$$
\left{\begin{aligned}&\mathop{\text{min }}\limits_{\omega,b} \frac{1}{2}\omega^{T}\omega\&s.t.y_{i}(\omega^{T}x_{i}+b)\geq 1\Leftrightarrow 1-y_{i}(\omega^{T}x_{i}+b)\leq 0,\underbrace{i=1,2,\cdots,N}_{N个约束}\end{aligned}\right.
$$
构建拉格朗日函数
$$
L(\omega,b,\lambda)=\frac{1}{2}\omega^{T}\omega+\sum\limits_{i=1}^{N}\lambda_{i}[1-y_{i}(\omega^{T}x_{i}+b)]
$$
注意这里$L$括号里面的$\lambda$为$N \times 1$，等号右边的$\lambda_{i}$为$1 \times 1$
拉格朗日乘子法具体后面文章会解释
例如本题，如果$1-y_{i}(\omega^{T}x_{i}+b)>0$，
$$
\mathop{\text{max }}\limits_{\lambda}L(\lambda,\omega,b)=\frac{1}{2}\omega^{T}\omega+ \infty=\infty
$$
如果$1-y_{i}(\omega^{T}x_{i}+b)\leq 0$
$$
\mathop{\text{max }}\limits_{\lambda}L(\lambda,\omega,b)=\frac{1}{2}\omega^{T}\omega+0=\frac{1}{2}\omega^{T}\omega
$$
因此有
$$
\mathop{\text{min }}\limits_{\omega,b}\mathop{\text{max }}\limits_{\lambda}L(\lambda,\omega,b)=\mathop{\text{min }}\limits_{\omega,b}(\infty, \frac{1}{2}\omega^{T}\omega)=\mathop{\text{min }}\limits_{\omega,b} \frac{1}{2}\omega^{T}\omega
$$
因此该问题的无约束形式为
$$
\mathop{\text{min }}\limits_{\omega,b}\mathop{\text{max }}\limits_{\lambda}L(\omega,b,\lambda),s.t.\lambda_{i}\geq 0
$$
这里的有无约束指的是对$\omega$的约束（这里的$\omega$相当于模板中的$x$）。本来$1\Leftrightarrow 1-y_{i}(\omega^{T}x_{i}+b)\leq 0$是对$\omega$的约束，通过拉格朗日函数将约束条件转化为$\lambda_{i}\geq 0$，是对$\lambda_{i}$的约束，不再是对$\omega$的约束，因此称为无约束形式
由于不等式约束是仿射函数，对偶问题和原问题等价，因此该问题的对偶形式为
$$
\mathop{\text{max }}\limits_{\lambda}\mathop{\text{min }}\limits_{\omega,b}L(\omega,b,\lambda)s.t.\lambda_{i}\geq 0
$$
先看$\mathop{\text{min }}\limits_{\omega,b}L(\omega,b,\lambda)$，对于$b$
$$
\begin{aligned}
\frac{\partial L}{\partial b}&=\frac{\partial }{\partial b}\left[\sum\limits_{i=1}^{N}\lambda_{i}-\sum\limits_{i=1}^{N}\lambda_{i}y_{i}(\omega^{T}x_{i}+b)\right]\
&=\frac{\partial }{\partial b}\left(-\sum\limits_{i=1}^{N}\lambda_{i}y_{i}b\right)\
&=-\sum\limits_{i=1}^{N}\lambda_{i}y_{i}=0
\end{aligned}
$$
将其代入$L(\omega,b,\lambda)$
$$
\begin{aligned}
L(\omega,b,\lambda)&=\frac{1}{2}\omega^{T}\omega+\sum\limits_{i=1}^{N}\lambda_{i}-\sum\limits_{i=1}^{N}\lambda_{i}y_{i}(\omega^{T}x_{i}+b)\
&=\frac{1}{2}\omega^{T}\omega+\sum\limits_{i=1}^{N}\lambda_{i}-\sum\limits_{i=1}^{N}\lambda_{i}y_{i}\omega^{T}x_{i}+\sum\limits_{i=1}^{N}\lambda_{i}y_{i}b\
&=\frac{1}{2}\omega^{T}\omega+\sum\limits_{i=1}^{N}\lambda_{i}-\sum\limits_{i=1}^{N}\lambda_{i}y_{i}\omega^{T}x_{i}
\end{aligned}
$$
对于$\omega$
$$
\begin{aligned}
\frac{\partial L}{\partial \omega}&=\frac{1}{2} \cdot 2\omega- \sum\limits_{i=1}^{N}\lambda_{i}y_{i}x_{i}=0\
\omega&=\sum\limits_{i=1}^{N}\lambda_{i}y_{i}x_{i}
\end{aligned}
$$
将其代入$L(\omega,b,\lambda)$
$$
\begin{aligned}
L(\omega,b,\lambda)&=\frac{1}{2}\underbrace{\left(\sum\limits_{i=1}^{N}\lambda_{i}y_{i}x_{i}\right)^{T}\left(\sum\limits_{j=1}^{N}\lambda_{j}y_{j}x_{j}\right)}{\in \mathbb{R}}-\underbrace{\sum\limits{i=1}^{N}\lambda_{i}y_{i}\left(\sum\limits_{j=1}^{N}\lambda_{j}y_{j}x_{j}\right)^{T}x_{i}}{\in \mathbb{R}}+\sum\limits{i=1}^{N}\lambda_{i}\
&=- \frac{1}{2}\sum\limits_{i=1}^{N}\sum\limits_{j=1}^{N}\lambda_{i}\lambda_{j}y_{i}y_{j}x_{i}^{T}x_{j}+\sum\limits_{i=1}^{N}\lambda_{i}
\end{aligned}
$$
因此原问题转化为
$$
\mathop{\text{max }}\limits_{\lambda}- \frac{1}{2}\sum\limits_{i=1}^{N}\sum\limits_{j=1}^{N}\lambda_{i}\lambda_{j}y_{i}y_{j}x_{i}x_{j}+\sum\limits_{i=1}^{N}\lambda_{i},s.t.\lambda_{i}\geq 0,\sum\limits_{i=1}^{N}\lambda_{i}y_{i}=0
$$
定义该优化问题的KKT条件（由于原、对偶问题具有强对偶关系$\Leftrightarrow$满足KKT条件）
$$
\left{\begin{aligned}&\frac{\partial L}{\partial \omega}=0,\frac{\partial L}{\partial b}=0\&\lambda_{i}[1-y_{i}(\omega^{T}x_{i}+b)]=0\&\lambda_{i}\geq 0\&1-y_{i}(\omega^{T}x_{i}+b)=0\end{aligned}\right.
$$
其中$\lambda_{i}[1-y_{i}(\omega^{T}x_{i}+b)]=0$叫做互补松弛条件，因为对于支持向量$y_{i}(\omega^{T}x_{i}+b)= 1$，对于其他数据点$\lambda_{i}=0$（根据拉格朗日函数的定义），即二者中一定至少有一个为$0$
在之前的$\begin{aligned} \frac{\partial L}{\partial \omega}\end{aligned}$中我们可以得到
$$
\omega ^{*}=\sum\limits_{i=1}^{N}\lambda_{i}y_{i}x_{i}
$$
其中$\lambda_{i}$通过求解$\begin{aligned} \mathop{\text{max }}\limits_{\lambda}- \frac{1}{2}\sum\limits_{i=1}^{N}\sum\limits_{j=1}^{N}\lambda_{i}\lambda_{j}y_{i}y_{j}x_{i}x_{j}+\sum\limits_{i=1}^{N}\lambda_{i},s.t.\lambda_{i}\geq 0,\sum\limits_{i=1}^{N}\lambda_{i}y_{i}=0\end{aligned}$可以得到
有看到怎么求，但是还是看不懂，反正就是能得到$\lambda_{i}$就对了
对于$b ^{*}$，我们假设
$$
\exists (x_{k},y_{k}),s.t.1-y_{k}(\omega^{T}x_{k}+b)=0
$$
显然$(x_{k},y_{k})$，就是所谓的支持向量（在最开始$\begin{aligned} \omega^{T}=\frac{\hat{\omega}^{T}}{a},b=\frac{\hat{b}}{a}\end{aligned}$那里设的）。因此
$$
\begin{aligned}
y_{k}(\omega^{T}x_{k}+b)&=1\
y_{k}^{2}(\omega^{T}x_{k}+b)&=y_{k}\
&y_{k}\in \left{+1,-1\right},y_{k}^{2}=1\
b ^{*}&=y_{k}-\omega^{T}x_{k}=y_{k}-\sum\limits_{i=1}^{N}\lambda_{i}y_{i}x_{i}^{T}x_{k}
\end{aligned}
$$
这里的$(x_{k},y_{k})$就是$\lambda_{i}$对应的向量，至于$\lambda_{i}$怎么求，还是那句话我不会，非常抱歉