# Derivative/Gradient of log $l_1$ norm

As derivative of $l_{p}$- norm is \begin{align*} \frac{\partial}{\partial\mathbf{x}}{||\mathbf{x}||}_{p} &= \frac{\mathbf{x} |\mathbf{x}|^{p-2}}{{||\mathbf{x}||}_{p}^{p-1}} \end{align*}

I want to find $\nabla log(||H||_{1})$, where $H$ is positive matrix. So, the chain rule is \begin{align*} \nabla log(||H||_{1}) &= log(||H||_{1})' (||H||_{1})'\\ &= \frac{1}{||H||_{1}}\frac{H |H|^{1-2}}{{||H||}_{1}^{1-1}}\\ &= \frac{1}{||H||_{1}}\\ \end{align*}

But the answer seems to be

\begin{align*} \nabla log(||H||_{1}) &= &= \frac{H}{||H||_{1}}\\ \end{align*}

What am I missing here? Where is $H$ at numerator from?

• "Seems to be" answer can't be right even for reasons of homogeneity. The logarithm is essentially 0-degree homogeneous (up to an additive constant) so its derivative is homogeneous of degree $-1$. – user357151 Jun 27 '18 at 13:26

Apply the sign function element-wise to the matrix $H$ $$S = {\rm sign}(H)$$ Use this to write the Manhattan norm as \eqalign{ \|H\|_1 &= S:H \cr } where the colon denotes the trace/Frobenius product, i.e. $\,\,A:B={\rm tr}(A^TB).$
\eqalign{ \Omega &= \log(\|H\|_1) \cr d\Omega &= \frac{d\|H\|_1}{\|H\|_1} = \frac{S:dH}{\|H\|_1} \cr \frac{\partial\Omega}{\partial H} &= \frac{S}{\|H\|_1} \cr\cr } If all elements of $H>0\,$ then the numerator is the matrix $S=1$ (as you expected).
In any case, the numerator is definitely not $H$.
• If the log function is $\nabla log(||H||_{1} + \epsilon)$ (where $\epsilon$ is a small positive number), I think the answer does not change? – Jan Jun 28 '18 at 4:50
• @Jan The gradient will change (the "smallness" of $\epsilon$ is irrelevant) becoming $$\frac{\partial\Omega}{\partial H} = \frac{S}{\|H\|_1+\epsilon}$$ – greg Jun 28 '18 at 16:10