What rescaling the parameters does and does not change

Write each quantity out with the $c$ in place and see whether it cancels.

The boundary. The set $\{\vec{x} : c\langle\vec{w},\vec{x}\rangle + cb = 0\}$ equals $\{\vec{x} : c(\langle\vec{w},\vec{x}\rangle + b) = 0\}$, and since $c \neq 0$ this is the same set as before. Unchanged.

The labels. $\mathrm{sign}(c\,s) = \mathrm{sign}(s)$ for $c > 0$. Unchanged.

The functional margin. $m = y(\langle\vec{w},\vec{x}\rangle + b)$ becomes

$y\big(c\langle\vec{w},\vec{x}\rangle + cb\big) = c\,m$

Scaled by $c$. It can be made as large as one likes by taking $c$ large, without moving the boundary an inch.

The geometric margin. Both numerator and denominator scale:

$\frac{y(c\langle\vec{w},\vec{x}\rangle + cb)}{\lVert c\vec{w}\rVert} = \frac{c\,m}{c\lVert\vec{w}\rVert} = \frac{m}{\lVert\vec{w}\rVert}$

Unchanged. The $c$ cancels exactly.

Which one can appear in a theorem. Only the geometric margin. A theorem whose hypothesis was “the functional margin is at least $\gamma$” would be vacuous: given any separating hyperplane, multiply its parameters by $c = \gamma / m_{\min}$ and the hypothesis is satisfied, with no change to the classifier at all. The statement would constrain nothing.

This is why I.2.T1 is stated with a unit vector $\vec{u}$, and why the proof in I.2.B03 uses $\lVert\vec{u}\rVert = 1$ in its Cauchy–Schwarz step. Normalising is not tidiness; it is what makes the quantity a property of the data rather than of an arbitrary scaling.

A consequence worth carrying forward. Any quantity you plan to threshold, compare across models, or put in a bound must be checked for this kind of spurious freedom first. The same question recurs for attention scores in II.3 — where the answer is that $\sqrt{d_k}$ removes a scale that would otherwise grow with width — and for logits in IV.8, where a temperature does exactly what $c$ does here.

One asymmetry. For $c < 0$ the boundary is still unchanged but every label flips, since $\mathrm{sign}(cs) = -\mathrm{sign}(s)$. So the invariance is to positive rescaling only, and the sign of $\vec{w}$ carries the orientation — which side is which.