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.


1 Answer 1


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 $F_1(x_1)F_2(x_2) = \sqrt{\|x_1\|\|x_2\|} \; \sqrt{\|x_1\|\|x_2\|} = \|x_1\|\|x_2\|$.

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    $\begingroup$ Thanks sir, It's a very elegant proof! $\endgroup$ Commented Nov 30, 2020 at 8:04

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