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$.
 A: 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.
Now we want to reduce the general case to the toy case. For this we just consider the function along rays (this is inspired by the notion of weakly analytic functions, which is quite useful when one works with analytic functions in general Banach spaces).
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.
