proof of calculus theorem used in the Lagrange method

Let $f(𝐱)$ and $g_i(𝐱),i=0,\mathrm{\dots },m$be differentiable scalar functions; $𝐱\in R^n$.

We will find local extremes of the function $f(𝐱)$ where$\nabla f=0$. This can be proved by contradiction:

$$\nabla f\ne 0$$

but then $f(𝐱)$ is not a local extreme.

Now we put up some conditions, such that we should find the $𝐱\in S\subset R^n$ that gives a local extreme of $f$. Let $S={\bigcap }_{i=1}^m{S}_i$, andlet $S_i$ be defined so that $g_i(𝐱)=0\forall 𝐱\in S_i$.

Any vector $𝐱\in R^n$ can have one component perpendicular tothe subset $S_i$ (for visualization, think $n=3$ and let$S_i$ be a flat surface). $\nabla g_i$ will be perpendicular to$S_i$, because:

But $g_i(𝐱)=0$, so any vector $𝐱+ϵ\nabla g_i$ must be outside $S_i$, and also outside $S$.(todo: I have proved that there might exist a component perpendicular to each subset $S_i$, but not that there exists only one; this should be done)

By the argument above, $\nabla f$ must be zero - but now we can ignoreall components of $\nabla f$ perpendicular to $S$. (todo: this should be expressed more formally and proved)

So we will have a local extreme within $S_i$ if there exists a${\lambda }_i$ such that

$$\nabla f={\lambda }_i\nabla g_i$$

We will have local extreme(s) within $S$ where there exists a set${\lambda }_i,i=1,\mathrm{\dots },m$such that

$$\nabla f=\sum {\lambda }_i\nabla g_i$$