Theorem 9.17 from Baby Rudin

Question

I'm a bit confused about the final step in the proof the the theorem below from Baby Rudin.

This theorem tells precisely that the value of the linear transformation $f'(x)$ at $e_j$ is the column-vector in $R^m$ which is the partial derivative of the vector $f=(f_1,\dots,f_m)^t$ with respect to the $j^{th}$ variable.

Now, doesn't Eq. $(28)$ already imply this claim? The left-hand side of this equation is precisely the directional derivative of $f$ at $x$ in the direction of $e_j$, which is precisely what we need. Do I miss something?

Furthermore, it seems to me that Rudin's reference to Theorem 4.10 is imprecise. This theorem says the following:

However, I don't understand how this theorem is used to conclude the existence of the limits of each quotient.

Answer 1

To answer your first question: It's true that equation $(28)$ implies the result, but the rest of the proof explains why this yields the result. That the equation $$\lim_{t\to 0}\frac{f(x+te_j)-f(x)}{t}=\lim_{t\to 0}\sum_{i=1}^m\frac{f_i(x+te_j)-f_i(x)}{t}u_i$$ holds may not be clear to someone without an understanding of basic linear algebra, and that interchanging the limit and the sum on the right hand side is legal because of Theorem $4.10$ (or equation $(24)$, I don't have a copy of the book readily available).

Now for your second question: Rudin uses the "if" statement in part $(a)$ of Theorem $4.10$ to show that

$$\lim_{t\to 0}\sum_{i=1}^m\frac{f_i(x+te_j)-f_i(x)}{t}u_i=\sum_{i=1}^m\lim_{t\to 0}\frac{f_i(x+te_j)-f_i(x)}{t}u_i.$$

Again, he's just fleshing out the details for a first-time reader.

Answer 2

I know this is a few years after @Aweygan's answer, but I feel after looking at various posts here on Math Stackexchange asking this same question, no one has provided (to me at least), a satisfactory explanation of how continuity is used to justify the equality $$\lim_{t\to 0}\sum_{i=1}^m\frac{f_i(x+te_j)-f_i(x)}{t}u_i=\sum_{i=1}^m\lim_{t\to 0}\frac{f_i(x+te_j)-f_i(x)}{t}u_i$$ mentioned in proof of 9.17 in Baby Rudin. Instead, I present a modified version of the proof starting from Rudin having established the validity of equation $(28)$: $$\lim_{t \to 0} \frac{f(x + te_j) - f(x)}{t} = f'(x)e_j$$

Since $f'(x)e_j \in \mathbb{R}^m$, we may present it as an $m$-tuple: $f'(x)e_j = (p_1, \ldots, p_m)$. Therefore, if we fix $\epsilon > 0$, equation $(28)$ tells us there exists $\delta > 0$ such that if $0 < |t| < \delta$ then $$ \left\Vert \frac{f(x+te_j) - f(x)}{t} - f'(x)e_j \right\Vert < \epsilon\ \ \ \ (*)$$

Expressing everything inside the norm as an $m$-tuple, $(*)$ says that $$ \left\Vert \left(\frac{f_1(x+te_j) - f_1(x)}{t}, \ldots, \frac{f_m(x+te_j) - f_m(x)}{t} \right) - (p_1, \ldots, p_m) \right\Vert = \left\Vert \left( \frac{f_1(x+te_j) - f_1(x)}{t} - p_1, \ldots, \frac{f_m(x+te_j) - f_m(x)}{t} - p_m\right)\right\Vert < \epsilon $$

By the definition of the norm on $\mathbb{R}^m$, this says $$\left(\sum_{i=1}^m \left|\frac{f_i(x+te_j)-f_i(x)}{t} - p_i \right|^2\right)^{1/2} < \epsilon$$

Since both sides of the inequality above are non-negative, we can square both sides while preserving the inequality to obtain: $$\sum_{i=1}^m \left|\frac{f_i(x+te_j)-f_i(x)}{t} - p_i \right|^2 < \epsilon^2$$ Since each term in the sum on the LHS is non-negative, it must be the case that $$\left|\frac{f_i(x+te_j)-f_i(x)}{t} - p_i \right|^2 < \epsilon^2 \ \text{for each }i \implies \left|\frac{f_i(x+te_j)-f_i(x)}{t} - p_i \right| < \epsilon$$ Thus, for each $i=1, \ldots, m$, we have shown that $$\lim_{t \to 0} \frac{f_i(x+te_j) - f_i(x)}{t} = p_i \ \ (**)$$ But the LHS of $(**)$ is the definition of $D_jf_i(x)$ whenever said limit exists, so we conclude $f'(x)e_j = (D_jf_1(x), \ldots, D_jf_m(x))$, as desired.