Taylor series approximation of function under norm I am reading this paper. At page number $4$, term $||Au - f||^{2}$  is 
approximated by taylor series approximation around $u^{k}$. The resulting approximations are 
$$\|Au - f\|^{2} \approx \|Au^{k}- f\|^{2}  +  2 \langle u, A^{T}(Au^{k}- f)\rangle  +  \frac{1}{\delta} \| u - u^{k}\|^{2} $$
I am finding it difficult to figure out this step.
Background: This paper discusses the convergence of the linearized Bregman iteration for solving the following
minimization problem:
$$\min \{ \|u\|_1: Au = f\}$$
where matrix $ A \in \mathbb{R}^{m\times n}$, with $n > m$ and $f \in \mathbb{R}^{m}$. 
I am aware of the Taylor series of a real valued function $f(x)$ about a point. However, I haven't read about finding the Taylor series of normed function. I need help to understand this. Any reference regarding this theory will also be helpful. 
 A: In general, if you want to approximate $F$ to the first order around some point $u^k$, Taylor's formula says
$$F(u) = F(u^k) + \Bbb d F (u^k) (u - u^k) + \frac 1 2 \Bbb d ^2 F (v) (u - u^k, u - u^k) ,$$
with $v$ some point on the line segment of endpoints $u$ and $u^k$. As you see, there are three terms showing up; let us analyze them one by one.
The first one is clear: just replace $u$ by $u^k$ and you're done.
In the second, you have to compute $\Bbb d (Au-f) (u^k)$, the derivative in the point $u^k$ of the function
$$u \mapsto \|Au - f \|^2 = \langle Au-f, Au-f \rangle .$$
This is a (scalar) product, so derive it like a product according to Leibniz's formula (yes, it is correct!). Remember that the differential of a linear map, applied in some point, is that linear map itself, so
$$\Bbb d (Au-f) (u^k) = A ,$$
(because $\Bbb d f (u^k) = 0$, $f$ being a constant vector with respect to $u$). Next, you have to apply this $\Bbb d (Au-f) (u^k)$ to the vector $u-u^k$, therefore getting $A (u-u^k)$. Putting everything together,
$$\Bbb d (Au-f) (u^k) (u-u^k) = A(u-u^k)$$
and the second term in the above Taylor expansion becomes
$$\langle A(u-u^k), A u^k - f \rangle + \langle A u^k - f, A(u-u^k) \rangle = 2 \langle A(u-u^k), A u^k - f \rangle$$
because the scalar product is symmetric. Now remember that, in general, $\langle Au, v \rangle = \langle u, A^T v \rangle$ (in fact, this is the definition itself of the transposition operation), therefore the above becomes equal to
$$2 \langle u-u^k, A^T (A u^k - f) \rangle = 2 \langle u, A^T (A u^k - f) \rangle \color{red} {- 2 \langle u^k, A^T (A u^k - f) \rangle} .$$
The first of the above two terms is exactly the second one in your formula. The one in red will be absorbed, at the end, in $\frac 1 \delta$.
Finally, the third term can be rewritten as
$$\frac 1 2 \Bbb d ^2 F (v) \Big( \frac {u - u^k} {\| u - u^k \|}, \frac {u - u^k} {\| u - u^k \|} \Big) \cdot \| u - u^k \|^2$$
because $d ^2 F (v)$ is linear in each argument. Remember that, in general, if $G(x) = Ax : \Bbb R ^n \to \Bbb R $ is linear, and $p,u,v \in \Bbb R ^n$, then $\Bbb d ^2 G (p) (u,v) = u A v^T$ (where I take vectors to be rows, thus their transposed are columns). Since $F(u) = Au-f$ (and $f$ is constant with respect to $u$) then
$$\Bbb d ^2 (Au-f) (u^k) \Big( \frac {u - u^k} {\| u - u^k \|}, \frac {u - u^k} {\| u - u^k \|} \Big) = \frac {u - u^k} {\| u - u^k \|} \cdot A \cdot \Big( \frac {u - u^k} {\| u - u^k \|} \Big) ^T .$$
Collecting everything, Taylor's formula looks like
$$\|Au-f\|^2 = \| A u^k -f \|^2 + 2 \langle u, A^T (A u^k - f) \rangle + \Bigg( \color{blue} {\frac 1 2 \frac {u - u^k} {\| u - u^k \|} \cdot A \cdot \Big( \frac {u - u^k} {\| u - u^k \|} \Big) ^T} \color{red} {- 2 \frac {\langle u^k, A^T (A u^k - f) \rangle} {\| u - u^k \|^2} } \Bigg) \| u - u^k \|^2 .$$
Now the authors note (in the paragraph between formulae (2.3) and (2.4)) that for large enough $k$ (i.e. after sufficiently many iterations), the quantity $\| A u^k - f \|$ becomes small enough, so $A u^k - f$ becomes small enough (as a vector), so that it and the term in red that contains it may be ignored. Furthermore, for $u$ sufficiently close to $u^k$ you may approximate $\frac {u - u^k} {\| u - u^k \|}$ by some fixed vector $v_k$ (on the unit sphere, but this is not important), so take $\delta = \frac 2 {v_k A v_k ^T}$, insert this back into the above formula and you're done. Note that all the approximations made in this last paragraph have transformed the mathematically correct equality in Taylor's formula into an approximate equality, which explains why the authors switch from $=$ to $\approx$.
