I got stuck on proving this exercise for quite a while and found the proof of this statement Positive operator is bounded :

For a real Banach space $E$ let $T:E\rightarrow E'$ be a positive operator in the sense that $(Tx)(x)\geq 0$ for all $x\in E$. Show that $T$ is bounded.

My question is, how exactly did we even come up with this proof construction at the first place? What I did was starting with $x_n \to x_0 \in E$ and $T(x_n) \to f \in E'$ and attempt to show $T(x_0) = f$. Namely, $T(x_0)(y) = f(y)$ for all $y \in E$. And then I literally just got stuck. The relation of the positiveness identity does not seem to "naturally" arise from this point (except from I could recognize that this might help us to show two-sided inequality). But how could we proceed beyond this point naturally? Or should I just take this proof as one of the proofs that does not have intuition at all and move on.

  • 1
    $\begingroup$ A heads up: There is no proof which "does not have intuition at all"! $\endgroup$
    – Qi Zhu
    Feb 9 at 11:17
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    $\begingroup$ Globally defined linear operators are either very inaccessible or continuous. Here you see that the algebraic property of positivity of the associated quadratic form is enough for continuity. The intuition is that it is very hard to find a globally defined (on a Banach space) unbounded operator that has any nice or useful properties. Every unbounded operator I could ever work with has a dense domain. $\endgroup$
    – s.harp
    Feb 9 at 23:25


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