Show that $\exists$ an inner product in $X$ such that $\langle x,x\rangle =\|x\|^2$ for all $x \in X$

Question

Problem: Let $X$ be normed space. If on every two dimensional subspace $Y$ of $X$, there is an inner product $\langle \cdot,\cdot\rangle_Y$ such that $\langle y,y\rangle_Y=\|y\|^2$ for all $y\in Y$. Then there is an inner product $\langle \cdot,\cdot\rangle$ on X such that $\langle x,x\rangle =\|x\|^2$ for all $x \in X$.

My attempt:

Fix any $y_0 \in Y$

Define $f: Y \to \mathbb{C}$

$f(y)=\langle y,y_0\rangle, y\in Y$

Clearly, $f$ is bounded linear functional.

Using Hahn Banach Extension Theorem,

$\exists g: X \to \mathbb{C} \ni g|_Y =f$ and $\|g\|=\|f\|$

But I am not able to define an inner product using $g$

Can anyone help me out with this?

Is this approach ok?

Answer 1

The inner product is already determined by the norm using the Polarization identity so no need to try and build it in a non-constructive way. If you define a function $\left< \cdot, \cdot \right>$ on $X \times X$ by

$$ \left< u, v \right> := \frac{||u + v||^2 + ||u - v||^2}{4} $$

then clearly $\left<u, u\right> = ||u||^2$ for all $u \in X$ and you need to check that $\left< \cdot, \cdot \right>$ is an inner product. To do that, note that the norm satisfies the parallelogram law (because it is induced from an inner product on each two-dimensional subspace and the parallelogram law can be checked on each two-dimensional subspace separately) and hence it is indeed an inner product.