Let $T: X \to Y$ be a bounded linear map between Banach spaces $X,Y$. Generally, the operator norm is something that is notoriously difficult to compute, as we have two things to show. First we need: $$ \|Tx\|_Y \leq \|T\|_{\text{op}}\|x\|_X $$ I.e, $\|T\|$ is an upper bound on the amount a vector $x$ can be "stretched". We must also show that it is a least upper bound/supremum. Generally, the strategy to do so is to find $x'$ such that: $$ \|Tx'\|_Y \geq \|T\|_{\text{op}}\|x'\|_X $$ My professor referred to this as "saturating" the inequality, meaning we have found some $x'$ for which equality holds. If there were a strictly lower upper bound than $\|T\|$, we would have $\|Tx'\| > \|T\|\|x'\|$, meaning that the norm was not actually an upper bound, and violating our initial assumption.
Now, recently, I read an article regarding James' Theorem, which stated that bounded linear functionals attained their norm (i.e saturated the inequality $|\langle f,x\rangle| \leq \|f\|\|x\|$) if and only if the space $X$ is reflexive.
My question then is, does the saturation method not account for all cases? I.e, can we find linear operators which have a certain norm, but do not satisfy the "reverse" inequality? Does anyone know of any examples? And if they do exist, how is one to verify that they have computed the correct norm?