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Let $X$ be a Banach space and let $A\colon X\to X^*$ be a linear operator satisfying $$(Ax)(x)\geq 0\quad\forall x\in X.$$ Show that $A\in L(X,X^*)$.

Here, $L(X,X^*)$ stands for the set of bounded linear transformations from $X$ to $X^*$ (the set of bounded linear functionals on $X$). By considering the identity $(A(x+y))(x+y) \geq 0$ and $(A(x-y))(x-y) \geq 0$ (I think) I was able to show that $A$ was bounded if and only if the expression $(Ax)(x)$ was bounded (across $|x| \leq 1$). But this doesn't seem to help much as I don't know how to show this expression is bounded. Does anyone have any ideas of how I could solve this one?

EDIT:

I asked this question on Reddit as well and someone proposed the following counter example that I cannot find fault in.

Let bₙ, n ∈ ℕ, be a countable set of linear independent vectors and bⁿ their duals. Complete bₙ to an algebraic basis of X. Set Abₙ = n⋅bⁿ and 0 on the other basis vectors

It'd be nice if someone could help explain what's wrong with this logic (or confirm that this question is in fact faulty as stated).

Rioghasarig
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    Please don’t use images; they aren’t searchable, and some screen readers cannot handle them, making your post inaccessible to users who use screen readers. – Arturo Magidin Dec 20 '20 at 01:14
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    Hint: apply the closed graph theorem. – Matematleta Dec 20 '20 at 02:22
  • The counterexample seems to assume that $b^n\in X^\ast$ and $|b^n|=1$, which is not the case. – MaoWao Dec 20 '20 at 11:05
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    See the answer to this question: https://math.stackexchange.com/q/3558088/152424 for a proof – s.harp Dec 20 '20 at 14:18
  • @MaoWao

    But what if we take for example $X = \ell^2$ and $b_n = e_n$ where $e_n$ is the $n$-th standard basis vector. Then isn't $b^n = e_n^T$. That is $b^n(x)$ just returns the $i$-th coordinate of $x$. Doesn't this work as a counterexample with $|b^n| = 1$?

     – Rioghasarig Dec 20 '20 at 16:53
  • Oh, wait I'm dumb. I just realized the "counter-example" won't necessarily satisfy the stated condition at all. – Rioghasarig Dec 20 '20 at 17:58

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