Đang chuẩn bị liên kết để tải về tài liệu:
Spectral Theory

Không đóng trình duyệt đến khi xuất hiện nút TẢI XUỐNG Tải xuống

Example 1.1 Define, for h ∈ R, the operator τh on L2(R) by τhf(x) = f(x − h). Show that τh is bounded. Obviously, τh is linear, and it follows from τhf 22 = +∞ −∞ |f(x − h)|2dx = +∞ −∞ |f(x)|2dx = f 22 , that Tf 2 = f 2 for all f ∈ L2(R), hence T = 1. Remark 1.1 Here we add that τh is also regular. In fact, if τhf = 0, then f(x−h) = 0 for all x ∈ R, thus f ≡ 0. This shows that τh is injective, hence the inverse operator exists. Then we get by the change of variable y = x − h, i.e