We set :

  • $\phi_1, \phi_2 : \mathbb{R} \to \mathbb{R}^M$ with compact support
  • $\theta \in \mathbb{R}^M$ non-zero, $\theta_{p-1} = (\operatorname{sign}(\theta_i)|\theta_i|^{p-1})_{1 \leq i \leq M} \in \mathbb{R}^M$ with $p \geq 2$.

We assume that the matrix

$$ \int_\mathbb{R} \phi_1 \phi_1^T - \int_\mathbb{R} \phi_2 \phi_2^T $$

is positive definite. I then want to show that (if it is true) :

$$ \int_\mathbb{R} \theta_{p-1}^T \phi_1(t) \phi_2(t)^T \theta < \int_\mathbb{R} \theta_{p-1}^T \phi_1(t) \phi_1(t)^T \theta $$

If $p=2$ the problem is easy with Cauchy-Schwarz on the second term + assumption. I struggle to prove the general form (which seems to be true according to some code but I am not sure).


The proof of the case $p=2$ :

$$ \int_\mathbb{R} \theta^T \phi_1(t) \phi_2(t)^T \theta \leq \sqrt{\int_\mathbb{R} (\theta^T \phi_1(t))^2 \int_\mathbb{R} (\theta^T \phi_2(t))^2 } $$ using Cauchy-Schwarz and $$ \int_\mathbb{R} (\theta^T \phi_2(t))^2 < \int_\mathbb{R} (\theta^T \phi_1(t))^2 $$ Finally : $$ \int_\mathbb{R} \theta^T \phi_1(t) \phi_2(t)^T \theta < \sqrt{\int_\mathbb{R} (\theta^T \phi_1(t))^2 \int_\mathbb{R} (\theta^T \phi_1(t))^2 } = \int_\mathbb{R} (\theta^T \phi_1(t))^2 $$

Attempt of a general proof (false !!!)

We write : $$ \theta_{p-1} = D \theta $$ with $$D = \operatorname{DiagonalMatrix}(|\theta_i|^{p-2})$$

From now (third step is false):

$$\begin{aligned} \int_\mathbb{R} \theta_{p-1}^T \phi_1(t) \phi_2(t)^T \theta &= \int_\mathbb{R} \phi_1(t)^T \theta_{p-1} \theta^T \phi_2(t) \\ &= \int_\mathbb{R} \phi_1(t)^T D \theta \theta^T \phi_2(t) \\ &= \int_\mathbb{R} \phi_1(t)^T \sqrt{D} \theta \theta^T \sqrt{D} \phi_2(t) \\ &\leq \sqrt{ \int (\phi_1(t)^T \sqrt{D} \theta)^2 \,dt \cdot \int (\phi_2(t)^T \sqrt{D} \theta)^2 dt } \\ &< \sqrt{ \int (\phi_1(t)^T \sqrt{D} \theta)^2 \,dt \cdot \int (\phi_1(t)^T \sqrt{D} \theta)^2 \,dt } \end{aligned} $$

Then $$ \begin{aligned} \int_\mathbb{R} \theta_{p-1}^T \phi_1(t) \phi_2(t)^T \theta &< \sqrt{ \int (\phi_1(t)^T \sqrt{D} \theta)^2 \, dt \cdot \int (\phi_1(t)^T \sqrt{D} \theta)^2 \, dt } \\ &= \int (\phi_1(t)^T \sqrt{D} \theta)^2 \, dt \\ &= \int \theta^T \sqrt{D} \phi_1(t) \phi_1(t)^T \sqrt{D} \theta \, dt \\ &= \int \theta^T D \phi_1(t) \phi_1(t)^T \theta \, dt \\ &= \int \theta_{p-1}^T \phi_1(t) \phi_1(t)^T \theta \, dt \end{aligned} $$

  • $\begingroup$ What does it mean that the inequality holds in the sense of positive definiteness? $\endgroup$ Jun 7 at 11:28
  • $\begingroup$ It means that the matrix ∫ϕ1ϕ1T - ∫ϕ2ϕ2T is positive definite. I modified the post to make it clearer, thank you. $\endgroup$ Jun 7 at 11:31
  • $\begingroup$ In the last formula, there should be $\phi_2(t)\phi_2(t)^T$ in the first integral, right? Also, the whole integral business seems kind of redundant, no? $\endgroup$
    – Hannes
    Jun 7 at 13:40
  • $\begingroup$ Yes, the last formula is true : it is $\phi_1 \phi_2^T$ and not $\phi_2 \phi_2^T$. I edited the post to put the case $p=2$ and I think I found the general case proof. Concerning the integrals, I think they are mandatory because the matrix $\phi_1 \phi_1^T$ with $t$ fixed has rank 1 and $\int \phi_1 \phi_1^T$ can have any rank. $\endgroup$ Jun 7 at 14:38

1 Answer 1


$\newcommand\th\theta\newcommand\R{\mathbb R}$This is false for any real $p>2$ (actually, this is false for any real $p>1$ such that $p\ne2$).

Indeed, if this were true, then, by continuity, we could replace "positive definite" with "positive semi-definite", at the same time replacing $<$ in the desired inequality by $\le$.

Next, assuming that each of the functions $\phi_1$ and $\phi_2$ takes only finitely many values, the positive answer to your question would imply the positive answer to the following question:

Suppose $A$ and $B$ are $m\times n$ real matrices such that
$$\|A\th\|\le\|B\th\|\quad\text{for all $\th=[\th_1,\dots,\th_n]^\top\in\R^n=\R^{n\times1}$ },\tag{1}\label{1}$$ where $\|\cdot\|$ is the Euclidean norm. Let $p\in(1,\infty)$. Does it then necessarily follow that $$(B\th^{[p-1]})\cdot(A\th)\le(B\th^{[p-1]})\cdot(B\th)\tag{2}\label{2}$$ for all $\th=[\th_1,\dots,\th_n]^\top\in\R^n$, where $\cdot$ is the dot product, $\th^{[p-1]}:=[\th_1^{[p-1]},\dots,\th_n^{[p-1]}]^\top$ and $u^{[p-1]}:=|u|^{p-1}\text{sign}\, u$ for real $u$?

(See the detail on this reformulation at the end of this answer.)

However, it is easy to find (say) $2\times2$ real matrices $A$ and $B$ and some $\mu\in\R^2$ such that \eqref{1} holds, $\|A\mu\|=\|B\mu\|$, $A\mu$ is in the same direction with $B\mu^{[p-1]}$, but $B\mu$ is not in the same direction with $B\mu^{[p-1]}$. Then \eqref{2} will fail to hold for $\th=\mu$.

E.g., suppose that $B=I_2$ and $\mu=[2,1]^\top$, so that $B\mu=\mu$ is not in the same direction with $B\mu^{[p-1]}=\mu^{[p-1]}$. Let now $A$ be the rotation matrix such that $A\mu$ is in the same direction with $B\mu^{[p-1]}=\mu^{[p-1]}$. Then \eqref{1} will hold and, in particular, we will have $\|A\mu\|=\|B\mu\|$, whereas \eqref{2} will fail to hold for $\th=\mu$.

Detail on the highlighted reformulation: This follows because $$\int_\R\rho^\top\phi\psi^\top\th \\ =\sum_{j,k}\rho_j\th_k\int_\R\phi^{(j)}\psi^{(k)} =\sum_{j,k}\rho_j\th_k\sum_i a_{ij}b_{jk} =(A\rho)\cdot(B\th),$$ where $A:=[a_{ij}]$, $B:=[b_{ik}]$, $\rho=[\rho_1,\dots,\rho_M]^\top$, $\th=[\th_1,\dots,\th_M]^\top$, $\phi=[\phi^{(1)},\dots,\phi^{(M)}]^\top$, $\psi=[\psi^{(1)},\dots,\psi^{(M)}]^\top$, $\phi^{(j)}=\sum_{i=1}^m a_{ij}1_{[i,i+1)}$, and $\psi^{(k)}=\sum_{i=1}^m b_{ik}1_{[i,i+1)}$.

  • $\begingroup$ Thank you for your answer ! It seems to be accurrate unfortunately for me. I will spend some time about it $\endgroup$ Jun 7 at 15:31
  • $\begingroup$ I would be curious if you have an idea about where i am cheating in my "proof" $\endgroup$ Jun 7 at 15:58
  • $\begingroup$ @OrsoForghieri : I don't like to read proofs. However, I think the equality on the first line where $\sqrt D$ appears is not true in general. Have you checked my proof? $\endgroup$ Jun 7 at 16:44
  • $\begingroup$ thank you for asking, my proof is false and I found where. Your example seems to works. Thank you ! $\endgroup$ Jun 23 at 9:08

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