If two maps have the same dual map, do the two maps coincide?

Question

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?

Answer 1

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 $L^* \ne 0$.

Answer 2

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.

Answer 3

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.