How is the Representer theorem used in the derivation of the SVM dual form?

Question

This is the primal form of the SVM hypothesis : $$ h _{\mathbf{\vec w}, b}(\mathbf{\vec x}^{(i)}) = \mathbf{\vec w}\cdot \mathbf{\vec x}^{(i)} + b $$ The Representer theorem as formulated here expresses the optimal weight vector as a linear combination of the training examples : $$ \mathbf{\vec w}^{*} = \sum_{i=1}^{m}\mu_{i}^{*}y ^{(i)}\mathbf{\vec x}^{(i)} $$ This can be derived using the Lagrange multiplier method.
Now, the dual form of the SVM hypothesis is given as : $$ h _{\boldsymbol{\vec \mu}, b} = \sum_{i=1}^{m}\mu_{j}^{*}y ^{(i)} \langle\mathbf{\vec x}^{(j)}, \mathbf{\vec x}^{(i)}\rangle +\, b ^{*} $$ Here, we have clearly substituted the weight vector $\mathbf{\vec w}$ using the Representer theorem.

However, the weight vector in the original SVM hypothesis is not the optimal weight vector. So, why can we substitute the weight vector in the primal form using the Representer theorem?

Answer 1

The representer theorem isn't needed for SVM. The solution form comes out on its own just by solving for the dual (i.e., introducing variables for the constraint, and through the KKT conditions).

Answer 2

The optimal weight vector $\mathbf{\vec w}^*$ is for the training dataset. We are essentially assuming that we have found the optimal weight vector for the training dataset, then substituting it using the Representer Theorem and then finding the optimal values of the Lagrange multipliers.

This assumption holds true since the optimal values of the Lagrange multipliers always yield the optimal weight vector. This way we only have to find the optimal values of the Lagrange multipliers, allowing us to use the dual form and kernel functions.