I asked this question on math.stack but I got no answer, so I try here.


Let $\phi(t)$ be a solution for the nonlinear Schroedinger equation\begin{equation}
i\partial\_t\phi(t)=-\Delta\phi(t)+(V\*|\phi|^2)\phi(t)
\end{equation}
inside the Hilbert space $L^2(\mathbb{R}^d)$. I know $\phi(t)$ is also supposed to live in $L^\infty\cap L^2$, however I'm not sure this is really important for the problem. Let's define the projector onto $\phi(t)$ as
\begin{equation}
p(t):=\left|\phi(t)\right\rangle\left\langle\phi(t)\right|.
\end{equation}
On the paper <http://arxiv.org/abs/0907.4313> I found a formula stating, without proof, that
\begin{equation}
\|\nabla p(t)\|=\|\nabla\phi(t)\|.
\end{equation}
I was trying to figure out how to prove this, but I got really confused. I think by definition and integration by part I can write
\begin{equation}
\|\nabla\phi\|^2=\sum\_{i=1}^d\|\nabla^i\phi(t)\|^2=\int dx^d\,\phi\_t(x)(-\Delta)\phi\_t(x),
\end{equation}
even though I'm not completely sure I'm using the correct formula for a vector-valued $L^2$ function.
But I really can't figure out how to write the norm of the vector of operators $\|\nabla p(t)\|$ to try to compare them. Could anybody help me?



Norm of derivative of rank one projector
I haven't checked the paper, but $\nabla p(t)$ most likely mean the operator $\psi \mapsto \nabla \phi \cdot \langle \phi | \psi\rangle$ and since it looks like by assumption $\phi$ has $L^2$ norm one (since you are using it do define a projector), the equality follows. 




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To be more detailed, $\|\nabla p(t) \psi\|^2 = \|\nabla\phi\|^2 \langle \phi | \psi\rangle|^2 \leq \|\nabla \phi\|^2 \|\phi\|^2 \|\psi\|^2$ so the operator norm is bounded above by $\|\phi\|^2 \|\nabla\phi\|^2 = \|\nabla\phi\|^2$. To show that this is attained just plug in $\phi = \psi$. 


