 Solution

For $x\in X$, define $e_x\in X^{\ast \ast }$ by $e_x(\varphi )=\varphi (x)$. The map $J:x\mapsto e_x$ is linear, andshows $\Vert Jx\Vert \le \Vert x\Vert$, so $J$ is continuous. The granted equality $\Vert Jx\Vert =\Vert x\Vert$ is the canonical embedding into the bidual.

Suppose, contrary to the claim, that exactly the same linear functionals are continuous for $\Vert \cdot \Vert$ and $\Vert \cdot {\Vert }^{\prime }$. Their common continuous dual has two operator norms, say $\Vert \cdot {\Vert }_{\ast }$ and $\Vert \cdot {\Vert }_{\ast }^{\prime }$. Both dual spaces are Banach, even if the second normed space $X$ is not complete. The identity maphas a closed graph: convergence in either operator norm implies pointwise convergence on $X$, so two limits must agree. The closed graph theorem makes the two dual norms equivalent.

Applying the isometric bidual formula for each primal norm then gives constants $c,C>0$ such thatThis says the two norms are equivalent norms, contrary to the hypothesis. Their continuous duals must therefore differ as sets, so a linear functional belonging to one and not the other is continuous for exactly one of the two norms.