On a manipulation of little oh in $R^n$.

Question

Take $x \in R^n$. and take $ f(x) = o(||x||)$, where $f: R^n \rightarrow R^m$ is a linear function.

This means by definition that $$\lim_{x \rightarrow 0} \frac{f(x)}{||x||} = 0 $$

Now we can find $t \in R$ and $k \in R^n$ s.t. $x =kt$ then we obtain that $tf(k) = o(||tk||)= o(t)$.

Now My doubt is: why is $o(||tk||)= o(t)$? I tried to prove it by taking $$\lim_{kt \rightarrow 0} \frac{o(||tk||)}{t} =\lim_{kt \rightarrow 0} \frac{to(||k||)}{t} = \lim_{kt \rightarrow 0} \frac{o(||k||) ||k||}{||k||}$$

But $k$ could go to infinity (then $t$ would need to go to zero faster) so this limit is undecided to me. Where is my mistake kind answerers? how could I prove this?

Edit(providing more context): We had just given the definition of differentiable function in $R^n$, that is, given a set $A \subset R^m$ and $x \in Int(A)$ and a function $f: A \rightarrow R^n$ we will say that $f$ is differentiable in $x_0$ if there exist a $L \in Hom(R^n, R^m)$ s.t. $$f(x_0 +h) - f(x_0) - L(h) = o(||h||)$$

We then wanted to prove that there is only one differential so we supposed there where two linear functions $L_1, L_2$ that both satisfied the above equation then subtracting one equation from the other we have $$(L_1-L_2)(h) = o(||h||)$$ at this point we want to show that $(L_1 - L_2)(h) = 0$ and here is the point that we make our substitution $h = tk$.

The reference book for the course is by Lanconelli and is called Analisi Matematica 2. (Italian university). The course is called mathematical analysis 2.

Answer 1

Note we may always (for $||x|| \neq 0$) find the unit vector of $x$ by setting $\hat x = x / ||x||$. Then calling $t = ||x||$, you also have that $ x = t \hat x$. Now your question of why $o ( ||t \hat x ||) = o ( t)$ should be obvious since $t \hat x = x$, and $t = ||x||$.

Answer 2

If the goal is to prove that $f(x)=o(||x||)$, $f$ is linear $\implies f\equiv0$:

Start by assuming we have such a linear function which is nonzero, i.e. $f(u)\neq0$. Without a loss of generality, we can take $||u||=1$ and $f(u)=1$. Then for any $t\in\mathbb{R}$, $f(tu)=t$.

But, as we stipulated that $f(x)=o(||x||)$, we have $t=f(tu)=o(||tu||)=o(||t||)$. This provides a contradiction, as $t$ is not $o(||t||)$.

So, by contradiction, any such linear function must be identically zero.

Answer 3

You know that $f(x)=o(||x||)$ means $$\lim_{x \rightarrow 0} \frac {\|f(x)\|}{\|x\|}=0$$ Consider the restriction of $\dfrac {\|f(x)\|}{\|x\|}$ to the "punctured" line $\{kt\,|\,t\in\mathbb R\setminus \{0\}\}$ with $k \neq 0\,$ fixed.

Now the limit of a restriction is the same, so $$\lim_{t \rightarrow 0} \frac {\|f(kt)\|}{\|kt\|}=0$$ Since $f$ is linear, you have $$\frac {\|f(kt)\|}{\|kt\|}= \frac {|t|\|f(k)\|}{|t|\|k\|} = \frac {\|f(k)\|}{\|k\|}$$ so $\,\dfrac {\|f(kt)\|}{\|kt\|}\,$ is constant and then $\,\dfrac {\|f(k)\|}{\|k\|}=0\,$ that is $$f(k)=0$$ Since $k$ is arbitrary, the theorem is proved.

You can look at p. 213 of baby Rudin (3rd ed.).