I'm studying real analysis in $\mathbb{R}^n$ and there is one step on the proof of the unicity of the derivatives that I'm not getting.
Theorem: If there exists $ T \in \mathbb{L}(\mathbb{R}^n,\mathbb{R}^m) $ as in the definition of differenciability, then $T$ is unique.
Proof:
Let $ T,S \in \mathbb{L}(\mathbb{R}^n,\mathbb{R}^m) $ such that
$ f(a+h) = f(a) + T(h) + r_{T}(h) = f(a) + S(h) + r_{S}(h)$
with
$ \lim_{||h||\rightarrow 0}\frac{||r_{T}(h)||}{||h||} = \lim_{||h||\rightarrow 0}\frac{||r_{S}(h)||}{||h||} = 0 $
Then $ \lim_{||h||\rightarrow 0} \frac{||T(h)-S(h)||}{||h||} = 0 $
//Here is where I'm lost:
Let's fix $ h \in \mathbb{R}^n , h \neq 0 $.
Then:
$ ||T(h) - S(h)|| = ||h|| \frac{||T(th)-S(th)||}{||th||} , \forall t \in \mathbb{R} - \{ 0 \} $
So, when $t \rightarrow 0$,
$||T(h)-S(h)||=0$, then $T=S$. $\square$
What does it mean fixing $h$, and how could the next equation be true for every $t \in \mathbb{R} - \{0\} $? I really can't see what was done there.
Thanks.
If we multiply and divide $||T(h)-S(h)||$ by $||h||$, which is a non zero real number, we get
$$||T(h)-S(h)||=||h||\frac{||T(h)-S(h)||}{||h||}$$
Now, the $t\neq 0$ comes from linearity of both $T$ and $S$. So, it is true that $\forall\;t\neq 0$ we have
$$||T(h)-S(h)||=||h||\frac{||T(th)-S(th)||}{||th||}$$
But we know that if $||a||\to 0$, then $\frac{||T(a)-S(a)||}{||a||}\to 0$ also. So, once $h$ (and hence $||h||$) is fixed, we have that $||t||\to 0\Rightarrow ||th||\to 0$, and then the result follows, i.e., for all $||h||\neq 0$, we have $||T(h)-S(h)||=0$ which implies $T(h)=S(h)$ for all $||h||\neq 0$, and then $T=S$ (if $h=0$ there is no problem beacause $T$ and $S$ coincide).