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

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

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

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

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

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:

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

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

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

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

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

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