0
$\begingroup$

This question is an exact duplicate of:

Let $T : V \rightarrow U$ be a bounded map between two normed spaces. Let $T^* : U^* \rightarrow V^*$ be defined by $T^*(f) = f\circ T$ for all $f\in U^*$ (the adjoint map).

My Question: If we identify $V$ and $U$ with their canonical images in $V^{**}$ and $U^{**}$ prove that the restriction of $T^{**}$ to $V$ coincides with $T$.

I think I want to try and use the canonical mappings $V \rightarrow V^{**}$ and $U \rightarrow U^{**}$ but I am not entirely sure what the canonical images would be. The restriction would be $T^{**}\cap V$ and I am not sure where that comes into play...

I would really appreciate some help on this proof. hank you.

$\endgroup$

marked as duplicate by Arnaud D., Lee David Chung Lin, user10354138, ArsenBerk, Don Thousand Nov 8 '18 at 17:07

This question was marked as an exact duplicate of an existing question.

1
$\begingroup$

I will instead write $v^{**}\in V^{**}$ and $u^{**}\in U^{**}$ as the images of $v\in V$ and $u\in U$, respectively, under the duality maps $V\to V^{**}$ and $U\to U^{**}$. We note that $$T^{**}v^{**}(\varphi)=v^{**}\circ T^*(\varphi) =v^{**}(\varphi \circ T)=\varphi\circ T(v)=\varphi(Tv)=(Tv)^{**}(\varphi)$$ for all $v\in V$ and $\varphi \in U^*$. Therefore, $T^{**}v^{**}=(Tv)^{**}$ for every $v\in V$, and the conclusion follows.

$\endgroup$
  • $\begingroup$ Thank you very much. I really appreciate the help on this one. $\endgroup$ – MathIsHard Nov 7 '18 at 23:58

Not the answer you're looking for? Browse other questions tagged or ask your own question.