Help with derivative on $\mathbb{R}^{n}$

Question

I'm stuck in this question, any help would be appreciated!

Let $\Omega \subset \mathbb{R}^{m}$ and $\textbf{x},\textbf{y} \in \Omega$ such as $S=\{t\textbf{x}+(1-t)\textbf{y}:t \in (0,1)\} \subset \Omega$. Let $f:\Omega \to \mathbb{R}^{n}$ differentiable in $\Omega$, and $T \in \mathcal{L}(\mathbb{R}^{m},\mathbb{R}^{n})$. Show that $||f(\textbf{x})-f(\textbf{y})-T(\textbf{x}-\textbf{y})|| \leq \sup\limits_{\textbf{z} \in S}||f'(\textbf{z})-T||_{\mathcal{L}(\mathbb{R}^{m},\mathbb{R}^{n})}||\textbf{x}-\textbf{y}||$

I don’t know where to start. My intuition says to create a function defined on S in a way that I could use the unidimensional Mean Value Theorem, however, I don’t know how to go further.

Answer 1

Let $u=f(x)-f(y)-T(x-y)$ and $\varphi(t)=u\cdot[f(tx+(1-t)y)-T(tx+(1-t)y)]$, then \begin{align*} \|u\|^{2}&=\varphi(1)-\varphi(0)\\ &=D\varphi(c)\\ &=u\cdot[f'(cx+(1-c)y)(x-y)-T(x-y)]\\ &\leq\|u\|\|f'(cx+(1-c)y)-T\|\|x-y\|\\ &\leq\|u\|\sup_{z\in S}\|f'(z)-T\|\|x-y\|, \end{align*} the result follows.