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Let $F:\mathbb{R}^3\setminus\{0\}\to\mathbb{R}^3$ be a function of class $C^1$ such that $$(x_1, x_2, x_3)\in \mathbb{R}^3\setminus\{0\}\mapsto F_i(x_1, x_2, x_3)\in\mathbb{R}, \quad \forall i\in\{1, 2, 3\}.$$

As an exercise, I have to find an upper bound (better if as tight as possible) for $$\left\Vert\left(\frac{\partial F_i}{\partial x_j}(y)\right)_{i, j=1, 2, 3}\right\Vert $$ where the above notation refers to the norm of the matrix $\left(\frac{\partial F_i}{\partial x_j}(y)\right)_{i, j=1, 2, 3}$ as it is defined in https://en.wikipedia.org/wiki/Matrix_norm (the norm defined through the supremum).

$\textbf{EDIT:}$ By following the idea given in the answer by @kieransquared, I checked again all the computations. The only information I have is that $$ \frac{\partial F_i}{\partial x_j}(y) \le\begin{cases} \displaystyle\frac{1}{\alpha^{\beta +1}} -\frac{y_i^2 (\beta +1)}{\gamma^{\beta +3}} &\hbox{ if } i=j\\[10pt] \displaystyle -\frac{y_j^2 (\beta +1)}{\gamma^{\beta +3}} &\hbox{ if } i\neq j, \end{cases}$$ where $a, \beta\ge 1$, $\alpha,\gamma >0$ constants and $|x|$ denotes the euclidean norm of the vector $(x_1, x_2, x_3)$.

Hence, by using the triangle inequality, it follows that $$ \left\vert\frac{\partial F_i}{\partial x_j}(y)\right\vert \le\begin{cases} \displaystyle\frac{1}{\alpha^{\beta +1}} +\frac{y_i^2 (\beta +1)}{\gamma^{\beta +3}} &\hbox{ if } i=j\\[10pt] \displaystyle \frac{y_j^2 (\beta +1)}{\gamma^{\beta +3}} &\hbox{ if } i\neq j, \end{cases}$$

Anyway, I can not deduce the desired upper bound from that information.

I hope someone could help. Thank you.

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  • $\begingroup$ What is the desired upper bound, now that $|x|$ is no longer present in your estimates for the $|\partial F_i/\partial x_j|$? $\endgroup$
    – kieransquared
    Mar 26 at 21:19
  • $\begingroup$ I would say that $|\partial F_i /\partial x_j (y)|\le \frac{1}{\alpha^{\beta +1}} +\frac{y_i^2 (\beta +1)}{\gamma^{\beta +3}}$ and hence $\|\partial F_i /\partial x_j (y) \|\le C \max\left(\frac{1}{\alpha^{\beta +1}} +\frac{y_i^2 (\beta +1)}{\gamma^{\beta +3}}\right)$. Is that correct? $\endgroup$
    – C. Bishop
    Mar 26 at 21:32
  • $\begingroup$ Yes, although I don't think you need a max (over what set are you taking the max?) $\endgroup$
    – kieransquared
    Mar 27 at 0:58

1 Answer 1

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All norms on $\mathbb{R}^n$ are equivalent, meaning for any two norms $\|\cdot\|_1, \|\cdot\|_2$, there exist constants $A,B$ such that $A\|x\|_2 \leq \|x\|_1 \leq B\|x\|_2$. Since you can view a matrix as a vector in $\mathbb{R}^{n^2}$, this applies to matrix norms too. You haven't specified which matrix norm you're using, but for each norm, there will be a $C$ such that

$$\|M\|\leq C\max_{i,j}|M_{ij}|$$

For any matrix $M$. So your stated bound indeed holds, but the constant will depend on the norm you use (see the end of the Wikipedia article on matrix norms for precise constants, which are often sharp).

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  • $\begingroup$ kieransquared, thank you for the answer. I thought about something like that but actually I have an estimate on (what you called) $M_{ij}$ and not on $|M_{ij}|$, hence I am not sure how to apply the inequality involving the maximum. $\endgroup$
    – C. Bishop
    Mar 26 at 17:50
  • $\begingroup$ Well, if the $M_{ij}$ can take arbitrarily large negative values, there’s no hope for a bound of the form you’re trying to prove. However, your function is $C^1$, which means you have a uniform bound on the absolute value of the derivatives: $|\partial F_i/\partial x_j| \leq C$. If this is an exercise, my guess is that you do have an estimate on the absolute values, but without more context it’s hard to say whether that’s something that needs to be proved or whether there’s a typo somewhere. $\endgroup$
    – kieransquared
    Mar 26 at 19:32
  • $\begingroup$ kieransquared, I edited the question by adding more details and checking my computations. I hope you could help again. $\endgroup$
    – C. Bishop
    Mar 26 at 20:27

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