Is there a Fréchet Derivative for the Norm on a Complex Hilbert Space?

Let $$\mathcal H$$ be a complex Hilbert space and consider the function $$\langle\cdot,\cdot \rangle = \|\cdot\|^2:\mathcal H \to \mathbb R \subset \mathbb C$$, which is the square of the norm induced by the inner product $$\langle\cdot,\cdot \rangle$$. Is there a Fréchet derivative of this function ?

My Attempt

The Fréchet derivative of a function $$f:V \to W$$ at a point $$v \in V$$ is a linear map $$Df(v):V \to W$$. In our case we have $$V = \mathcal H$$ and $$W = \mathbb C$$. Assuming that this linear map is bounded, then $$Df(v) \in \mathcal H^*$$, i.e. in the topological dual of $$\mathcal H$$, and by the Reisz representation Theorem there is $$u \in \mathcal H$$ (that depends on $$v$$) such that $$Df(v)(x) = \langle u,x \rangle,\ \forall x \in \mathcal H$$. The task reduces to finding the vector $$u \in \mathcal H$$ such that the limit (in the definition of the Fréchet derivative) $$\lim_{h\to 0} \frac{| \|v+h\|^2 - \|v\|^2 - \langle u,h \rangle |}{\|h\|}$$ vanishes. Now by reforming the expresseion in the numerator I get \begin{align} \lim_{h\to 0} \frac{| \|v+h\|^2 - \|v\|^2 - \langle u,h \rangle |}{\|h\|} &= \lim_{h\to 0} \frac{| \langle v+h,v+h \rangle - \langle v,v \rangle - \langle u,h \rangle |}{\|h\|} \\ &= \lim_{h\to 0} \frac{| \langle h,v \rangle + \langle v,h \rangle - \langle u,h \rangle |}{\|h\|} \end{align} In the last line I used $$\lim_{h\to 0} \frac{\langle h,h \rangle}{\|h\|} = \lim_{h\to 0} \frac{\|h\|^2}{\|h\|} = \lim_{h\to 0} \|h\| =0$$ and that is as far as I could go with the complex case, because $$\langle x, \cdot\rangle + \langle \cdot ,x\rangle$$ is not $$\mathbb C$$-linear for $$\langle x, \alpha h\rangle+\langle \alpha h ,x\rangle = \alpha \langle x, h\rangle + \bar \alpha \langle h ,x\rangle \neq \alpha (\langle x, h\rangle + \langle h ,x\rangle)$$ and so there is no $$u \in \mathcal H$$ such that $$D(\|v\|^2)(\cdot) = \langle u,\cdot\rangle$$

For the real case we have instead a real Hilbert space with real inner it is straightforward to find $$u \in \mathcal H$$ such that the limit vanishes. Indeed, since $$\langle v,h \rangle = \langle h,v \rangle$$, then \begin{align} \lim_{h\to 0} \frac{| \langle h,v \rangle + \langle v,h \rangle - \langle u,h \rangle |}{\|h\|} &= \lim_{h\to 0} \frac{| 2\langle v,h \rangle - \langle u,h \rangle |}{\|h\|} \\ &= \lim_{h\to 0} \frac{| \langle 2v - u,h \rangle |}{\|h\|} \end{align} and it is clear that $$u = 2v$$ serves for that, and so $$D(\|v\|^2)(\cdot) = 2\langle v,\cdot \rangle$$.

No, there is in general no Fréchet derivative (unless $$\mathcal{H}$$ has dimension $$0$$).
We will first consider the toy case $$\mathcal{H}=\mathbb{C}$$. Then for the standard scalar product we are asking whether the map $$f:\mathbb{C} \rightarrow \mathbb{C}, \ z \mapsto \vert z \vert^2$$ is complex differentable (that is what it means for the Fréchet derivative to exist). The open mapping theorem from complex analysis tells us that a nonconstant holomorphic (aka complex differentiable map) maps open sets to open sets. Thus, if $$f$$ was complex differentiable, then $$f(\mathbb{C})=\mathbb{R}$$ was open in $$\mathbb{C}$$ which is a contradiction.
Assume that $$g: \mathcal{H} \rightarrow \mathbb{C}, g(v)=\Vert v\Vert^2$$ is complex differentiable. Fix some $$v\in \mathcal{H} \setminus \{ 0\}$$ (here we use that $$\mathcal{H}\neq \{0\}$$). Then $$h:\mathbb{C}\rightarrow \mathbb{R}, z\mapsto \Vert z v\Vert^2=\vert z\vert^2 \Vert v\Vert^2$$ is complex differentiable by the chain rule (the map $$\mathbb{C}\rightarrow \mathcal{H}, z\rightarrow zv$$ is complex differentiable and hence $$h(z)=g(zv)$$ is too). However, now we are exactly in the toy case again and we get the desired contradiction.
We can directly compute that the Fréchet derivative at the origin exists. We have for $$A: \mathcal{H} \rightarrow \mathbb{C}, A(v)=0$$ $$\frac{\Vert h + 0 \Vert^2 - \Vert 0 \Vert^2 - A(h)}{\Vert h \Vert} = \Vert h \Vert \rightarrow 0$$ for $$\Vert h \Vert \rightarrow 0$$. Thus, the Fréchet derivative is simply the zero map.
If you prefer, you can also translate this into picking $$v\in \mathcal{H}\setminus \{ 0\}$$ and considering the complex Gateaux derivative of $$g$$ at $$v$$ in direction of $$v$$ and to see that this does not exist. Namely we would need to consider ($$z\in \mathbb{C}$$) $$\frac{\Vert v + zv \Vert^2 - \Vert v \Vert^2}{z} = \frac{(\vert 1+z\vert^2-1)}{z} \Vert v \Vert^2.$$ However, the limit does not exist for $$z\rightarrow 0$$. To see this, we simply compute (now $$\varepsilon \in \mathbb{R}$$) the limit twice, first parallel to the real axis and then parallel to the imaginary axis. $$\lim_{\varepsilon \rightarrow 0} \frac{(\vert 1+\varepsilon\vert^2-1)}{\varepsilon} \Vert v \Vert^2 = 2 \Vert v \Vert^2.$$ On the other hand, $$\lim_{\varepsilon \rightarrow 0} \frac{(\vert 1+i\varepsilon\vert^2-1)}{i\varepsilon} \Vert v \Vert^2 = \lim_{\varepsilon \rightarrow 0} \frac{\varepsilon^2}{i\varepsilon} \Vert v \Vert^2 =0.$$ This gives us the even stronger result that $$g$$ is not even complex Gateaux differentiable away from the origin and so surely not Fréchet differentiable away from the origin.