Given $X,Y$ be normed spaces over $\mathbb k$ (where $\mathbb k$ could be $=\mathbb C$ or $\mathbb R$). Let $T\in L(X,Y) $ (a continuous and linear operator) then we define $T':Y'\to X'$ by $T'y'=y'T$, it's called the transpose of $T$.

Given a Banach space $X$ a projection from $X$ to a closed subspace $M\le X$ is a continuous linear operator $P:X\to X$ such that $Range(P)=M$ and $P^2=P$ thus $P^2$ is the identity on $M$.

Important: When we say that $P$ is a projection to $M$ we consider $M$ closed for some important reasons, that $P$ is surjective over $M$, and that $P$ is the identity only on $M$ in fact $Ker(I-P)=M$.

Prove that if $X$ is a Banach space, then there exist a projection of norm 1, from $X'''$ to $X'$, when we consider that $X'\subset X'''$ under the canonical injection $Jx' (x'')=x''(x')$.

I have no idea how to attack this problem, I don't know any non-zero map either.

  • $\begingroup$ Can you name a (nonzero) linear map $X''' \to X'$? $\endgroup$ – Daniel Fischer Nov 1 '13 at 21:26


$$\Phi \colon X \to X'';\qquad \Phi(x)(\lambda) = \lambda(x)$$

is an isometry (in particular has norm $1$). What does that tell you about $\Phi'$?

  • $\begingroup$ In the same way, we can define $J:X'\to X'''$ by $J(x')(x'')=x''(x')$ thus $J(x')\in X'''$ and in fact $J$ in an isometry. But how can I use this? $\endgroup$ – Castaroth Nov 1 '13 at 22:03
  • $\begingroup$ How can you use $\Phi$ to get a map $X''' \to X'$? $\endgroup$ – Daniel Fischer Nov 1 '13 at 22:04

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