# Cauchy-Schwarz-like inequality with a power $p$ term

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}

• What does it mean that the inequality holds in the sense of positive definiteness? Jun 7 at 11:28
• It means that the matrix ∫ϕ1ϕ1T - ∫ϕ2ϕ2T is positive definite. I modified the post to make it clearer, thank you. Jun 7 at 11:31
• 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? Jun 7 at 13:40
• 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. Jun 7 at 14:38

$$\newcommand\th\theta\newcommand\R{\mathbb R}$$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 $$\psi^{(k)}=\sum_{i=1}^m b_{ik}1_{[i,i+1)}$$.

• Thank you for your answer ! It seems to be accurrate unfortunately for me. I will spend some time about it Jun 7 at 15:31
• I would be curious if you have an idea about where i am cheating in my "proof" Jun 7 at 15:58
• @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? Jun 7 at 16:44
• thank you for asking, my proof is false and I found where. Your example seems to works. Thank you ! Jun 23 at 9:08