This is my second attempt, following the first question here: Hilbert space with a "norm" that does not respect the triangle inequality?
I believe this inner product produces a Hilbert space, except its norm does not respect the triangle inequality. All axioms of the Hilbert space are respected. I prove this explicitly here. Is the space still a Hilbert space?
$u$ and $v$ are vectors of $2\times 2$ matrices $G$ from the orientation-preserving general linear group $GL^+(2,\mathbb{R})$ : i.e. $\det G > 0$.
$$ \langle u,v\rangle=\frac{1}{4}(||u+v||^2-||u-v||^2) $$
where
$$ ||x||^2=\sum_{G \in x}\det G $$
Where the notation $G\in x$ designate picking each element of the vector $x$ (which are $n\times n$ matrices).
Here, for the proofs, I am assume vectors have only one elements, but it generalizes to multiple elements. According to https://mathworld.wolfram.com/InnerProduct.html, the axioms of an inner product are as follows.
- $\langle u+v, w \rangle = \langle u,w\rangle + \langle v, w\rangle$
$$ \begin{align} &>u=\left( \begin{array}{cc} \text{u1} & \text{u2} \\ \text{u3} & \text{u4} \\ \end{array} \right) \\ &>v=\left( \begin{array}{cc} \text{v1} & \text{v2} \\ \text{v3} & \text{v4} \\ \end{array} \right) \\ &> w=\left( \begin{array}{cc} \text{w1} & \text{w2} \\ \text{w3} & \text{w4} \\ \end{array} \right)\\ &>\text{InnerProduct}(\text{a$\_$},\text{b$\_$})=\frac{1}{4} (| a+b| -| a-b| )\\ &>\text{Simplify}[\text{InnerProduct}(u+v,w)=\text{InnerProduct}(u,w)+\text{InnerProduct}(v,w)]\\ &out = true \end{align} $$
My apologies for using Mathematica for a "proof", but the notation would be substantial otherwise.
- $\langle \alpha v, w \rangle = \alpha \langle v,w \rangle$
$$ \begin{align} &>\text{Simplify}[\text{InnerProduct}(a u,v)=a \text{InnerProduct}(u,v)]\\ &out =true \end{align} $$
- $\langle v, w \rangle = \langle w,v \rangle$
$$ \begin{align} &>\text{Simplify}[\text{InnerProduct}(u,w)=\text{InnerProduct}(w,u)]\\ &out=true \end{align} $$
- $\langle u,u \rangle \geq 0 \text{ iff } u=0$
Here we have a bit of a pickle, but it's resolvable. Since, I have stated we only have $\det G > 0$, we can either state that the space has no $0$ in it which is still compliant with the axiom (it doesn't require a zero, it just takes that if there is one, then its inner product must be $0$). Or we can just manually add the zero-valued matrix to complete the space with a $0$.
$$ \langle u,u\rangle=||u||^2=\sum_{G \in u} \det G $$
Since the matrices $G$ have $\det g>0$, then it follows that $\langle u,u\rangle>0$.
Now, my concern is that the norm induced by the inner product does not seem to respect the triangle inequality (but maybe I am wrong and it does?).
$$ ||u||=\sqrt{\langle u,u \rangle} $$
And the triangle inequality is:
$$ ||u+v||\leq ||u||+||v|| $$
Yielding:
$$ \begin{align} &>\text{ExpandAll}[\text{InnerProduct}(u+v,u+v)]\\ &out=\text{u1} \text{u4}+\text{u1} \text{v4}-\text{u2} \text{u3}-\text{u2} \text{v3}-\text{u3} \text{v2}+\text{u4} \text{v1}+\text{v1} \text{v4}-\text{v2} \text{v3}\\ &>\text{ExpandAll}[\text{InnerProduct}(u,u)+\text{InnerProduct}(v,v)]\\ &out=\text{u1} \text{u4}-\text{u2} \text{u3}+\text{v1} \text{v4}-\text{v2} \text{v3} \end{align} $$
Can we show that:
$$ \begin{align} \text{u1} \text{u4}+\text{u1} \text{v4}-\text{u2} \text{u3}-\text{u2} \text{v3}-\text{u3} \text{v2}+\text{u4} \text{v1}+\text{v1} \text{v4}-\text{v2} \text{v3}&\leq \text{u1} \text{u4}-\text{u2} \text{u3}+\text{v1} \text{v4}-\text{v2} \text{v3}\\ \text{u1} \text{v4}-\text{u2} \text{v3}-\text{u3} \text{v2}+\text{u4}\text{v1} &\leq0 \end{align} $$
It appears to me that because the inequality contains a mixture of positive and negative signs, then it must that that the inequality fails. However, since it adheres to the axioms of the inner product, it must be that this is a norm. Are the restriction to the orientation-preserving general linear group "conspire" in some subtle manner to guarantee this inequality holds?