Can we have something like an orthonormal basis for a finite dimensional normed space?

Question

So I proved a certain theorem about finite dimensional inner-product spaces, but after completing the proof, I realized the only point where I used the idea of orthogonality was the construction of an orthonormal basis. Everywhere else, I just used the norm instead, which leads me to wonder if I could generalize the theorem to finite dimensional normed spaces as well. The proof would be the same everywhere except for the point where I use the orthonormal basis. I'd have to use a basis such that the norm of every member is $1$ and for two distinct vectors $x$ and $y$, $$||x+y||^2=||x||^2+||y||^2$$ Which is the analog of orthogonality in a normed space. So my question is, does such a basis exist for all finite dimensional normed spaces and if yes, how do we show the existence? I'd be much obliged if you could give me some pointers.

Answer 1

As a very beginning, this is true for all real normed spaces of dimension 2. (For dimension 0 and 1 it is trivial.)

Let $\|\cdot\|$ be any norm on $\mathbb{R}^2$ and choose any $x$ with $\|x\|=1$. Let $\gamma : [0,1] \to \mathbb{R}^2 \setminus \{0\}$ be any continuous path connecting $x$ to $-x$ that avoids 0. Set $f(t) = \left\| x + \frac{\gamma(t)}{\|\gamma(t)\|}\right\|^2$, so that $f(0)=\|x+x\|^2 = 4$ and $f(1) = \|x-x\|^2=0$. Now $f$ is continuous, so by the intermediate value theorem there exists $t_0$ such that $f(t_0) = 2$. Set $y = \frac{\gamma(t_0)}{\|\gamma(t_0)\|}$. Then $\|y\|^2 = 1$ so $\|x+y\|^2 = \|x\|^2 + \|y\|^2$ as desired. Moreover, clearly $y$ is neither $x$ nor $-x$ so $\{x,y\}$ is a basis for $\mathbb{R}^2$.

Note that if you want to use this to prove something, you should be extremely careful that the only property of "orthonormal basis" you use is that $\|x+y\|^2 = \|x\|^2 + \|y\|^2 = 2$. Many other properties of orthonormal bases won't hold in general. For example, the "Pythagorean theorem" $\|ax+by\|^2 = a^2 + b^2$ need not hold. (Consider $\mathbb{R}^2$ with the sup norm, $x = (1,0)$, $y=(\sqrt{2}-1, 1)$, $a=2$, $b=1$. Then $\|ax+by\|^2 = \|(1+\sqrt{2}, 1)\|^2 = (1+\sqrt{2})^2 \ne 2^2 + 1^2$.)

Answer 2

This addresses a slightly different question, but I think it is the case that the only real normed spaces in which the parallelogram law $\|x+y\|^{2} + \|x-y\|^{2} = 2(\|x\|^{2} + \|y\|^{2})$ holds are inner product spaces.