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  • $\begingroup$ This will give you an upper bound, but will not in general give the maximum. Also, to be clear, are we treating $P$ as a function of $y$ and finding its maximum? $\endgroup$ Commented Mar 8, 2023 at 21:33
  • $\begingroup$ Yes, $P$ is a function of $y$. $\endgroup$ Commented Mar 8, 2023 at 22:15

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