# Maximum value of a complex integral

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)

• 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? Commented Mar 8, 2023 at 21:33
• Yes, $P$ is a function of $y$. Commented Mar 8, 2023 at 22:15