Quadratic approximation of L1 regularized cost function
The analysis becomes a little bit hairy because of the discontinuous derivative of the L1 penalty.  There may be a better way of phrasing the argument, but here's something.

Taking the derivative of $|w_i|$ gives $\operatorname{sgn}(w_i)$, unless $w_i=0$ which we will take as a special case.
In the main case, when $w_i\neq0$, taking the derivative w.r.t. $w_i$ equal to zero gives:
$$\begin{align*}
H_{ii}(w_i-w_i^*) + \alpha \operatorname{sgn}(w_i) &amp; = 0 \\
\Longrightarrow w_i &amp;= w_i^* - \frac{\alpha \operatorname{sgn}(w_i)}{H_{ii}}.
\end{align*}$$
Now for a trick.  From the original loss formulation, it's clear that the optimum $w_i$ will have the same sign as $w_i^*$ (or is zero): otherwise, switching signs decreases the first term and doesn't change the second term.  So we have
$$\begin{align*}
w_i &amp;= w_i^* - \frac{\alpha \operatorname{sgn}(w_i^*)}{H_{ii}} \\
 &amp;= \operatorname{sgn}(w_i^*) \left( |w_i^*| - \frac{\alpha}{H_{ii}} \right).
\end{align*}$$
And again, knowing that $w_i$ and $w_i^*$ must share sign, we must have that $|w_i^*|-\frac{\alpha}{H_{ii}}\geq0$.  If not, then this is not a location of zero-derivative, and we have to look to our other case.  In that case, simply $w_i=0$, and we can tack that on as a separate case here using the maximum given in the text.  (We should also confirm that $w_i=0$ is not the minimum when $|w_i^*|-\frac{\alpha}{H_{ii}}&gt;0$, which is easy to do.)