|
Post Closed as "too localized" by Yemon Choi, Bill Johnson, Andres Caicedo, Andrew Stacey, Mariano Suárez-Alvarez
|
||||
|
|
||||
|
2 | added 42 characters in body | ||
|
Let $B = l_\infty$ be the Banach space of bounded sequences and the operator $$A: (x_1,x_2,x_3,\dots) \mapsto (x_1,x_2/2,x_3/3,\dots).$$ This operator is bounded. Therefore its image $B':=A(l_\infty)$ B':=A(l_\infty)$, with the norm inherited from $l_\infty$, is a Banach space and $A:B\to B'$ is a bounded bijection. By the bounded inverse theorem, $A^{-1}$ is bounded, which is obviously not true. What is the mistake in this reasoning? |
||||
|
|
1 |
|
||
question on the bounded inverse theorem for banach spacesLet $B = l_\infty$ be the Banach space of bounded sequences and the operator $$A: (x_1,x_2,x_3,\dots) \mapsto (x_1,x_2/2,x_3/3,\dots).$$ This operator is bounded. Therefore its image $B':=A(l_\infty)$ is a Banach space and $A:B\to B'$ is a bounded bijection. By the bounded inverse theorem, $A^{-1}$ is bounded, which is obviously not true. What is the mistake in this reasoning?
|
||||
