Proof intuition of Positive Linear Operators are Bounded. 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.
 A: Perhaps writing the proof in a slightly different form might offer a better insight:
Define $P:X\times X\rightarrow \mathbb R$ as
$$ P(x,y) := T(x)(y) + T(y)(x)$$
Observe that this is a symmetric, bilinear form that is positive semi-definite. Thus Cauchy-Schwarz gives us
$$ |P(x,y)| \leq \sqrt{ P(x,x)\cdot
P(y,y)}$$
Our aim is to show $T$ is continuous. This means $T(x)(y)$ is continuous in both $x$ and $y$. We are already given that $T(x)(y)$ is continuous in $y$ for a fixed $x$. What about the reverse?
Given $x_n \rightarrow 0$ do we have $T(x_n)(y)\rightarrow 0$ for a fixed $y$? Note
$$|T(x_n)(y)+T(y)(x_n)|=|P(x_n,y)|\leq \sqrt{ P(x_n,x_n)\cdot
P(y,y)}$$
If we knew $P(x_n,x_n) \rightarrow 0$ then we could conclude $T(x_n)(y)\rightarrow 0$ for a fixed $y$.
But it is unclear how to do this. This is where the power of the closed graph theorem comes in.

Closed Graph theorem: Let $L:X\rightarrow Y$ is a linear map between Banach spaces. If for all $x_n \rightarrow 0$ such that $L(x_n)\rightarrow y\in Y$ we have $y=0$ then $L$ is bounded and continuous.

Take $L=T$ and given $x_n\rightarrow 0$ such that $T(x_n)\rightarrow T_0$, observe that $$P(x_n,x_n) \rightarrow 2T_0(0)=0$$
Hence,  $T(x_n)(y)\rightarrow 0$ for any $y$ which means $T_0(y)=0$ for any $y$ and hence by closed graph theorem we get $T$ is continuous and bounded.
