I have this definition in my book:
Definition:
Let X,Y be normed linear spaces. An operator $T \in B(X,Y)$ is said to be invertible if there exists $S \in B(Y,X)$ such
that $ST=I_X, TS=I_Y$, in which case S is the inverse of T and is
denoeted $T^{-1}$.
B is the set of bounded linear operators from the respective space to the other.
One problem with this definition is that it assumes the existence of S in $B(X,Y)$. I was thinking is it enough to assume the existence of S in the set of all operators from Y to X, and then show that if $ST=I_x, TS=I_Y$ it is in $B(X,Y)$? I was able to show that if T is linear, then S is also linear, but I am not able to show that S is bounded, is it nececarrily bounded or must we assume that it is bounded(and hence in $B(Y,X$))?
That is, if $\|T(x)\|\le \|T\|\|x\|, T(ax_1+bx_2)=aT(x_1)+bT(x_2)$. I can then show that $S(ay_1+by_2)=aS(y1)+bS(Y_2)$, but can I show that there exist an M such that:
$\|S(y)\|\le M\|y|, \forall y \in Y$?
Best Answer
If an inverse of any kind exist, $T$ is a bijection. As a consequence of the open mapping theorem, a bijective operator is bounded from below, meaning that there is $c>0$ such that $\|Tx\|\ge c\|x\|$ for all $x$. This and the property $TS=I$ imply that $S$ is bounded.