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Let $X$ and $Y$ be normed spaces. And $L_1 : X \to Y$ and $L_2 : X \to Y$ be bounded linear maps. If the two maps have the same dual map, that is $L_1^*=L_2^*$, then is it possible to conclude that $L_1=L_2$? I cannot find a way to prove it, but am quite suspicious. Could anyone help me?

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First of all, it is a linear problem. So it is the same as asking if $L^* = 0$ is $L = 0$? If $L \ne 0$, then there is an $x \in X$ such that $Lx \ne 0$. And since $Lx \ne 0$, there is (using Hahn Banach) a $\phi \in Y^*$ such that $\langle Lx, \phi\rangle \ne 0$. Then $\langle x, L^* \phi\rangle \ne 0$, so $L^* \ne 0$.

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  • $\begingroup$ I only assume normed spaces. But you used inner product. $\endgroup$ – Keith Mar 11 '18 at 16:44
  • $\begingroup$ No, I just wrote $\phi(Lx)$ as $\langle Lx, \phi \rangle$. It's a common notational convention, but you can rewrite if you prefer. $\endgroup$ – fredgoodman Mar 11 '18 at 16:45
  • $\begingroup$ $\langle \cdot, \cdot \rangle$ here refers to the duality pairing of a normed space with its dual. So for example if $\phi \in Y^\ast$ and $y \in Y$ then $\langle y, \phi \rangle := \phi(y)$. $\endgroup$ – Rhys Steele Mar 11 '18 at 16:45
  • $\begingroup$ Oh I see.... Ok $\endgroup$ – Keith Mar 11 '18 at 16:46
  • $\begingroup$ And nobody is correct in pointing out the existence of $\phi$ as claimed depends on Hahn Banach. So the whole thing is Hahn Banach corollary. $\endgroup$ – fredgoodman Mar 11 '18 at 16:48
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As a corollary of the Hahn-Banach Theorem, for any distinct points $x,y \in Y$ there is a bounded functional $\phi_{x,y} \in Y^\ast$ such that $\phi_{x,y}(x) \neq \phi_{x,y}(y)$.

If $L_1 \neq L_2$ then there is a point $x \in X$ such that $L_1 x \neq L_2 x$. Then, by the above there is a $\phi \in Y^\ast$ such that $\phi(L_1 x) \neq \phi(L_2 x)$. Therefore $L_1^\ast \phi \neq L_2^\ast \phi$. This is the contrapositive of your desired result.

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Of course this requires Hahn-Banach (or some other assumption beyond ZF) ... cannot be proved in ZF itself.

For nonzero Banach space $X = l^\infty/c_0$ it is consistent that $X^* = \{0\}$. Then the zero and identity operators on $X$ both have the same dual.

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  • $\begingroup$ Excellent way to totally confuse a beginner in functional analysis. $\endgroup$ – fredgoodman Mar 11 '18 at 16:52

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