Showing bound on operator $||Tx|| \ge \frac{||x||}{||(T')^{-1}||}$

Question

T is an bounded linear operator on a Banach space, X, and I'm given that it's adjoint $T'$ is invertible, I'm trying to show: $||Tx|| \ge \frac{||x||}{||(T')^{-1}||}$. I feel like this should be simple but I'm finding it very frustrating!

My ideas so far: $T'$ is invertible so we have that for any $x \in X$, $(T')^{-1} T g (x) = g(x)$ and then taking norms of this, and using the fact everything is bounded we find: $||g(x)|| \le ||(T')^{-1}|| ||T|| ||g|| ||x||$. I'm having trouble then deducing the inequality from this even though I feel it should follow?

Thaanks

Answer 1

Fix $x\in X$. If $f\in X'$ then $|f(x)|=|[(T')^{-1}T'f](x)|=|(T')^{-1}f(Tx)|\leq\|(T')^{-1}\|\|f\|\|Tx\|$. Hence $$\|Tx\|\geq\frac{|f(x)|}{\|f\|}\cdot\frac1{\|(T')^{-1}\|}.$$ By a corollary to the Hahn-Banach theorem, there exists $f\in X'$ with $\|f\|=1$ and $f(x)=\|x\|$, completing the proof. This method manages to avoid any spectral theory.

Answer 2

Since $T'$ is invertible, $\operatorname{spec}(T')=\operatorname{spec}(T)$ is bounded away from zero, and we can define: $$\lambda^- = \min_{\lambda\in\operatorname{spec}(T)}|\lambda|.$$ Obviously we have: $$ \| T x\| \geq \lambda^{-}\|x\|,$$ hence it is sufficient to check that: $$ \lambda^{-} = \frac{1}{\|T'^{-1}\|} $$ that follows from: $$ \|T^{-1}\| = \max_{\lambda\in\operatorname{spec}{T'^{-1}}}|\lambda| =\frac{1}{\min_{\lambda\in\operatorname{spec}{T'}}|\lambda|}=\frac{1}{\lambda^-}.$$