Let f:U⊂ $\mathbb{R}^n$--->$\mathbb{R}$ a $C^1$ function with U being convex and an open set. Let g: U ⊂ $\mathbb{R}^n$ ---> $\mathbb{R}^m$ (with m smaller than n) an affine application. Let M={x $ \in$ U | g(x)=0 and rank(Jg(x))=m} Let $x_o$ in M and $\lambda $ in $\mathbb{R^m}$ such that $\bigtriangledown f(x_o)= \lambda . \bigtriangledown g(x_o) $. Show that if f is convex, than $x_o$ is an absolute minimum of the function f on M. (Note : the point . represents the scalar product)
I know that if f is convex, than f(y) ≥ f(x) + f'(x) (y-x) for x,y in U . Can anyone help me complete this problem? thanks.
The Lagrange multiplier condition tells you that $\nabla f(x_0) = \sum_{i=1}^m \lambda_i\nabla g_i(x_0)$ is normal (orthogonal) to the affine subspace $M$. This means that $\nabla f(x_0)\cdot (y-x_0) = 0$ for any $y\in M$. Thus, since $f$ is convex, for any $y\in M$ we have $$f(y)\ge f(x_0)+\nabla f(x_0)\cdot (y-x_0) = f(x_0).$$ This means $x_0$ is a global minimum of $f$ on $M$.