Namgyun Lee edited DeclareMathOperator_argmin_argmin_newcommand_reals__.tex  about 8 years ago

Commit id: 9ade5c963eb11d69b508e85ce72790c1b643f830

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For the second term, we apply the chain rule to the composite function $h = g \circ f : \complex^{\nv} \to \reals$  defined by $h(m) = g(f(m))$, with $g(x) = \|x\|_{\ell_1} : \complex^{\nk} \to \reals$   and $f(m) = \Fu m - y : \complex^{\nv} \to \complex^\nk$.  By definition of the $\ell_1$-norm, We we  have \begin{eqnarray} \begin{equation}  \|x\|_{\ell_1} &=& \sum_{i=1}^n |x_i|^2 = \sum_{i=1}^n x_i^*x_i \nonumber\\  &=& \sum_{i=1}^n (a_i +jb_i)^*(a_i +jb_i) \nonumber\\  &=& \sum_{i=1}^n (a_i -jb_i)(a_i +jb_i) \nonumber\\  &=& \sum_{i=1}^n (a_i^2 +b_i^2). \nonumber  \end{eqnarray}   We first look at the first element of Equation \ref{eqn:derivative} with $f(x) = \|x\|_{\ell_2}^2$. Applying the definition of the complex derivative gives  \begin{eqnarray}  \frac{\partial \|x\|_{\ell_2}^2}{\partial x_1} &=& \frac{\partial \|x\|_{\ell_2}^2}{\partial a_1} +   j\frac{\partial \|x\|_{\ell_2}^2}{\partial b_1} \nonumber\\  &=& \frac{\partial }{\partial a_1} \left(\sum_{i=1}^n (a_i^2 +b_i^2)\right) +  j\frac{\partial }{\partial b_1} \left(\sum_{i=1}^n (a_i^2 +b_i^2)\right) \nonumber\\  &=& 2a_1 + j2b_1 \nonumber\\  &=& 2x_1. |x_i|  \nonumber \end{eqnarray} \end{equation}  The derivative of $g$ at $x$ is a $1 \times n$ matrix, given by  \begin{eqnarray}  Dg(x) &=& \left[ \frac{\partial g}{\partial x_1}(x), \dots, \frac{\partial g}{\partial x_{\nk}}(x) \right] \nonumber\\