# An interesting application of the Hahn-Banach Theorem.

Let $$X$$ be a normed space and $$x_1,x_2\in X$$ nonzero elements. Show that there are functionals $$F_1,F_2\in X'$$ such that $$F_1(x_1)F_2(x_2)=\lVert x_1\rVert \lVert x_2\rVert$$ and $$\lVert F_1\rVert \lVert x_1\rVert =\lVert F_2\rVert \lVert x_2\rVert$$.

My attempt:

My idea was to define a functional $$f_1:\langle \{x_1\}\rangle\to \mathbb{R}$$, by $$f_1(\alpha x_1)=\alpha \lVert x_2\rVert.$$ Then, by using Hahn-Banach, I extend $$f_1$$ to a functional $$F_1:X\to \mathbb{R}$$ such that $$F_1(x_1)=\lVert x_2\rVert$$ and $$\lVert F_1\rVert=\lVert f_1\rVert=\frac{\lVert x_2\rVert}{\lVert x_1\rVert}$$. After that, I tried to define a functional $$F_2:X\to \mathbb{R}$$ such that $$F_2(x_2)=\lVert x_1\rVert$$ and $$\lVert F_2\rVert=1$$, but I coundn't do that.

What you're doing is great! Observe the symmetry between the indices $$1$$ and $$2$$ in the question. I think that $$\lVert F_1\rVert=\lVert f_1\rVert=\frac{\lVert x_2\rVert}{\lVert x_1\rVert}\tag{*}\label{*}$$ breaks the symmetry. Your goal is $$\lVert F_1\rVert \lVert x_1\rVert =\lVert F_2\rVert \lVert x_2\rVert$$, so the denominator of \eqref{*} should be $$1$$ instead. To acheive this, replace $$f_1$$ with $$f_1(ax_1) = a$$ in your proof. Then $$\| F_1 \| = 1/\|x_1\|$$. Similarly, $$\|F_2\| = 1/\|x_2\|$$.
But $$F_1(x_1)F_2(x_2) = 1$$. To fix this, it's simple. Just multiply the everything by $$\sqrt{\|x_1\|\|x_2\|}$$.
To wrap up, define $$f_1(ax_1) = a \sqrt{\|x_1\|\|x_2\|}$$ and $$f_2(ax_2) = a \sqrt{\|x_1\|\|x_2\|}$$. Then $$\|F_1\| = (1/\|x_1\|) \; \sqrt{\|x_1\|\|x_2\|}$$ and $$\|F_2\| = (1/\|x_2\|)\sqrt{\|x_1\|\|x_2\|}$$, so that $$\|F_1\|\|x_1\| = \|F_2\|\|x_2\| = \sqrt{\|x_1\|\|x_2\|}$$ and $$F_1(x_1)F_2(x_2) = \sqrt{\|x_1\|\|x_2\|} \; \sqrt{\|x_1\|\|x_2\|} = \|x_1\|\|x_2\|$$.