Definition: $\|D\|_\mathrm{op}=\sup_{x\in\mathbb{R}^n/\{0\}} \frac {\|Dx\|} {||x||}$

Where $\|\cdot\|$ is the Euclidean norm. We have sub-multiplicativity, $\|AB\|\leq\|A\|\cdot\|B\|$.

Now say I have some $f:\mathbb{R}^m \to \mathbb{R}^n$ differentiable and every partial derivative is bounded. So I have $\|Dx\|<M $ for all $x\in\mathbb{R}^m$.

Now I`m interested in showing $\|D\|_\mathrm{op}$ exists , and moreover, bound it with $M$.

I have no idea how to approach this,since I don`t know much about operators and operators norms. I'd appriciate it if anyone can provide an elementary proof.

the motivation behind the question is to easily show that bounded partials implies local lipchitz: \begin{align} & \|f(x+h)-f(x)\|=\|D_f(x)+o(\|h\|)\|\leq\|D_f(x)h\|+\|o(\|h\|)\| \\[10pt] \leq {} & \|D_f(x)\|_\mathrm{op} \|h\|+||o(\|h\|)\leq \hat M\|h\| +o(\|h\|)\to 0 \end{align}

P.S : I wasn`t sure under what tags my question belongs, so I've entered the main topic I'm studying right now.


There is not much to prove here. Your troubles stem from the unprecise notation. I'm using $|x|$ for the euclidean norm of vectors $x\in {\mathbb R}^n$, whatever $n\geq1$. For a constant linear map $A:\>{\mathbb R}^n\to{\mathbb R}^m$ one then defines the operator norm $$\|A\|:=\sup_{x\ne0}{|Ax|\over |x|}\ .$$ Now, given a differentiable function $f:\>{\mathbb R}^n\to{\mathbb R}^m$, this $f$ has at every single point $x\in{\mathbb R}^n$ a derivative $$df(x):\quad{\mathbb R}^n\to{\mathbb R}^m\>,$$ which is an element of ${\cal L}({\mathbb R}^n,{\mathbb R}^m)$, and is characterized by $$f(x+X)-f(x)=df(x).X+o\bigl(|X|\bigr)\qquad(X\to0)\ .$$ The matrix of this $df(x)$ is the matrix of partial derivatives $f_{i.k}(x):={\partial f_i\over\partial x_k}(x)$. Therefore the operator norm $\|df(x)\|$ can be estimated in terms of these partial derivatives. A simple estimate is $$\|df(x)\|\leq\sqrt{\sum\nolimits_{i\in[m], \>k\in|n]}\bigl|f_{i.k}(x)|^2}\ .$$

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