Proving that the derivative is unique in higher dimensions

Question

Can someone help me prove this please? I am thinking of using triangle inequality. However, I feel as if I would be doing to much and there is a better way to prove the following. $\def\h{{\mathbf h}} \def\x{{\mathbf x}} \def\f{{\mathbf f}} \def\0{{\mathbf 0}} \def\R{{\mathbb R}} \def\L{{\mathcal L}}$

Let $\f\colon D\to \R^m$ where $D\subseteq\R^n$ is open. Let $\x_0\in D$ and suppose that $\f$ is differentiable at $\x_0$. Prove $T\in\L(\R^n,\R^m)$ satisfies the definition of derivative, such that $$\lim_{\h\to\0}\frac{\|\f(\x_0+\h)-\f(\x_0)-T(\h)\|}{\|\h\|} = 0$$ is unique. Essentially, I want to conclude $\|T-S\|<\epsilon$.

$\textit{Proof.}$ Suppose $T$ and $S$ are two linear transformations which satisfy our definition. By $\epsilon-\delta$ definition of limit, for any given $\epsilon >0$, there exists $\delta >0$ such that $0< \|\h\| < \delta$ then $\displaystyle{\frac{\|\f(\x_0+\h)-\f(\x_0)-T(\h)\|}{\|\h\|}<\frac{\epsilon}{2}}.$ Now, $0<\|\h\|<\delta$ implies $\displaystyle{\|\f(\x_0+\h)-\f(\x_0)-T(\h)\| < \frac{\epsilon}{2}\|\h\|}$

Answer 1

We also have for all $\epsilon>0$, there is $\delta'>0$ such that for all $0<\|h\|<\delta'$, $\|f(x_0+h)-f(x_0)-S(h)\|<\dfrac{\epsilon}{2}\|h\|$.

By the triangle inequality, for $0<\|h\|<\delta'':=\min\{\delta,\delta'\}$, $$\|(T-S)h\|=\|f(x_0+h)-f(x_0)-T(h)-(f(x_0+h)-f(x_0)-S(h))\|$$ $$\le\|f(x_0+h)-f(x_0)-T(h)\|+\|f(x_0+h)-f(x_0)-S(h)\|<\epsilon\|h\|$$

Now using the definition of the operator norm, $$\|T-S\|=\sup_{\|x\|\le1}\|(T-S)(x)\|$$ and so $$\|T-S\|\cdot\delta''=\sup_{\|x\|<\delta''}\|(T-S)(x)\|<\epsilon\delta'',$$ so $\|T-S\|<\epsilon$.

Answer 2

Let $E$ and $F$ be normed linear spaces, so $\mathbb{R}^n$ and $\mathbb{R}^m$ in your case. Let $T_1$ and $T_2$ be continuous linear maps satisfying: $$ f(x+h) = f(x)+T(h)+|h|\psi(h)$$ where $\psi:E\to F$ satisfies $\lim_{h\to 0}\psi(h)=0$ and that $\psi (0)=0$. Let $v\in E, t\in\mathbb{R}, t>0$ such that $x+tv$ lies in $D$. Let $h=tv$, then $$\begin{aligned} f(x+h)-f(x)&=T_1(h)+|h|\psi_1(h)\\ &=T_2(h)+|h|\psi_2(h) \end{aligned}$$ Where $\lim_{h\to 0}\psi_j(h)=0$ for $j=1,2$. Let $T=T_1-T_2$. Subtracting the two expressions for $f(x+tv)-f(x)$, we have that: $$T_1(h)-T_2(h)=|h|(\psi_2(h)-\psi_1(h)) $$ Setting $h=tv$ and using linearity of $T$ $$t(T_1(v)-T_2(v))=t|v|(\psi_2(tv)-\psi_1(tv)) $$ Dividing by $t$: $$T_1(v)-T_2(v)=|v|(\psi_2(tv)-\psi_1(tv)) $$ Take the limit as $t\to 0$, the limit of the right hand side is equal to $0$. This implies that $T_1(v)-T_2(v) = 0\implies T_1(v)=T_2(v)$.