# SVM: Would we care about the functional margin if maximizing only with geometric margin were convex?

I am reading Andrew Ng's SVM notes (https://see.stanford.edu/materials/aimlcs229/cs229-notes3.pdf) and am lacking the intuition for why we need the functional margin. As far as I understand we need it as the following optimization problem expressed only in terms of the geometric margin is not convex:

$$max_{\gamma, w, b} = \gamma$$

such that $$y_{i}(w^{T}x + b) \geq \gamma$$ $$i = 1 \ldots m$$

and $$\lvert \lvert w \rvert \rvert = 1$$

The constraint $$\lvert \lvert w \rvert \rvert = 1$$ is not convex. Suppose for the sake of understanding the importance of the functional margin that $$\lvert \lvert w \rvert \rvert = 1$$ is convex and we can directly optimize this problem. Could we then entirely forget about the concept of the functional margin? In other words, do we only need the notion of a functional margin, as it allows us to rewrite the above equation as a convex optimization problem?

Generally, the single constraint $$||w||^2 = 1$$ actually has zero-duality gap.
However, a simpler explanation can help for this situation. We can actually solve instead for $$||w||^2 \leq 1$$, which is convex. To see this, note though that if $$||w||^2 < 1$$ then if I multiply $$\gamma$$ and $$w$$ by some factor $$\beta > 1$$ that I maintain the constraint, and improve the objective. Thus, given any valid solution having $$0 < ||w||^2 < 1$$, I can simply multiply through by $$\beta = \frac{1}{||w||^2}$$ to maximize the objective, for $$w$$ pointing in "that direction". Thus, the constraint $$||w||=1$$ can actually be replaced by the convex $$||w||^2 \leq 1$$.