0

Let $K\in\mathbb{R}^*$ and $F:\mathbb{R}^3\to\mathbb{R}$ be of class $C^{\infty}$ be such that $$a+u\mapsto F(a+u),$$

where $a\in\mathbb{R}^3$ is a fixed nonnull vector and $u\in\mathbb{R}^3$ is such that $|u|\le K$.

(In case my notation should be ambiguous, I am just saying that when you read $F(a+u)$ it means that $F$ is evaluated in $a+u$ where $a$ is a fixed vector in $\mathbb{R}^3$ and $u\in\mathbb{R}^3$ satisfies $|u|\le K$).

I would like to evaluate the quantity $$\sum_{i=1}^3 \left\vert\frac{\partial F}{\partial u_i}(a+u) - \frac{\partial F}{\partial u_i}(a)\right\vert.$$ The only thing I am trying so far is the following. By means of Mean Value Theorem, it should be $$\sum_{i=1}^3\left\vert\frac{\partial F}{\partial u_i}(a+u) - \frac{\partial F}{\partial u_i}(a)\right\vert\le \sum_{i, j=1}^3\left\vert \frac{\partial^2 F}{\partial u_i \partial u_j} (z) \cdot u \right\vert\le K \sum_{i, j=1}^3\left\vert \frac{\partial^2 F}{\partial u_i \partial u_j} (z) \right\vert,$$ for a vector $z\in [a, a+u]$. My first question is: does my argument hold true? I do not feel confident about how I applied the MVT.

Arrived at this point (if everything is correct), I would please ask you to suggest a way to estimate the quantity $$\sum_{i, j=1}^3\left\vert \frac{\partial^2 F}{\partial u_i \partial u_j} (z) \right\vert.$$

My idea was to estimate with $$\max_{-K\le v\le K}\left\vert\frac{\partial^2 F}{\partial u_i \partial u_j} (a+v)\right\vert,$$ but I do not feel confident about that. Do you have something else to suggest?

I hope someone could answer both the questions.

Thank you for your time.

  • What's $[a,a+u]$? The segment ${a+\lambda u\mid\lambda\in[0,1]}$? – joriki Feb 03 '23 at 12:28
  • @joriki, yes, it is. I was "inspired " by the notation used here: https://math.stackexchange.com/questions/1592457/prove-multi-dimensional-mean-value-theorem which that I have understood as the way you have said. –  Feb 03 '23 at 12:30
  • I think the sum in the middle is also over both $i$ and $j$? – joriki Feb 03 '23 at 13:00
  • @joriki, you are right, it is an oversight, I am going to edit. Thank you. –  Feb 03 '23 at 13:01
  • No, that does not work in general. As the gradient is a true vector-valued function, each component can have a different midpoint in the mean-value theorem. Express the differences as integral, use integral inequalities to get one scalar integrand and apply the mean-value theorem of integration. – Lutz Lehmann Feb 03 '23 at 15:55
  • @LutzLehmann do you have any idea about how to estimate that difference then? –  Feb 03 '23 at 16:07

1 Answers1

0

For any vector valued function $G$, like here $G=\nabla F$, you can write $$ G(a+u)-G(a)=\int_0^1 G'(a+su)u\,ds $$ Now apply the 1-norm and the triangle inequality for integrals to get $$ \|G(a+u)-G(a)\|_1\le\int_0^1\|G'(a+su)\|_{1,op}\|u\|_1\,ds. $$ $G'$ is here the Hessian matrix of $F$. The operator norm $\|\cdot\|_{1,op}$ is the column-sum norm.

To this one can now apply the mean-value theorem for integration to get $$ \|G(a+u)-G(a)\|_1\le\|G'(a+s^*u)\|_{1,op}\|u\|_1,~~~s^*\in(0,1). $$ One can also write $a+s^*u=z$ with $z\in[a,a+u]$.

Lutz Lehmann
  • 131,652
  • Lutz Lehmann, thank you for your answer. Anyway, it is not so clear to me how should I modify my computations according to your suggestion. Could you please provide more details concerning my specific case? –  Feb 03 '23 at 16:21
  • In your bound you get the sum over the column norms, in my version it is the sum over the column norms of the Hessian of $F$. For your intended statement the constant is then of course the maximum of these operator norms over the full ball $B(a,K)$, similar to what you indicated. – Lutz Lehmann Feb 03 '23 at 16:26
  • In my case is simply $G=\frac{\partial F}{\partial u_i}$, isn't it? I think it should be $G^{\prime} =\frac{\partial^2 F}{\partial u_i \partial u_j}$ and I get the sum over the column norms? Finally, $|\cdot|_1$ means the $L^1$ norm? –  Feb 03 '23 at 16:38
  • As I said, the full gradient, the transpose of the derivative/Jacobian, so that $G'$ is the Hessian matrix. Of course, if you take it apart into single components, then you get a larger upper bound. – Lutz Lehmann Feb 03 '23 at 16:41
  • I took a look closer and it is (almost) all clear. I got 2 more questions. 1) could you please justify why $|G^{\prime}(a+su) u|1\le |G^{\prime}(a+su)|{1, op} |u|_1$? (if it is straightforward, I apologize). 2) in the last one inequality, why did you get rid of the integral? Thank you. –  Feb 03 '23 at 17:01
  • I think https://math.stackexchange.com/questions/519279/why-is-the-matrix-norm-a-1-maximum-absolute-column-sum-of-the-matrix answers my first question. –  Feb 03 '23 at 17:32
  • 1
    Yes, the operator norm associated to a vector space norm satisfies this by definition. For some vector norms the operator norm has an easy formula. You are right, one could also take the upper bound inside the integral. But then there would be no mean-value theorem. – Lutz Lehmann Feb 03 '23 at 18:54
  • Lutz Lehmann, thank you. I got one very last question, if you can/want. Could you please edit justifying $G(a+u)-G(a)=\int_0^1 G^{\prime}(a+su) u ds$? I am unsuccessfully trying to prove it. Thank you in advance. –  Feb 03 '23 at 19:50
  • This is just the fundamental theorem with chain rule. Set $g(s)=G(a+su)$ then compute what $g'(s)$ is and insert into the fundamental theorem for $g(1)-g(0)$. – Lutz Lehmann Feb 03 '23 at 19:58