Q:

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

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

A:

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.

