Implicit Function Theorem Proof (Rudin)

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Given $f\in\mathscr{C'}$ where $f$ is a map from the open set $E\subset R^{n+m}$ to $R^{n}$, $f(a,b)=0$ for some point $(a,b)\in E$.

In the second part of the proof it says:

Finally, to compute $g'(b)$, put $(g(y),y)=\phi(y)$. Then $$ \phi'(y)k=(g'(y)k,k)$$ for $y\in W=\{y\in R^m : (0,y)\in V\}, k\in R^m$

What $V$ or $W$ is isn't really important. My question is where does the k come from and why does it simply disappear later in the proof?

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$k$ is an arbitrary tangent direction introduced to better describe the derivative in the form of a directional derivative. The author could also have written $$ ϕ'(y)=(g'(y),I)\text{ or }ϕ'(y)= g'(y)\oplus I $$ but probably wanted to avoid this concatenation of matrices.