Suppose $(X, \mathcal M, \mu)$ is a fixed measure space. For $f \in L^2(X)$, i.e., $f : X \to \mathbb C$ is measurable and $\int |f(x)|^2 d\mu(x) < \infty$, we define a functional $\phi : L^2(X) \to \mathbb R$ by \begin{align*} f \mapsto \int |f|^2 d \mu = \int f^* fd\mu. \end{align*} I am wondering whether or not $\phi$ is Frechet differentiable with respect to the norm topology on $L^2(X)$.
If we consider some small $g \in L^2(X)$, then \begin{align*} \phi(f+g) &= \int (f+g)^*(f+g) d \mu = \int |f|^2 d \mu + \int (f^*g+g^* f)d\mu + \int |g|^2 d \mu \\ &= \phi(f) + \int Re(f^* g) d \mu + \int|g|^2 d\mu. \end{align*} It seems that the middle term defines a bounded linear functional by $g \mapsto \int Re(f^* g) d \mu$ but I am not sure this makes sense.
Suppose $A$ is a set of measure one.
Pick $z \in \mathbb{C}$ and define the function $z 1_A$. Note that $\phi(z 1_A) = |z|^2$ and define $f: \mathbb{C} \to \mathbb{C}$ by $f(z) = \phi(z 1_A)$. Note that $f$ is not differentiable. Hence $\phi$ is not differentiable.