Optimization equivalence Can someone help me with the step by step demonstration of the following equivalence used in SVM:
$$maximize: m = \frac{1} {\|w\|} \equiv minimize: m =\frac{1} {2}\|w\|^2 $$
I would be most grateful if related theorems are included or mentioned. 
 A: The argument goes like this. 
1) the value that maximizes $1/||w||$ is the value that minimizes $||w||$
2) the value that minimizes $||w||$ also minimizes $||w||^2$, 
3) which also minimizes $1/2 ||w||^2$
A: I presume that you are referring to the maximising/minimising values over some range $w \in \mathscr{W}$.  The reason this equivalence holds is that you are using a strictly decreasing transformation.  You therefore have:
$$\begin{aligned}
\underset{w \in \mathscr{W}}{\text{arg max }} \frac{1}{||w||}
&= \Big\{ w \in \mathscr{W} \Big| (\forall w'): \frac{1}{||w||} \geqslant \frac{1}{||w'||} \Big\} \\[6pt]
&= \Big\{ w \in \mathscr{W} \Big| (\forall w'): ||w'|| \geqslant ||w|| \Big\} \\[6pt]
&= \Big\{ w \in \mathscr{W} \Big| (\forall w'): \frac{1}{2} ||w'||^2 \geqslant  \frac{1}{2} ||w||^2 \Big\} \\[6pt]
&= \Big\{ w \in \mathscr{W} \Big| (\forall w'): \frac{1}{2} ||w||^2 \leqslant  \frac{1}{2} ||wl||^2 \Big\} \\[6pt]
&= \underset{w \in \mathscr{W}}{\text{arg min }} \frac{1}{2} ||w||^2. \\[6pt]
\end{aligned}$$
