Theorem: Let $X$ and $Y$ be normed spaces such that $X\cong Y$. Let $\phi:X\rightarrow Y$ be an isometric isomorphism. Then the dual map ${\phi}^*:{Y}^*\rightarrow{X}^{*},\lambda\mapsto\lambda\circ\phi$ is an isometric isomorphism.

I have already shown that ${\phi}^*$ is surjective and here is my attempt to show ${\phi}^*$ is isometric:

Claim that $\{\phi(x):\|x\|=1\}=\{y\in Y:\|y\|=1\}$. Let $y\in\{\phi(x):\|x\|=1\}$. Then $y\in Y$ and there exists $x\in X$ such that $\phi(x)=y$ and $\|x\|=1$. Since $\phi:X\rightarrow Y$ is isometric, $\|x\|=\|\phi(x)\|=\|y\|=1$. Hence, $y\in\{y\in Y:\|y\|=1\}$. Conversely, let $y\in\{y\in Y:\|y\|=1\}$. Then there exists $x\in X$ such that $\phi(x)=y$ since $\phi:X\rightarrow Y$ is surjective. Since $\phi:X\rightarrow Y$ is isometric, $\|x\|=\|\phi(x)\|=\|y\|=1$. Hence, $y\in\{\phi(x):\|x\|=1\}$. Therefore, $\{\phi(x):\|x\|=1\}=\{y\in Y:\|y\|=1\}$. By the claim above, for all $f\in Y^*$, we have $$\|\phi^*(f)\|=\|f\circ\phi\|=\sup\{|f(\phi(x))|:\|x\|=1\}=\sup\{|f(y)|:\|y\|=1\}=\|f\|.$$

But my professor told me I should directly show ${\phi}^*$ is isometric. I think my attempt is direct enough. Is there any other way to show ${\phi}^*$ is isometric? Thank you in advance!

  • 1
    $\begingroup$ Maybe (s)he meant that your argument is too verbose? I would accept just the last line by itself as a valid proof (maybe with an explanation that the last equality follows since $\phi$ is a isometry and a bijection). $\endgroup$ – J.R. May 25 '18 at 21:00

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