Inequality of a linearized function The following inequality is used in one of the steps for the proof of the centre manifold theorem. Consider the function $$f(x)=Ax+\tilde{f}(x)$$
where $x\in \mathbb{R}^n$, $f:\mathbb{R}^n \rightarrow \mathbb{R}^n$ is of class $C^k$ for some $k\geq 1$ and $f(0)=0$, $Df(0)=A$.
The author during the proof of a lemma proceeds to state the following result.
$$\|f(x)\|\leq\|x\|\sup_{s\in[0,1]}\|Df(sx)\|$$
The author has not defined how the norm of $Df(sx)$ is defined and I am not sure if that is required. I tried proving this for the one dimensional case by replacing all the norms with moduli but couldn't come up with a simplification for the RHS.
The text which this is taken from is A. Vanderbauwhede, "Centre manifolds, normal forms and elementary bifurcations", Dynamics Reported, Vol 2 (1989), pg 95.
 A: The result you are refering to is a version of the Mean value theorem (cf. the last proposition of that subsection). In your case, it yields
$$f(x) - f(0) = \int_0^1 Df(sx) x\, \mathrm d s. $$
Now take any norm $\lVert \, \cdot \, \rVert$ on $\mathbb R^n$. Then you can consider its associated operator norm
$$\lVert \, \cdot \, \rVert_{\text{op}} \colon \mathbb R^{n \times n} \to [0, \infty), \quad \lVert A \rVert_{\text{op}} := \sup_{\lVert x \rVert = 1} \lVert A x \rVert. $$
Then one has $\lVert A x \rVert \leq \lVert A \rVert_{\text{op}} \cdot\lVert x \rVert $ for all $x \in \mathbb R^n$: Indeed, this is evident for $x = 0$ and for $x \neq 0$ set $y := \lVert x \rVert^{-1} x$, notice that $\lVert y \rVert = 1$ and thus $\lVert A y \rVert \leq \lVert A \rVert_{\text{op}}$. Therefore,
$$ \lVert A x \rVert = \lVert A y \rVert \cdot \lVert x \rVert \leq \lVert A \rVert_{\text{op}} \lVert x \rVert.$$
Turning back to your problem, the identity above yields
\begin{align*}
\lVert f(x) \rVert &= \lVert f(x) - f(0) \rVert = \bigg\lVert \int_0^1 Df(sx) x\, \mathrm d s \bigg\rVert \leq \int_0^1 \lVert Df(sx) x \rVert\, \mathrm d s\\
&\leq \int_0^1 \lVert Df(sx) \rVert_{\text{op}} \lVert x \rVert\, \mathrm d s \leq \sup_{s \in [0, 1]} \lVert Df(sx) \rVert_{\text{op}} \lVert x \rVert.
\end{align*}
Note that you can choose any norm on $\mathbb R^n$ to begin with as long as you define the operator norm with respect to that norm. The usual choice might be the Euclidean norm. Otherwise, on might have to include suitable constants but for most applications I can think of that should be irrelevant.
