0
$\begingroup$

Let ${(\phi_n(.),\psi_n(.))}_{n\geq 1}$ be a Riesz basis in $H^1_0(0,1) \times L^2(0,1)$. My question is the following: If we multiply the basis by the matrix $e^{Mx}$, $x \in (0,1)$ where $M$ is a constant matrix of dimension $2$, is this still a Riesz basis? I can prove that the operator $T: H^1_0(0,1) \times L^2(0,1) \rightarrow H^1_0(0,1) \times L^2(0,1)$ taking $(f,g)\rightarrow e^{Mx}(f,g)$ is bounded from above, but not from below. Is this statement true?.

Thank you in advance.

Improve this question
$\endgroup$
4
  • 2
    $\begingroup$ What is $x$ here? A number? If so, then $T$ will not be an operator into $H^1_0(0,1) \times L^2(0,1)$ unless $x=0$ or the matrix $M$ is lower triangular. $\endgroup$
    – Iosif Pinelis
    Commented Apr 5, 2021 at 19:59
  • 1
    $\begingroup$ I am surprised at what you can prove. Given an arbitrary pair $(\phi, \psi)\in H^0_1 \times L^2$, and consider the matrix valued function $e^{Mx}$. If you write $(\tilde{\phi}, \tilde{\psi})$ for $e^{Mx}(\phi, \psi)$, generally $\tilde{\phi}$ is not going to be vanishing on the boundary. $\endgroup$
    – Willie Wong
    Commented Apr 5, 2021 at 20:00
  • $\begingroup$ thank you. I can see it now. $\endgroup$
    – Gustave
    Commented Apr 5, 2021 at 20:11
  • $\begingroup$ And what if I consider the space $H^1_0(0,1)^2$ instead of $H^1_0(0,1)\times L^2(0,1)$?. I think that this is true. Please, correct me if I'm mistaken. Thank you. $\endgroup$
    – Gustave
    Commented Apr 6, 2021 at 13:48

0

Reset to default

You must log in to answer this question.