Let $\phi$ entire and such that $|\phi(0)|=1$. I am wondering if it is true that the function $\Phi$ given by $\Phi(t)=|\phi(t)|^2$ is real differentiable in some interval of the form $(p, q)$, where $p<q\in\Bbb R$ (maybe in $(0, +\infty)$ or $(1,+\infty)$?). Is that true of false?
If it is false, could you give me a counterexample please?
If it is false, could you tell me what is $\Phi'(t)$ please?
(I am looking for $\phi$ among those with "moderate growth" but I can't find a counterexample)
If $\phi$ is holomorphic on $\mathbb{C}$ (and hence, in particular, smooth on $\mathbb{R}^2$), then $\overline{\phi}$ is antiholomorphic on $\mathbb{C}$ (and hence, in particular, smooth on $\mathbb{R}^2$), so that $\Phi = \lvert \phi \rvert^2 = \phi \overline{\phi}$ is the product of two smooth functions on $\mathbb{R}^2$, and hence itself smooth on $\mathbb{R}^2$. In particular, since $$ \frac{\partial}{\partial x} = \frac{\partial}{\partial z} + \frac{\partial}{\partial \bar{z}}, \quad \frac{\partial}{\partial y} = i\left(\frac{\partial}{\partial z} - \frac{\partial}{\partial \bar{z}}\right), $$ and since $$ \frac{\partial\phi}{\partial z} = \phi^\prime, \quad \frac{\partial\phi}{\partial\bar{z}} = 0, \quad \frac{\partial\bar{\phi}}{\partial z} = \overline{\frac{\partial\phi}{\partial\bar{z}}} = 0, \quad \frac{\partial \bar{\phi}}{\partial \bar{z}} = \overline{\frac{\partial\phi}{\partial z}} = \overline{\phi^\prime}, $$ it follows that $$ \frac{\partial \phi}{\partial x} = \phi^\prime, \quad \frac{\partial\phi}{\partial y} = i\phi^\prime, \quad \frac{\partial\bar{\phi}}{\partial x} = \overline{\phi^\prime}, \quad \frac{\partial\bar{\phi}}{\partial y} = -i\overline{\phi^\prime}, $$ and hence that $$ \frac{\partial\Phi}{\partial x} = \phi^\prime\bar{\phi}+\phi\overline{\phi^\prime}, \quad \frac{\partial\Phi}{\partial y} = i\left(\phi^\prime\bar{\phi}-\phi\overline{\phi^\prime} \right), $$ where $\frac{\partial\Phi}{\partial x}$ is the derivative you want.