$X$ is a Banach space, $T \in B(X)$ and $T^*$ be its adjoint, show that $ \lVert Tx \rVert \geq \lVert (T^*)^{-1}\rVert^{-1}\lVert x \rVert$ 
Let X be a complex Banach space. Let $T\in B(X)$ be a bounded linear operator on $X$. Let $T^*\in B(X^*)$ be the adjoint of $T$.
Prove: If $T^*$ is invertible, then for all elements $x\in X$,
$$ \|Tx \| \geq \| (T^*)^{-1}\|^{-1}\| x \|$$
and use the inequality to prove that $T$ is invertible

 A: Let $x\in X$. By Hahn-Banach there is $f\in X^\ast$ with $\|f\|=1$ and
$|f(x)|=\|x\|$. Then we obtain
$$\begin{align}
\|x\|&=|f(x)| \\
&=|(T^\ast)^{-1}(T^\ast(f))(x)|\\
&\le \|(T^\ast)^{-1}\||(T^\ast(f))(x)|\\
&=\|(T^\ast)^{-1}\||(f\circ T)(x)|\\
&\le\|(T^\ast)^{-1}\|\|T(x)\|,
\end{align} $$
which is equivalent to the inequality of the statement. This immediately implies that $T$ is injective and that if $(Tx_n)$ is a Cauchy sequence, also $(x_n)$ must be Cauchy. Thus, $T(X)$ is complete. 
It remains to be shown that $T(X)=X$. Suppose this is not the case. Then choose $x\in X\setminus T(X)$. Again by Hahn-Banach and the fact that $T(X)$ is closed, there is a functional $f\in X^\ast$ such that $f$ is zero on $T(X)$, but $f(x)\neq 0$. This is a contradiction, since $T^\ast$ is injective, but $T^\ast(f) = f\circ T = 0$.
A: Obviously we have, 
\begin{split}
\|x\| &=& \sup\{|\langle x,y\rangle| ; \|y\|=1\}\\
&=& \sup\{|\langle T^{-1}Tx,y\rangle| ; \|y\|=1\}\\
&=& \sup\{|\langle T^{-1}Tx,(T^{-1})^*y\rangle| ; \|y\|=1\}\\
&\le& \|Tx\|\sup\{\|T^{-1})^*y\| ; \|y\|=1\}\\
& = &\|Tx\|\|(T^{-1})^*\|
\end{split}
that is $$\|x\| |(T^{-1})^*\|^{-1}\le \|Tx\|$$
