Derivative on Hilbert space Please, on a Hilbert space what is the  derivative of  $\displaystyle\frac{x}{||x||}$ ?
I know that it's equal to $\displaystyle \frac{1}{||x||}-\frac{\langle x,\cdot\rangle}{||x||^3} x$ but can I write it as $\displaystyle\frac{1}{||x||}-\frac{\langle x,x\rangle}{||x||^3}$?
thank you 
 A: For each $x_0$, the derivative of $x\mapsto\displaystyle\frac{x}{||x||}$ in the point $x_0$ is the linear map
$$h\mapsto\displaystyle\frac{1}{||x_0||}h-\frac{\langle x_0,h\rangle}{||x_0||^3}x_0.$$
A: You want a linear approximation to the function $F(x)=\frac{1}{\|x\|}x$. The usual notation for a linear operator $L$ acting on a vector $y$ is $Ly$. And $F'(x)$ is a linear map from $H$ to $H$, where $H$ is the underlying Hilbert space. So $F'(x)y$ means the linear operator $F'(x)$ applied to the vector $y$.
I assume you're working with a real Hilbert space. Then $N(x)=(x,x)$ is a function from $H$ to $\mathbb{R}$, and has derivative $N'(x)y=2(x,y)$. Therefore the function $Q : X\rightarrow \mathbb{R}$ defined as
$$
                 Q(x) = \frac{1}{N(x)^{1/2}}
$$
has derivative given by the chain rule:
$$
                   Q'(x)y=-\frac{1}{2}\frac{1}{N(x)^{3/2}}N'(x)y=-\frac{1}{\|x\|^{3}}(x,y).
$$
Finally, using the product rule, the derivative of the expression $F(x)=\frac{1}{\|x\|}x$ is
$$
                     F'(x)y = \frac{1}{\|x\|}y-\frac{(x,y)}{\|x\|^{3}}x.        
$$
The derivative $F'(x)$ is a linear operator at each $x$. So it cannot be written as you have done. In the original statement, you need a placeholder $(\cdot)$ multiplying $1/\|x\|$. In the second case, you have incomplete $h$ terms.
What you can write is
$$
      F'(x) = \frac{1}{\|x\|^{2}}I - P_{x}
$$
where $I$ is the identity linear operator and $P_{x}$ is the linear projection of a vector onto $x$. That is,
$$
                          P_{x}y = \left(y,\frac{1}{\|x\|}x\right)\frac{1}{\|x\|}x.
$$
