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).


***Edit***


**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
$$ \begin{aligned} 
\int\_\mathbb{R} \theta\_{p-1}^T \phi\_1(t) \phi\_2(t)^T \theta &< \sqrt{ \int (\phi\_1(t)^T \sqrt{D} \theta)^2 \, dt \cdot \int (\phi\_1(t)^T \sqrt{D} \theta)^2 \, dt } \\
&= \int (\phi\_1(t)^T \sqrt{D} \theta)^2 \, dt \\
&= \int \theta^T \sqrt{D} \phi\_1(t) \phi\_1(t)^T \sqrt{D} \theta \, dt \\
&= \int \theta^T D \phi\_1(t) \phi\_1(t)^T \theta \, dt \\
&= \int \theta\_{p-1}^T \phi\_1(t) \phi\_1(t)^T \theta \, dt
\end{aligned} $$



Cauchy-Schwarz-like inequality with a power $p$ term
$\newcommand\th\theta\newcommand\R{\mathbb R}$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 $\psi^{(k)}=\sum\_{i=1}^m b\_{ik}1\_{[i,i+1)}$.


