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Let $X$ be a normed space. Let the function $J:X \rightarrow X'' $ be defined by $$J(x)(x')=x'(x)\ \forall\ x'\in X' $$

where $X''=\{f:X'\rightarrow\mathbb{C} \mid \hbox{$f$ is bounded linear} \}$ and $X'$ is the dual of $X$.

Is $J$ injective?

I was trying this,

Suppose we have $x,y\in X $such that $$J(x)=J(y)$$ $$\Rightarrow J(x)(x')=J(y)(x') $$ $$\Rightarrow x'(x)=x'(y)$$ $$\Rightarrow x'(x-y)=0$$ But to get $x=y $ we must have $x'$ to be injective. I am stuck here. Is this the correct way to go about it?

Siminore
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Epsilon
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  • Note that you have the relation $x'(x-y)=0$ for all $x'\in X'$. – Deliasaghi Sep 16 '15 at 09:49
  • It's important to see that this function is well-defined, and so you can see its linearity and injectivety. – Deliasaghi Sep 16 '15 at 10:10
  • Maybe I presume too much, but before tackling this sort of question, it could be a good idea to review basic facts about linear algebra, like "a linear function is injective iff its kernel is trivial". – Najib Idrissi Sep 16 '15 at 13:55
  • @Najib I think both the criterias are equivalent. – Epsilon Sep 17 '15 at 07:33
  • Yes of course, but my point is that it's needlessly complicated to take two elements $x$, $y$ st $J(x) = J(y)$, then re-arranging as $J(x-y) = 0$ and eventually proving the kernel is trivial anyway, then $x-y \in \ker J \implies x=y$. Simply start with $J(x) = 0$ and prove $x=0$. – Najib Idrissi Sep 17 '15 at 07:36

2 Answers2

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Please pay attention: you end up with the statement

$x'(x-y)=0$ for all $x' \in X'$.

This implies $x-y=0$, as a corollary of the Hahn-Banach theorem.

Siminore
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The mapping $J$ defined as above is always injective. And it is onto only if the space $X$ is reflexive (this is the definition for reflexive Banach space actually). That from $x'(x-y)=0\Rightarrow x=y$ is because the dual space is a total set and it follows from Hahn-Banach Theorem.

The corollary is:

If $x\in X\,\,$ and $x'(x)=0,\quad\forall x'\in X'$ then $x=0$

Svetoslav
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