We set :
 - $\phi_1, \phi_2 : \mathbb{R} \mapsto \mathbb{R}^M$ with compact support
 - $\theta \in \mathbb{R}^M$ non-zero, $\theta_{p-1} = (sign(\theta_i)|\theta_i|^{p-1})_{1 \leq i \leq M} \in \mathbb{R}^M$ 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**

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

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

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