So we were shown this problem from our first functional analysis lecture and I was wondering if someone could help or give a hint:
Show that the unit sphere $\mathbb{S}^{n-1} := \{ x \in \mathbb{R}^n: \| x \| = 1 \}$ is a complete metric space equipped with $d(x,y):= \arccos \langle x,y \rangle_{\mathbb{R}^n}$ where $\langle x,y \rangle_{\mathbb{R}^n}$ denotes the standard dot product.
We're having no trouble showing positivity and symmetry but could use help showing completeness and the triangle inequality.
Kind regards
Edit:
So with MatiasHeikkilä's help, what's left to show is that $\theta_{x,y} \leq \theta_{x,z} + \theta_{y,z}$ if I'm not mistaken. Do I need to make a case-by-case proof to show this inequality or is there an easier way? I would argue something along the lines of: "$z$ lies between $x$ and $y$ implies the equality", "$z$ does not lie on the (smallest) path between $x$ and $y$ but $z$ lies on the half-sphere with $x$ and $y$ on it implies the inequality (since $\theta_{x,y} < \theta_{x,z}$) and lastly $z$ does neither lie on the (smallest) path between $x$ and $y$ nor does $z$ lie on the half-sphere implies the inequality (since $\theta_{x,z} + \theta_{y,z} = 2 \pi – \theta_{x,y} \geq \pi $)
Best Answer
Let us first show the triangle inequality for $d$, i.e. $d\left(x,z\right)\leq d\left(x,y\right)+d\left(y,z\right)$. Since we have $\left\langle Ux,Uy\right\rangle =\left\langle x,y\right\rangle $ for every orthogonal matrix, by choosing $U$ with $Uy=e_{1}=\left(1,0,\dots,0\right)$, we can assume $y=e_{1}$ (in short, we used that $d$ is invariant under orthogonal transformations). Since the cosine is strictly decreasing on $\left[0,\pi\right]$ (and since $\arccos$ takes values in $\left[0,\pi\right]$), the desired inequality is equivalent to \begin{align*} \left\langle x,z\right\rangle & \overset{!}{\geq}\cos\left(\arccos\left\langle x,e_{1}\right\rangle +\arccos\left\langle e_{1},z\right\rangle \right)\\ & =\cos\left(\arccos x_{1}+\arccos z_{1}\right)\\ \left(\text{by trigonometric formulas}\right) & =\cos\left(\arccos x_{1}\right)\cos\left(\arccos z_{1}\right)-\sin\left(\arccos x_{1}\right)\sin\left(\arccos z_{1}\right)\\ \left(\text{since }\sin\geq0\text{ and hence }\sin=\sqrt{1-\cos^{2}}\text{ on }\left[0,\pi\right]\right) & =x_{1}z_{1}-\sqrt{1-\cos^{2}\left(\arccos x_{1}\right)}\sqrt{1-\cos^{2}\left(\arccos z_{1}\right)}\\ & =x_{1}z_{1}-\sqrt{1-x_{1}^{2}}\sqrt{1-z_{1}^{2}}. \end{align*} By rearranging, we see that this inequality is equivalent to \begin{align*} -\sum_{i=2}^{n}x_{i}z_{i} & =x_{1}z_{1}-\left\langle x,z\right\rangle \\ & \overset{!}{\leq}\sqrt{1-x_{1}^{2}}\sqrt{1-z_{1}^{2}}\\ \left(\text{since }1=\left|x\right|^{2}=\sum_{i=1}^{n}x_{i}^{2}\right) & =\sqrt{\sum_{i=2}^{n}x_{i}^{2}}\sqrt{\sum_{i=2}^{n}z_{i}^{2}}. \end{align*} But by Cauchy Schwarz, we have $$ -\sum_{i=2}^{n}x_{i}z_{i}\leq\left|\sum_{i=2}^{n}x_{i}z_{i}\right|\leq\sqrt{\sum_{i=2}^{n}x_{i}^{2}}\sqrt{\sum_{i=2}^{n}z_{i}^{2}}, $$ so that the inequality above holds. We have thus verified the triangle inequality. By what you have already shown, $d$ is a metric.
For completeness, I leave it to you to verify that $$ \Phi:\left(S^{n-1},\mathcal{T}\right)\to\left(S^{n-1},d\right),x\mapsto x $$ is continuous, where $\mathcal{T}$ denotes the usual topology. Thus, $\left(S^{n-1},d\right)$ is compact. Now, it is well known that every compact metric space is complete, so that we are done.