Why is minimizing the norm of an vector ($|w|$) the same as minimizing its sum of squares ($0.5 |w|^2$)? Like the title says, im not sure why some popular machine learning textbooks say this. Is there something i am missing. I can elaborate if needed.
Similar to this question: Minimize the norm of $w$.
But i was not able to follow the answer.
 A: In general, if $f$ is a strictly increasing function then minimizing $f(g(x))$ is equivalent to minimizing $g(x)$. In your case, set $f: \mathbb{R}_{\geq 0} \rightarrow \mathbb{R}$ so that $f(x) = .5x^2$ and set $g: V \rightarrow \mathbb{R}$ so that $g(x) = |x|$. Then $f(x)$ is strictly increasing, so minimizing $|x|$ is equivalent to minimizing $.5|x|^2$.
Intuitively, just imagine if you're playing a game where there is some dollar payout for getting a high score, and the higher your score the more the payout.  No matter what the relationship is between the score and the payout, the strategy to make the most money is to get the highest score.
More formally:
Lemma. Let $g: A \rightarrow B$ be a function, with $A$ any set and $B \subseteq \mathbb{R}$ and let $f: B \rightarrow \mathbb{R}$ be an increasing function, so that $x \leq y \implies f(x) \leq f(y)$. Then, if $x_0 \in A$ is such that $g(x_0)$ minimizes $g(x)$ then $f(g(x_0))$ minimizes $f(g(x))$.
Proof: Suppose $g(x_0)$ is a minimum.  Then for any $x$, $g(x_0) \leq g(x)$, and so since $f$ is increasing, $f(g(x_0)) \leq f(g(x))$.  $\square$
Note also that if $f$ is strictly increasing, with $x < y\implies f(x) < f(y)$, then $f$ "preserves uniqueness" in the sense that if $x_0$ is the unique minimizer of $g$ then $x_0$ is also the unique minimizer of $f\circ g$. [exercise]
