Understanding Rudin's PMA Theorem 9.17

Question

$9.17$ Theorem Suppose f maps an open set $E\subset R^n$ into $R^m,$ and f is differentiable at a point $\mathbf{x}\in E.$ Then the partial derivatives $D_jf_i(\mathbf{x})$ exist, and \begin{align} \mathbf{f'(x)e_j}=\sum\limits_{i=1}^m(D_jf_i)\mathbf{(x)u_i} &&(1\le j\le n)\tag{27} \end{align}

$\mathbf{e_i}'$s and $\mathbf{u_j}'$s are standard bases of $R^n$ and $R^m$ respectively.

Proof Fix $j.$ Since f is differentiable at $\mathbf{x},$ $$\mathbf{f(x+}t\mathbf{e_j)}-\mathbf{f(x)}=\mathbf{f'(x)(}t\mathbf{e_j)}+\mathbf{r(}t\mathbf{e_j)}\tag{28}$$ where $\left|\mathbf{r(}t\mathbf{e_j)}\right|/t\to 0$ as $t\to 0.$ The linearity of $\mathbf{f'(x)}$ shows therefore that

$$\lim_{t\to 0}\sum\limits_{i=1}^m\frac{f_i(\mathbf{x}+t\mathbf{e_j})-f_i(\mathbf{x})}{t}\mathbf{u_i}=\mathbf{f'(x)e_j.}\tag{29}$$

It follows that each quotient in this sum has a limit, as $t\to 0$(see Theorem $4.10$), so that each $(D_jf_i)(\mathbf{x})$ exists, and then $(27)$ follows from $(29).$

link to theorem $4.10$

Question Suppose I denote $g_i(t) =\frac{f_i(\mathbf{x}+t\mathbf{e_j})-f_i(\mathbf{x})}{t}\mathbf{u_i}.$ Then $(29)$ is just saying that $\lim_{t\to 0}\sum g_i(t) =L.$ Then why can't I just conclude $\lim_{t\to 0}g_i(t)=l$ for some $l\in R^m.$ Why do I need Theorem $4.10$ here?

Answer 1

Theorem 4.10 is necessary to ensure that each quotient in the sum in (29) has a limit, as 𝑡→0, and that the limit exists for each (𝐷𝑗𝑓𝑖)(𝐱). The existence of these limits is not guaranteed without Theorem 4.10, as there may be cases where the limit does not exist or is infinite.

While it is true that (29) can be rewritten as lim𝑡→0∑𝑔𝑖(𝑡)=𝐿, where 𝐿 is the limit of the sum, this does not necessarily imply that lim𝑡→0𝑔𝑖(𝑡) exists for each i. In other words, the existence of the limit of the sum does not imply the existence of the limits of each individual quotient.

Therefore, Theorem 4.10 is needed to ensure the existence of each limit, which is necessary to establish the partial derivatives and ultimately prove (27).

Answer 2

Your question may be answered by a simple example. Suppose $g_1(t) = e^t$ and $g_2(t) = -e^t$

Clearly $\lim_{t \to \infty} \sum_{i=1}^2 g_i(t) = \lim_{t \to \infty} e^t + -e^t = \lim_{t \to \infty} 0 = 0$

However, $\lim_{t \to \infty} g_1(t) = \infty$ and $\lim_{t \to \infty} g_2(t) = -\infty$

I present a more detailed proof of the actual theorem in this post.

Answer 3

The key is to understand the notations here. Each quotient in this sum without $u_i$(i.e. $\frac{f_i(x + te_j)-f_i(x)}{t}$) is the ith coordinate in $R^m$ with respect to $u_i$. So basically, (29) is telling you that a linear combination of $u_i$ converges to a vector in $R^m$. By the theorem 4.10, we can argue in the same manner that each coordinate converges to $f'(x)e_j\cdot u_i$, from which we get the $(D_jf_i)(x)$ we want.

Proof: $t \to 0,|\sum\frac{f_i(x + te_j)-f_i(x)}{t}u_i -f'(x)e_j| \to 0 \iff \frac{f_i(x + te_j)-f_i(x)}{t} \to f'(x)e_j\cdot u_i,\forall i$ (theorem 4.10)