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In the question Multivariate Mean Value Theorem Reference was written the following statement for $x,y\in \mathbb{R}^{n}$ \begin{equation} ||f(x) - f(y)||_q \leq \sup_{z\in[x,y]}||f'(z)||_{(q,p)}||x-y||_p, \end{equation}

where $z∈[x,y]$ denotes a vector $z$ contained in the set of points between $x,y\in \mathbb{R}^{n}$, and $||f′(z)||_{(q,p)}$ is the $L(p,q)$ norm of the derivative matrix of $f:\mathbb{R}^{n}→\mathbb{R}^{m}$ evaluated at $z$.

I have not found proof of this statement anywhere, including the link provided in the question above.

But I've found the following proof of the classical Mean Value Theorem ($p=q=2$) here Theorem 5.4.

My question is can this proof be used for arbitrary norms?

For cases of norm $p=q=1$ or $p=q=\infty$ it is obvious that the same proof can be used since it is easily can be obtained the following estimations for integrable vector-valued functions

$$ \|\int_0^1f(t)dt\|_1\le\int_0^1\|f(t)\|_1\,dt, $$ $$ \|\int_0^1f(t)dt\|_\infty\le\int_0^1\|f(t)\|_\infty\,dt. $$

But I don't know about such estimates for more general norms.

If such estimates are valid, then the proof should also be true in the more general case.

It would be great if someone could help me check this point or provide a link to the full proof of the fact above, or maybe provide their own proof.

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  • $\begingroup$ Have you tried? Where do you get stuck? $\endgroup$
    – Deane
    Commented Mar 17 at 12:39
  • $\begingroup$ @Deane, if $\|\int_0^1f(t)dt\|_p\le\int_0^1\|f(t)\|_p\,dt$ is valid for any $p\in [1,+\infty],$ then the statement above can be obtained exactly as in the proof of Theorem 5.4 above (only the norm changes). Am I right? $\endgroup$ Commented Mar 17 at 14:22
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    $\begingroup$ Could you tell me what you mean by $\|\int_0^1f(t)\,dt\|_p$? $\endgroup$
    – Deane
    Commented Mar 18 at 0:15
  • $\begingroup$ @Deane I've meant Norm $\sqrt[p]{(\int_0^1f_1(t)dt)^p+...+(\int_0^1f_m(t)dt)^p}$ of the integral of a vector-valued function of a one-dimensional argument $t$ $\endgroup$ Commented Mar 18 at 9:16
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    $\begingroup$ @Deane Yes, by applying Hölder's inequality I obtained these estimates for the integrals. As a result, the above proof of Theorem 5.4 can be used for the general case. The question can be closed $\endgroup$ Commented Mar 20 at 5:18

2 Answers 2

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By @Deane 's suggestion I've posted the proof of the statement above here. In essence, this proof simply repeats the one given in the link above with the norm $l_2$ replaced by a more general one.

Let $f:\mathbb{R}^{n}→\mathbb{R}^{m}$ be continuously differentiable on a domain containing the two points $x,y\in \mathbb{R}^{n}$ with a segment connecting them. Let $p,q\in [1,\infty]$

By theorem 5.3 in the link above we have the following representation for $f(x) - f(y)$ $$f(x) - f(y) = \begin{pmatrix}\int_0^1\nabla f_1(x+t(y-x))^T(y-x)dt\\...\\\int_0^1\nabla f_m(x+t(y-x))^T(y-x)dt\end{pmatrix}=\int_0^1\nabla f(x+t(y-x))^T(y-x)dt,$$ where $\nabla f_k$ is the gradient of $k$th component of $f$. Then

$$||f(x) - f(y)||_q = ||\int_0^1\nabla f(x+t(y-x))(y-x)dt||_q=\otimes$$ To estimate $\otimes$ we can apply $\|\int_0^1f(t)dt\|_p\le\int_0^1\|f(t)\|_p\,dt$ (to prove this inequality, it suffices apply Hölder's inequality for finite sums for the $||\sigma_f||^p_p,$ where $\sigma_f$ is the (finite) Riemann sum of $f$). Then we have

$$\otimes \leq \int_0^1||\nabla f(x+t(y-x))(y-x)||_qdt\leq$$ applying the estimate for the operator norm

$$ \leq \int_0^1||\nabla f(x+t(y-x))||_{(q,p)}||(y-x)||_pdt\leq\sup_{z\in[x,y]}||f'(z)||_{(q,p)}||x-y||_p,$$

where $f'(z)=\nabla f(z)$ is Jacobian matrix of $f$. The theorem is proven.

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  • $\begingroup$ Theorem 5.3 only applies if you assume some integrability condition on $f'$, which isn't the case in the statement of the mean value inequality. $\endgroup$ Commented Mar 21 at 14:08
  • $\begingroup$ @LázaroAlbuquerque You're right. I've added continuous differentiability to my statement here $\endgroup$ Commented Mar 21 at 15:06
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By @LázaroAlbuquerque 's suggestion I've posted the one more proof of the statement above here for differatiable function (more general case than the previous one).

Let $F:\mathbb{R}^{n}→\mathbb{R}^{m}$ be differentiable on a domain containing the two points $x,y\in \mathbb{R}^{n}$ with a segment connecting them. Let $p,q\in [1,\infty].$

By restriction of $F$ on $[x,y]$ we can consider parameterization $$f(t) = F(x+t(y-x)),$$ where $t\in[0,1].$ So, $f:[0,1]→\mathbb{R}^{m}$ is differentiable on $[0,1]$ with derivative $$f'(t)=\begin{pmatrix}\nabla F_1(x+t(y-x))^T(y-x)\\...\\\nabla F_m(x+t(y-x))^T(y-x)\end{pmatrix}=\nabla F(x+t(y-x))(y-x),$$

where $\nabla F(z)$ is Jacobian matrix of $F.$ Then, by applying second statement/proof from this (which can be proven in the same manner for $l_q$-norms with $q\in[1;\infty]$)

$$||F(x) - F(y)||_q = ||f(0) - f(1)||_q \leq \sup_{t\in[0,1]}||\nabla F(x+t(y-x))(y-x)||_q=\otimes$$

applying the estimate for the operator norm

$$\otimes \leq \sup_{t\in[0,1]}||\nabla F(x+t(y-x))||_{(q,p)}||(y-x)||_p=\sup_{z\in[x,y]}||F'(z)||_{(q,p)}||x-y||_p.$$

Thus we have a simpler proof for the more general case.

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  • $\begingroup$ Yup, that's it. $\endgroup$ Commented Mar 21 at 22:09
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    $\begingroup$ @LázaroAlbuquerque thanks for the help $\endgroup$ Commented Mar 22 at 4:47

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