Question about Bounded operator in a Normed Linear space Let $X$ be a normed space and $A \in B(X) $ be a bounded operator, Then $A$ is invertible if and only if $A$ is bounded below and Surjective.
I let $A$ be invertible then $A$ will be bounded below since
$\|x\|$= $\|A^-1(Ax)\| \leq \|A^{-1}\|\,\|Ax\|$ but how can we say that it is surjective?

If $A$ is invertible then for any $x\in X$ you have $x=A(A^{-1}x)$. So $A$ is surjective.
Conversely, if $A$ is bounded below you know $\|x \|≤c\|Ax \|$ for some $c$ and all $x$. So if $Ax=0$ then $\|x\|=0$, so $x=0$ and $A$ is injective. At this stage you are not entirely done because to show that $A$ is invertible not only you need it bijective, but you also need the inverse to be bounded. In general one uses the Inverse Mapping Theorem for these things, but we're you simply have
$$
\|A^{-1}x \|≤c\|AA^{-1}x\|=c\|x \|,
$$
and so $A^{-1}$ is bounded.
