Proof of g in Lagrangian problem as a lower bound of true objective

We have original problem:

p∗=min⁡f0(x⃗)s.t.∀i∈[1,m],fi(x⃗)≤0∀i∈[1,p],hi(x⃗)=0p^* = \min f_0(\vec{x}) \\ \\ \text{s.t.} \\ \forall i \in [1,m], f_i(\vec{x}) \le 0 \\ \forall i \in [1,p], h_i(\vec{x}) = 0

Define the Lagrangian function:

L(x⃗,λ⃗,ν⃗)=f0(x⃗)+∑i=1mλifi(x⃗)+∑i=1pνihi(x⃗)L(\vec{x},\vec{\lambda}, \vec{\nu}) = f_0(\vec{x}) + \sum_{i=1}^m \lambda_i f_i(\vec{x}) + \sum_{i=1}^p \nu_i h_i(\vec{x})

So, the Lagrangian problem:

min⁡x⃗L(x⃗,λ⃗,ν⃗)=g(λ⃗,ν⃗)s.t. λ⃗≥0\min_{\vec{x}} L(\vec{x},\vec{\lambda},\vec{\nu}) = g(\vec{\lambda},\vec{\nu}) \\ \text{s.t. } \vec{\lambda} \ge 0

We want to prove:

∀λ⃗≥0,ν⃗→g(λ⃗,ν⃗)≤p∗\forall \vec{\lambda} \ge 0, \vec{\nu} \rightarrow g(\vec{\lambda},\vec{\nu}) \le p^*

Consider x~⃗\vec{\tilde{x}} feasible for the primal

fi(x~⃗)≤0,hi(x~⃗)=0f_i(\vec{\tilde{x}}) \le 0, h_i(\vec{\tilde{x}}) = 0

So

L(x~⃗,λ⃗,ν⃗)=f0(x~⃗)+∑λifi(x~⃗)undefined≤0+∑νihi(x~⃗)undefined=0≤f0(x~⃗)L(\vec{\tilde{x}}, \vec{\lambda}, \vec{\nu}) = f_0(\vec{\tilde{x}}) + \underbrace{\sum \lambda_i f_i(\vec{\tilde{x}})}_{\le 0} + \underbrace{\sum \nu_i h_i(\vec{\tilde{x}})}_{=0} \le f_0(\vec{\tilde{x}})

Since g(λ⃗,ν⃗)=min⁡x~⃗L(x~⃗,λ⃗,ν⃗)g(\vec{\lambda}, \vec{\nu}) = \min_{\vec{\tilde{x}}} L(\vec{\tilde{x}}, \vec{\lambda}, \vec{\nu}),

∀x⃗,g(λ⃗,ν⃗)≤f0(x⃗)→g(λ⃗,ν⃗)≤p∗\forall \vec{x}, g(\vec{\lambda}, \vec{\nu}) \le f_0(\vec{x}) \rightarrow g(\vec{\lambda}, \vec{\nu}) \le p^*