Simplifying this inequality with orthonormality relation

57 Views Asked by At

I am trying to simplify the following inequality

\begin{equation} \langle (a_0e_0+a_1e_1)|a_0e_0 +a_1e_1|^2, \overline{e}_1\rangle \end{equation}

where $a_ie_i$ are eigenvalue-eigenfunction pairs, and we use the $L^2$ inner product. Both $a_i$ and $e_i$ are complex and the orthogonality relation is such that $\langle e_i, \tilde{e}_j \rangle = \delta_{ij}$ where $\delta_{ij}$ is the Kronecker delta. It is that the eigenfunctions depend only on space, and that the eigenvalues depend only on time.

Also, with regard to the $L_2$ norm, I also want to know how to show that

\begin{equation} \| (|u| |v| w )\|_{L^2} \leq \|u\|_{L^2} \| v \|_{L^2} \|w\|_{L^2} \end{equation} for all $u,v,w \in H^1_0(\mathbb{R},\mathbb{C}) \cap H^2( \mathbb{R},\mathbb{C})$.

Any resources as to whether this is possible for the nonlinearity for the Navier-Stokes flow would be appreciated! I feel as though I need a crash course in inequalities!

Kindest regards,

Catherine

1

There are 1 best solutions below

4
On

You can do \begin{align*} \langle (a_0e_0+a_1e_1)|a_0e_0 +a_1e_1|^2, \overline{e}_1\rangle&= \left[\langle a_0e_0|a_0e_0 +a_1e_1|^2, \overline{e}_1\rangle+\langle a_1e_1|a_0e_0 +a_1e_1|^2,\overline{e}_1\rangle\right]\\ &=\left[a_0\langle e_0|a_0e_0 +a_1e_1|^2, \overline{e}_1\rangle+a_1\langle e_1|a_0e_0 +a_1e_1|^2,\overline{e}_1\rangle\right], \end{align*} but I think that's as far as you can go. Alternatively, you could try to write out $|a_0e_0+a_1e_1|^2=(a_0e_0+a_1e_1)(\overline{a}_0\overline{e}_0+\overline{a}_1\overline{e}_1),$ but it's not clear where that would lead you. It is not the case that $\langle e_i,e_j\rangle=\delta_{ij}$ implies $\langle e_i,\overline{e}_j\rangle=\delta_{ij}$. Moreover, all integrals in sight have four functions multiplied together inside! It would be impossible to use orthogonality to simplify any of them.

[EDIT] Reevaluating the following reasoning...

For your inequality there, you can simply apply Cauchy-Schwarz twice: $$\|uvw\|\le \|u\| \|vw\|\le \|u\| \|v\| \|w\|$$