Understanding theorem 9.12 in Rudin's PMA

Question

$9.11$ Definition Suppose $E$ is an open set in $R^n,$ f maps $E$ into $R^m,$ and x $\in E.$ If there exists a linear transformation $\mathbf{A}$ of $R^n$ into $R^m$ such that

$$\lim_{h\to 0}\frac{\left|\mathbf{f(x +h)-f(x)-Ah}\right|}{|\mathbf{h}|}=0,\tag{14}$$ then we say $\mathbf{f}$ is differentiable at $\mathbf{x},$ and we write $\mathbf{f'(x)=A}$

$9.12$ Theorem Suppose $E$ and f are as in Definition $9.11,$ x $\in E,$ and $(14)$ holds with $\mathbf{A=A_1}$ and with $\mathbf{A=A_2}.$ Then $\mathbf{A_1=A_2}.$

So the idea is to consider $\mathbf{B= A_1-A_2}$ and show that $\left|\mathbf{Bh}\right|\le \epsilon$ for every $\epsilon >0.$ But I failed to understand Rudin's argument :

If $\mathbf{B=A_1-A_2},$ the inequality $$\left|\mathbf{Bh}\right|\le\left|\mathbf{f(x+h)-f(x)-A_1h}\right|+\left|\mathbf{f(x+h)-f(x)-A_2h}\right|$$ shows that $\frac{|\mathbf{Bh}|}{|\mathbf{h}|}\to 0$ as $\mathbf{h}\to 0.$ For fixed $\mathbf{h\ne 0},$ it follows that $$\frac{|\mathbf{B(th)}|}{|\mathbf{th}|}\to 0 \text{ as } t\to 0.\tag{16}$$

The linearity of $\mathbf{B}$ shows that the left side of $(16)$ is independent of $t.$ Thus $\mathbf{Bh}=0$ for every $\mathbf{h}\in R^n.$ Hence $\mathbf{B}=0.$

I understood each and every step but I am not able to see the link showing $\left|\mathbf{Bh}\right|\le \epsilon$ for every $\epsilon >0.$

Can someone help me with this?

Answer 1

$$\frac{|B\mathbf{h}|}{|\mathbf{h}|} = \lim_{t\to 0} \frac{|B\mathbf{h}|}{|\mathbf{h}|}\cdot\frac{t}{t} = \lim_{t\to 0} \frac{|B(t\mathbf{h})|}{|t\mathbf{h}|} = \lim_{\mathbf{u} \,=\, t\mathbf{h} \to 0} \frac{|B\mathbf{u}|}{|\mathbf{u}|} = 0$$

So $|B\mathbf{h}| = 0$ for all $\mathbf{h}$, hence $B=0$.

Answer 2

Rudin doesn't actually show that $$|B\mathbf{h}|\le\epsilon$$ for every $\epsilon>0$. Instead, he directly shows that $$|B\mathbf{h}|=0$$ which is an equivalent statement.

Answer 3

By Rubin's question is very important in all of calculus: [derivatives] are the unique transformation of input (displacement) that provides for a linear approximation that goes to zero [faster than the input does] -- his proof is general enough to cover single variable calculus as well as any other mapping $R^m$ to $R^n$, the problem is that he is cavalier regarding the last proposition:

I'll make minor assumption that we are dealing with R^m to R, you'll see that the proof below generalizes easily to n>1

$$\lim_{\vec{h} \to \vec{0}} \frac{ |\vec{B} \cdot \vec{h}| }{|h|}=0 \implies \vec{B} = \vec{0}$$

The reason why the above is true is because $\vec{h}/h$ is a unit vector, its length is always $1$, so the expression $\frac{ |\vec{B} \cdot \vec{h}| }{|h|}$ reduces to: $\frac{ ||\vec{B}|| . ||\vec{h}|| . |\cos(\theta)| }{|h|} = ||\vec{B}|| . |\cos(\theta)| $

$\theta$ is just an angle that does not approach any limit and so the limit can only hold if $\vec{B}$ is the zero vector:

We must have $\epsilon \gt ||\vec{B}|| . |\cos(\theta)|$ for all $\epsilon, \theta$ and since $\vec{B}$ is a constant vector, it must be the zero vector.

For n>1, we have B a matrix not a vector, and instead of modulus we will have a norm: $||B\vec{h}||/|h|$

$$\epsilon \gt \frac{||B\vec{h}||}{|h|} = \frac{\sqrt{(\vec{b}_{1} \cdot \vec{h})^2 + \ldots + (\vec{b}_{n}\cdot \vec{h})^2}}{|h|} = \sqrt{(|\vec{b}_{1}| \cdot \cos{\theta_{1}})^2 + \ldots + (|\vec{b}_{n}|\cdot \cos{\theta_{n}})^2}$$

...for all $\epsilon$

So this can only be possible if each row of $B$ ($\vec{b}_{i}$) is the zero vector in R^m (same reasoning as above).