Derivative of norm of difference between two vectors I want to project a vector $\tilde{x}$ onto a hyperplane,
which leads to the following optimization problem:
$\min_x \frac{1}{2} ||x - \tilde{x}|| \quad s.t. \quad w^{T} x + b=0$
Using Lagrangians I can write:
$\mathcal{L}(x,\lambda) =\frac{1}{2}  ||x - \tilde{x}|| + \lambda (w^{T} x + b)$
So to solve the problem I have to calculate the derivative of $\mathcal{L}(x,\lambda)$ w.r.t. to x.
However, I'm not quite sure how to correctly derive $||x - \tilde{x}||$ w.r.t. $x$.
Is it valid to rewrite the problem as following, without changing the result,
does this make the derivation easier?
$\mathcal{L}(x,\lambda) =\frac{1}{2}  ||x - \tilde{x}||^2 + \lambda (w^{T} x + b)$
At some places I saw the equivalience:
$||x - \tilde{x}||^2 = ||x||^2 + ||y||^2 + 2xy $
But I'm not sure if this is correct, 
neither it's obvious to me why it should be.
I'm especially confused because there is no p for the norm given, is there
any convention to just assume $p=2$ or any other arbitrary number?
 A: Define the variable $z=(x-\bar{x})$ and write the objective function as
$$\eqalign{
\phi &= \|x-\bar{x}\| = \|z\| \cr
}$$
Now write the general solution to the linear constraint as the least-squares solution plus an arbitrary contribution from the null space 
$$\eqalign{
 w^Tx &= -b \cr
 x &= (I-ww^+)y - (w^+)^Tb \cr
}$$
where $w^+$ denotes the pseudoinverse and $y$ is an arbitrary vector.
Note that $P=(I-ww^+)$ is an orthoprojector (i.e. $P^2=P=P^T$) into the nullspace of $w$. Therefore 
$$\eqalign{
 P(w^+)^T &= 0 \cr
 Px &= P^2y - P(w^+)^Tb \,\,= Py \cr
}$$
Substitute this into the objective function to obtain an unconstrained problem with respect to $y$.
$$\eqalign{
 \phi^2 &= \|z\|^2 = z:z \cr
 \phi\,d\phi &= z:dz \cr
 d\phi&= \phi^{-1}z:dx \cr
 &= \phi^{-1}z:P\,dy \cr
 &= \phi^{-1}Pz:dy \cr
\frac{\partial\phi}{\partial y} &= \phi^{-1}Pz \cr
}$$
Set the gradient to zero and solve 
$$\eqalign{
 Pz &= 0 \implies Px &= P\bar{x} \implies Py &= P\bar{x} \cr
}$$
Substitute this result into the parametric expression for $x$
$$\eqalign{
 x &= Py - (w^+)^Tb \cr
   &= P\bar{x} - (w^+)^Tb \cr
   &= (I-ww^+)\bar{x} - (w^+)^Tb \cr\cr
}$$
For vectors, there is a closed-form expression for the pseudoinverse
$$w^+ = \frac{w^T}{w^Tw}$$
In some of the intermediate steps above, a colon was used to denote the trace/Frobenius product 
$$A:B = {\rm tr}(A^TB)$$
A: In general, when projecting, we use the 2 norm. You're basically looking at a particular proximal operator. It is fine to use the square of the norm rather than just the norm, it doesn't make a difference as far as the projection is concerned. You can derive the formula for taking the gradient of the squared 2-norm explicitly component wise.
