Functional where only way to prove property is to make it "appear" after the property is proven I have $f : \ell^{\infty}_{\mathbb{R}} \to \mathbb{R}$ with a bunch of properties and want to extend to $g : \ell^{\infty}_{\mathbb{C}} \to \mathbb{C}$ while preserving the properties. If we define $g(x+iy) = f(x)+if(y),$ then all the properties get satisfied except $||g|| = 1,$ which appears impossible to prove. Try and try as I might, it can't be proven. Using $||f|| = 1$ and any other properties, the inequalities just don't line up to force $||g|| = 1.$
Cnsider $f$ as a function $ \ell^{\infty}_{\mathbb{R}} \to \mathbb{C},$ then there is an extension $g$ with $||g|| = ||f|| = 1$ and $g|_{\ell^{\infty}_{\mathbb{R}}} = f$ by the Hahn-Banach theorem. However, this means $g(x+iy) = g(x)+ig(y) = f(x)+if(y),$ the exact extension we had in mind earlier.
Suddenly, we get a proof of $||g|| = 1$ for the explicit extension despite the fact that defining the extension immediately and then trying to prove this was impossible. Is this a common scenario, or have I made a mistake?
 A: I don't think this is a valid proof. Namely, for $\ell^\infty_{\Bbb{R}}$ to be a subspace of $\ell^\infty_{\Bbb{C}}$, you have to consider both of them as real vector spaces. Then you have a real-linear map $f : \ell^\infty_{\Bbb{R}} \to \Bbb{R}$ with $\|f\|=1$ which by Hahn-Banach extends to a real-linear map $g : \ell^\infty_{\Bbb{C}} \to \Bbb{R}$ with $\|g\|=1$. Nowhere is it said that this $g$ will be complex-linear (in fact it certainly won't be since the codomain is again $\Bbb{R}$) so you cannot use $g(x+iy) = g(x)+ig(y)$.
Here is an elementary proof that $g : \ell^{\infty}_{\mathbb{C}} \to \mathbb{C}$ defined by $g(x+iy) = f(x)+if(y)$ has $\|g\|=1$. First notice that for any $z \in \ell^\infty_{\Bbb{C}}$ we have
$$\operatorname{Re}g(z) = g(\operatorname{Re}z) = f(\operatorname{Re}z).$$
Let $z \in \ell^\infty_{\Bbb{C}}$ be arbitrary and let $$g(z) = |g(z)|e^{i\phi}$$
be the polar form of the complex number $g(z)$. Since $g$ is complex-linear, we have
$$|g(z)| = e^{-i\phi}g(z) = g(e^{-i\phi}z) = \operatorname{Re} g(e^{-i\phi}z) = f(\operatorname{Re}(e^{-i\phi}z)) \le \|\operatorname{Re}(e^{-i\phi}z)\|_\infty \le \|z\|_\infty$$
and hence $\|g\|\le 1$. Since $g$ extends $f$ we get $\|g\| \ge \|f\| = 1$ as well.
