Derivative of L1 norm of Hadamard product I am trying to find the derivative of  $f(B)=\lambda\Vert W \bigodot B \Vert_1 + \frac{\rho}{2}\Vert A-B \Vert_F^2 + tr(\Delta^T(A-B))$ with respect to B.
where B is (n×n）matrix, W is (n×n）constant matrix, A is (n×n）constant matrix. $\lambda$ and $\rho$ are scalars. $tr$ is the trace of the matrix. $W \bigodot B$ is the Hadamard product of W and B.
I am troubled in finding the derivative involving Hadamard product and L-1 norm. Therefore, I first replaced $W \bigodot B$ with T.
$$T=W \bigodot B$$
$$B=W^{-1} \bigodot T$$
where $W^{-1}$ is the element-wise inverse. $W \bigodot W^{-1}=I$.
$$f(T)=\lambda\Vert T \Vert_1 + \frac{\rho}{2}\Vert A-W^{-1} \bigodot T  \Vert_F^2 + tr(\Delta^T(A-W^{-1} \bigodot T))$$
I do not konw what to do next.
Thank you in advance for any help you can provide.
 A: Let's use a colon to denote the trace/Frobenius product, 
i.e. $\;A:B = {\rm Tr}(A^TB)$
The cyclic property of the trace allows such products to be rearranged in many different ways, e.g.
$$\eqalign{
A:B &= A^T:B^T &= B:A \\
A:BC &= AC^T:B &= B^TA:C \\
}$$
Together, the Frobenius and Hadamard products form a scalar/triple product, whose terms commute.
$$A:(B\odot C) = (A\odot B):C$$
Using these products and the element-wise sign function, we can calculate the differential and subgradient of the troublesome term.
$$\eqalign{
\phi &= \lambda \|T\|_1 \\
 &= \lambda\operatorname{sign}(T):T \\
d\phi
 &= \lambda\operatorname{sign}(T):dT \\
 &= \lambda\operatorname{sign}(T):W\odot dB \\
 &= \lambda W\odot\operatorname{sign}(T):dB \\
 &= \lambda W\odot\operatorname{sign}(W)\odot\operatorname{sign}(B):dB \\
\frac{\partial\phi}{\partial B}
 &= \lambda W\odot\operatorname{sign}(W)\odot\operatorname{sign}(B) \\
}$$
Introducing an element-wise absolute value function, 
we can write this as 
$$\eqalign{
\frac{\partial\phi}{\partial B}
  &= \lambda\operatorname{abs}(W)\odot\operatorname{sign}(B) \\
}$$
