Theorem 9.12 Rudin

Question

9.11 Definition Suppose is an open set in $^{n}$, f maps into $^{m}$, and $x ∈$. If there exists a linear transformation of $^{n}$ into $^{m}$ such that $$\lim_{h\to 0}\frac{\left|\mathbf{f(x +h)-f(x)-Ah}\right|}{|\mathbf{h}|}=0,\tag{14}$$

9.12 Theorem Suppose and f are as in Definition 9.11, $x ∈$, and (14) holds with =1 and with =2. Then 1=2.

Here's Rudin's Proof

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 $h\to 0$. 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 shows that the left side of (16) is independent of $t$. Thus $=0$ for every $∈R^{n}$. Hence $=0$.

My question is that why do we need the step involving t? Can we directly claim $B=0$ when we know $\frac{|\mathbf{Bh}|}{|\mathbf{h}|}\to 0$ as $h\to 0$? Could someone please give me a counterexample?

Thanks in advance

Answer 1

You can claim that only when $B$ is linear, as per the explanation. If $B$ is not linear that is not true, with a simple example when $B(h)=(h^2,0,.....,0)$. Then:

$$ \dfrac{\vert B(h)\vert}{\vert h\vert}=\frac{\vert (h^2,0,.....,0) \vert}{\vert h \vert} = \vert(h,0,...,0)\vert =h \rightarrow 0$$

But $B(x)\neq 0$ for all fixed $x\neq 0$.

Answer 2

You want to show that $B=0$ which means that you need to prove that $Bh=0$ for $\textit{all possible values of}\ h$.

You know that $\frac{Bh}{h}\to 0$. Now, let $h_0\neq 0$ be a fixed but arbitrary vector. Then, $\frac{B(th_0)}{th_0}\to 0$ as $t\to 0$ simply because as $t\to 0,\ th_0\to 0.$

Now, linearity of $B$ implies that $\frac{B(th_0)}{th_0}=\frac{Bh_0}{h_0}\to 0$. But $\frac{Bh_0}{h_0}$ is $\textit{constant}$, so in fact $\frac{Bh_0}{h_0}=0$ and since $h_0\neq 0$ it must be that $Bh_0=0$. Therefore, $B$ is the zero transformation.