For some let:

If is defined and Convex Function,
then there is a with:

where is the Lagrangian
Furthermore, if is differentiable,
then can be taken to be

Hence if is convex,
we are sure to find in Lagrange sufficiency theorem.

Proof

Suppose is a Convex Function.
Then find Supporting hyperplane of at and write:

\phi(b)&=\inf_{c}(\phi(c)+\lambda^T(b-c))\\ &=\inf_c(\inf_{x\in X,\ g(x)=c}(f(x)+\lambda^T(b-c)))\\ &=\inf_c(\inf_{x\in X,\ g(x)=c}(f(x)+\lambda^T(b-g(x))))\\ &=\inf_{x\in X}(f(x)+\lambda^T(b-g(x)))\\ &=\inf_{x\in X}L(x,\lambda) \end{align*}$$ ## Helpful fact Suppose $X$ is [[Algebra/Vector Spaces/Convex Set]], $f$ is [[Analysis/Convex Function]], and all functional constraints are convex. Then $\phi$ is convex.