i am currently reading the book Convex Analysis. At "Theorem 7.8.1" there is given a sketch of a proof which seems faulty to me. I will briefly explain the Theorem and the proof and then my problem i have. Maybe you can hint me where i missed something.
We start out as usual with continuous differerentiable functions $f_0,\dots,f_m : U\to \mathbb R$ with $U\subset \mathbb R^n$ open.
If the following Problem has a minimum at $\hat x$:
$\min f_0(x), x\in \mathbb R^n,\ f_1(x) = \dots = f_m(x) =0$
Then there exists $\lambda = \lambda_0,\dots,\lambda_m$ , not all $0$ with:
$\mathcal L(\hat x,\lambda) = \lambda_0 f'_0(\hat x) + \dots + \lambda_m f'_m(\hat x) = 0$
The proof thats shown in the book goes like this: (you may skip this part at first glance)
Let $\hat x$ be a minimal point of the Problem. Define $F(x,y) = f_0(x) \text{ if } f_1(x) +y_1 = \dots = f_m(x) + y_m = 0$ and $\infty$ otherwise. For $$\min f(x) := F(x,0)$$ we get our original Problem then. Taking the solutionset of $(z,y) = (f_0(x),\dots, f_m(x))$ we can define a function $g(x)=(z,y)$ which is continuously differnentiable and its Graph ,,can be seen'' as the Graph $\Gamma_F$ of $F$ (This is sort of wrong already but since its a sketch i get what he means). Now the author says the Tangent space $T_{(\hat x, 0_Y, f(\hat x))}$ at the $(\hat x, 0 , f(\hat x))(\Gamma_F)$ then is the Graph of the derivative $g'(\hat x)x = (f_0'(\hat x)x, \dots, f_m'(\hat x)x)$ so we can describe it by an equation $$ \langle (\lambda_1,\dots,\lambda_m),y\rangle -\langle \lambda_0,z\rangle = 0 $$
And like this we get the multiplicators. Now comes the part where it seems wrong to me: We want $y$ to be $0$ to solve our problem $\min f(x) = F(x,0)$ and also $z$ to be $0$ for minimality sake which translates into $$ (\mathbb R^n \times 0_Y \times 0) + T_{(\hat x, 0_Y, f(\hat x))} \subsetneq \mathbb R^n \times Y\times \mathbb R $$ note the proper inclusion is equivalent to $(\mathbb R^n \times 0_Y \times 0)\subset T_{(\hat x, 0_Y, f(\hat x))}$. BUT why would $z$ be $0$?. The Author says because Tangent space of the Graph of $f$ at the point $(\hat x, f(\hat x))$ is contained in $\mathbb R^n\times 0$, ,,which follows immediatly from the local minimality of $f$''. But that thing is not even differentiable by definition. I mean okay one could go with the subgradient instead or something. But still its wrong in my opinion. I mean imagine $n=2$ and $f_0$ to be an inverse Paraboloid and $f_1 = 0$ is our only restriction which is $0$ on the unit ball, smooth and $>0$ outside the unit ball. Then the subgradient $\partial f(x)$ will never have an element $0$ inside its subgradient. Even in this more general case its wrong right?