I'm not in the mathematics field and not very comfortable with strict mathematical formalism. The information I find on the Internet includes so many technical terms that might take ages for me to understand, so I hope someone can explain the following in a intuitive manner
Suppose $T$ is a bounded, invertible linear transformation so that the following is true
$||T(x)||=||Tx||\leq M ||x||$
I want to prove that
$||Tx||\geq N ||x||$
When can I assume that $T^{-1}=-T$ in order to prove the above relation? I know this holds for Hilbert transformations, but I'm not sure I can assume that the transformation is of that kind.
In general, there is no reason that $T^{-1} = -T$, even if $T$ is a map between Hilbert spaces (or even between one-dimensional vector spaces). However, if $T \colon X \rightarrow X$ is bounded and invertible linear map when $X$ is a Banach space (in particular, this holds if $X$ is a Hilbert space) then the open mapping theorem guarantees that the inverse $T^{-1}$ is bounded. Thus, there exists $N > 0$ such that $||T^{-1}(y)|| \leq N||y||$ for all $y \in X$. If we plug in $y = T(x)$ then we get $||x|| \leq N||T(x)||$ or $||T(x)|| \geq \frac{1}{N} ||x||$ for all $x \in X$.