Find a tight frame of exponentials for $L^2(T)$, where $T \subset \mathbb{R}^2$ is a triangle with vertices $(0,1)$, $(1,0)$, and $(-1,0)$.

Question

Find a tight frame of exponentials for $L^2(T)$, where $T \subset \mathbb{R}^2$ is a triangle with vertices $(0,1)$, $(1,0)$, and $(-1,0)$.

Normally, I would do is find a matrix representation and then set the elements of the diagonal equal to each other and then set the other elements of the diagonal equal to zero. How would I do it in this case?

Answer 1

Recall that a Parseval frame is the image of an orthonormal basis under orthogonal projection from some larger Hilbert space. So, the following approach can be taken.

1. The exponentials $\{\exp(\pi i mx+2\pi i ny):m,n\in\mathbb Z\}$ form an orthonormal basis for the rectangle $R = [-1,1]\times [0,1]$.
2. There is a natural projection from $L^2(R)$ onto $L^2(T)$, namely restriction to $T$.