Derivative as a continuous mapping of a subset of $R^n$ into the set of all invertible elements

Question

I have been studying the proof of the inverse function theorem in the Rudin's book (Principles of mathematical analysis).

Near the end of the proof, he makes the following statement

"(...) observe that $f'$ is a continuous mapping from U into the set of all invertible elements of $L(R^n)$"

$U$ is a subset of $R^n$, and so is the image of $f$. $f'$ denotes the derivative of $f$.

Can someone explain to me what this phrase means? It does not make any sense.

Answer 1

The hypotheses of the theorem are that

$f: E \rightarrow \mathbb{R}^n, E \subseteq \mathbb{R}^n$ open, $f$ is $C^1$ and $a\in E$ is a point such that $f'(a)$ is invertible.

The derivative $f'(x)$ of $f$ at the point $x$ is, by definition, a linear transformation $f'(x) : \mathbb{R}^n \rightarrow \mathbb{R}^n$ (that satisfies certain properties which makes it the best linear approximation of $f$ at the point $x$, read the exact definition in Rudin). Thus $f'(x)$ can be represented as an $n\times n$ matrix, where the $ij$ entry is $\frac{\partial f_i}{\partial x_j}(x)$ (the partial derivative of the $i$th component function $f_i$ with respect to $x_j$, evaluated at $x = (x_1,x_2,\dots,x_n)).$

The hypothesis says that $f'(a)$ is an invertible matrix for some $a\in E$, which by the way is equivalent with the statement that the determinant of $f'(a) \neq 0$.

If you go back to the beginning of the proof, you will see that a neighborhood $U$ of $a$ is chosen in such a way (depending on the operator norm of $f'(a)$) so that for every $x\in U$, $f'(x)$ is invertible. See the reference back to Theorem 9.8 later in the proof.

Thus, $f'$ is a function defined on $U$ such that $f'(x)$ is an invertible $n\times n$ matrix for every $x\in U$. This explains the second half of the statement in your question: "mapping from U into the set of all invertible elements of $L(\mathbb{R}^n)$.

Now, how to understand the statement that $f'$ is continuous?. $f'$ is a function between two metric spaces. The domain, $U$, inherits its metric space structure from $\mathbb{R}^n$. The image space i.e. the invertible elements of $L(\mathbb{R}^n)$ inherits its metric space structure via the operator norm on the set $L(\mathbb{R}^n)$. That is, if $A\in L(\mathbb{R}^n)$, then

$$ ||A|| = \sup_{|x| = 1} |Ax|, $$

where $|x|$ and $|Ax|$ means the norm of the respective vectors in $\mathbb{R}^n$.

Now, you can interpret the statement that $f'$ is continuous on $U$ in the usual sense: for each $x\in U$ and for all $\epsilon > 0$, there exists $\delta > 0$ such that $|y-x| < \delta$ implies that $||f'(y) - f'(x)|| < \epsilon$, where the distance is determined by the norms I just described.

So the last thing to explain is not just what it means for $f'$ to be continuous, but to explain why in fact it is continuous on $U$. Well, this follows from the fact that $f$ is $C^1$, which means that its partial derivatives exist and are themselves continuous. That $f'$ is continuous easily follows from this.

Answer 2

You have $f'(x) \in L(\mathbb R^n)$ for every $x \in U,$ right? So there is a natural function from $U$ to $L(\mathbb R^n)$ given by $x \to f'(x).$ Since $L(\mathbb R^n)$ is a metric space, we can talk about whether this map is continuous. (That this map takes $U$ into the invertible elements of $L(\mathbb R^n)$ has presumably been shown already.)