3.3 Proof of Theorem 2.3
(a) For \(\theta \in(0,\frac{\pi}{4})\), we have \(0<\tan \theta < 1 < \cot \theta \) and \({\mathcal {L}}_{\theta }\subset {\mathcal {K}}^{n} \subset {\mathcal {L}}_{\theta }^{\ast}\). We discuss three cases according to \(x\in {\mathcal {L}}_{\theta }\), \(x\in-{\mathcal {L}}_{\theta }^{\ast}\), and \(x\notin {\mathcal {L}}_{\theta }\cup(-{\mathcal {L}}_{\theta }^{\ast})\).
Case 1. If \(x\in {\mathcal {L}}_{\theta }\), then \(Ax\in {\mathcal {K}}^{n}\), which clearly yields \(\operatorname{dist}(Ax, {\mathcal {K}}^{n}) = \operatorname{dist}(x,{\mathcal {L}}_{\theta }) =0\).
Case 2. If
\(x\in-{\mathcal {L}}_{\theta }^{\ast}\), then
\(x_{1}\cot \theta \leq-\|x_{2}\|\) and
$$\operatorname{dist}(x,{\mathcal {L}}_{\theta })=\|x\|=\sqrt{x_{1}^{2}+ \|x_{2}\|^{2}}. $$
In this case, there are two subcases for the element
Ax. If
\(x_{1}\cot \theta \leq x_{1}\tan \theta \leq-\|x_{2}\|\),
i.e.,
\(Ax\in-{\mathcal {K}}^{n}\), it follows that
$$\begin{aligned} \operatorname{dist}\bigl(Ax,{\mathcal {K}}^{n}\bigr) =& \|Ax\|=\sqrt{x_{1}^{2} \tan^{2}\theta +\|x_{2}\|^{2}} \leq\sqrt {x_{1}^{2}+\|x_{2}\| ^{2}} \\ =& \operatorname{dist}(x,{\mathcal {L}}_{\theta }) \leq\sqrt{x_{1}^{2}+ \|x_{2}\|^{2} \cot^{2}\theta } \\ =& \cot \theta \cdot\sqrt{x_{1}^{2}\tan^{2}\theta + \|x_{2}\|^{2}} \\ =& \cot \theta \cdot \operatorname{dist}\bigl(Ax,{\mathcal {K}}^{n}\bigr), \end{aligned}$$
where the first inequality holds since
\(\tan \theta < 1\) (it becomes an equality only in the case of
\(x=0\)), and the second inequality holds since
\(\cot \theta > 1\) (it becomes an equality only in the case of
\(x_{2}=0\)). On the other hand, if
\(x_{1}\cot \theta \leq-\|x_{2}\| < x_{1}\tan \theta \leq0\), we have
$$\begin{aligned} \operatorname{dist}\bigl(Ax,{\mathcal {K}}^{n}\bigr) =& \bigl\| Ax-(Ax)_{+}\bigr\| \\ =& \sqrt{\frac{1}{2}\bigl(x_{1}\tan \theta -\|x_{2} \|\bigr)_{-}^{2}+\frac{1}{2}\bigl(x_{1}\tan \theta +\|x_{2} \|\bigr)_{-}^{2}} \\ =& \sqrt{\frac{1}{2}\bigl(x_{1}\tan \theta -\|x_{2} \|\bigr)^{2}}\\ \leq&\sqrt{x_{1}^{2}\tan ^{2}\theta +\|x_{2}\|^{2}} < \sqrt{x_{1}^{2}+ \|x_{2}\|^{2}} \\ =& \operatorname{dist}(x,{\mathcal {L}}_{\theta }) \leq \sqrt{x_{1}^{2}-x_{1} \|x_{2}\|\cot \theta } \\ < & \sqrt{\frac{1}{2}x_{1}^{2}+ \frac{1}{2}\|x_{2}\|^{2}\cot^{2} \theta -x_{1}\|x_{2}\| \cot \theta } \\ =& \cot \theta \cdot \operatorname{dist}\bigl(Ax,{\mathcal {K}}^{n}\bigr), \end{aligned}$$
where the third inequality holds because
\(\|x_{2}\| \leq-x_{1}\cot \theta \), and the fourth inequality holds since
\(\|x_{2}\| > -x_{1}\tan \theta \geq0\). Therefore, for the subcases of
\(x \in-{\mathcal {L}}_{\theta }^{\ast}\), we can conclude that
$$\operatorname{dist}\bigl(Ax,{\mathcal {K}}^{n}\bigr) \leq \operatorname{dist}(x,{\mathcal {L}}_{\theta }) \leq\cot \theta \cdot \operatorname{dist}\bigl(Ax,{\mathcal {K}}^{n}\bigr), $$
and
\(\operatorname{dist}(x,{\mathcal {L}}_{\theta }) = \cot \theta \cdot \operatorname{dist}(Ax,{\mathcal {K}}^{n})\) holds only in the case of
\(x_{2}=0\).
Case 3. If
\(x\notin {\mathcal {L}}_{\theta }\cup(-{\mathcal {L}}_{\theta }^{\ast})\), then
\(-\|x_{2}\|\tan \theta < x_{1}<\|x_{2}\|\cot \theta \), which yields
\(x_{1}\tan \theta <\|x_{2}\|\) and
\(x_{1}\cot \theta >-\|x_{2}\|\). Thus, we have
$$\begin{aligned} \operatorname{dist}(x,{\mathcal {L}}_{\theta }) =& \bigl\Vert x-x^{\theta }_{+} \bigr\Vert \\ =& \sqrt{\frac{\cot^{2}\theta }{1+\cot^{2}\theta }\bigl(x_{1}\tan \theta -\|x_{2} \|\bigr)_{-}^{2} + \frac{\tan^{2}\theta }{1+\tan^{2}\theta }\bigl(x_{1}\cot \theta +\|x_{2} \|\bigr)_{-}^{2}} \\ =& \sqrt{\frac{\cot^{2}\theta }{1+\cot^{2}\theta }\bigl(x_{1}\tan \theta -\|x_{2} \|\bigr)^{2}}. \end{aligned}$$
On the other hand, it follows from
\(-\|x_{2}\|\tan \theta < x_{1}<\|x_{2}\|\cot \theta \) and
\(\theta \in(0,\frac{\pi}{4})\) that
$$-\|x_{2}\| < -\|x_{2}\|\tan^{2}\theta < x_{1} \tan \theta < \|x_{2}\|. $$
This implies that
$$\begin{aligned} \operatorname{dist}\bigl(Ax,{\mathcal {K}}^{n}\bigr) =& \bigl\Vert Ax-(Ax)_{+} \bigr\Vert \\ =& \sqrt{\frac{1}{2}\bigl(x_{1}\tan \theta -\|x_{2} \|\bigr)_{-}^{2}+\frac{1}{2}\bigl(x_{1}\tan \theta +\|x_{2} \|\bigr)_{-}^{2}} \\ =& \sqrt{\frac{1}{2}\bigl(x_{1}\tan \theta -\|x_{2} \|\bigr)^{2}}. \end{aligned}$$
From this and
\(\theta \in(0,\frac{\pi}{4})\), we see that
$$\begin{aligned} \operatorname{dist}\bigl(Ax,{\mathcal {K}}^{n}\bigr) =& \sqrt{\frac{1}{2}\bigl(x_{1} \tan \theta -\|x_{2}\|\bigr)^{2}}\\ < & \sqrt{\frac{\cot^{2}\theta }{1+\cot^{2}\theta }\bigl(x_{1} \tan \theta -\|x_{2}\|\bigr)^{2}} \\ =& \operatorname{dist}(x,{\mathcal {L}}_{\theta }) = \sqrt{\frac{2}{1+\tan^{2}\theta }}\sqrt{\frac {1}{2}\bigl(x_{1} \tan \theta -\|x_{2}\|\bigr)^{2}} \\ =& \sqrt{\frac{2}{1+\tan^{2}\theta }} \operatorname{dist}\bigl(Ax,{\mathcal {K}}^{n}\bigr). \end{aligned}$$
Therefore, in these cases of
\(x\notin {\mathcal {L}}_{\theta }\cup(-{\mathcal {L}}_{\theta }^{\ast})\), we can conclude
$$\operatorname{dist}\bigl(Ax,{\mathcal {K}}^{n}\bigr) < \operatorname{dist}(x,{\mathcal {L}}_{\theta }) = \sqrt{\frac{2}{1+\tan^{2}\theta }} \operatorname{dist}\bigl(Ax,{\mathcal {K}}^{n}\bigr). $$
To sum up, from all the above and the fact that
\(\max\{\cot \theta ,\sqrt{\frac{2}{1+\tan^{2}\theta }}\}=\cot \theta \) for
\(\theta \in(0, \frac{\pi}{4})\), we obtain
$$\operatorname{dist}\bigl(Ax,{\mathcal {K}}^{n}\bigr) \leq \operatorname{dist}(x,{\mathcal {L}}_{\theta }) \leq\cot \theta \cdot \operatorname{dist}\bigl(Ax,{\mathcal {K}}^{n}\bigr). $$
(b) For \(\theta \in(\frac{\pi}{4},\frac{\pi}{2})\), we have \(0<\cot \theta < 1 < \tan \theta \) and \({\mathcal {L}}_{\theta }^{\ast}\subset {\mathcal {K}}^{n} \subset {\mathcal {L}}_{\theta }\). Again we discuss the following three cases.
Case 1. If \(x\in {\mathcal {L}}_{\theta }\), then \(Ax\in {\mathcal {K}}^{n}\), which implies that \(\operatorname{dist}(Ax,{\mathcal {K}}^{n})= \operatorname{dist}(x,{\mathcal {L}}_{\theta })=0\).
Case 2. If
\(x\in-{\mathcal {L}}_{\theta }^{\ast}\), then
\(x_{1}\cot \theta \leq-\|x_{2}\|\) and
$$\operatorname{dist}(x,{\mathcal {L}}_{\theta })=\|x\|=\sqrt{x_{1}^{2}+ \|x_{2}\|^{2}}. $$
It follows from
\(x_{1}\cot \theta \leq-\|x_{2}\|\) and
\(\theta \in(\frac{\pi}{4},\frac{\pi}{2})\) that
\(x_{1}\tan \theta \leq x_{1}\cot \theta \leq-\|x_{2}\|\), which leads to
\(Ax\in-{\mathcal {K}}^{n}\). Hence, we have
$$\operatorname{dist}\bigl(Ax,{\mathcal {K}}^{n}\bigr) = \|Ax\|= \sqrt{x_{1}^{2} \tan^{2}\theta +\|x_{2}\|^{2}}. $$
With this, it is easy to verify that
\(\operatorname{dist}(Ax,{\mathcal {K}}^{n}) \geq \operatorname{dist}(x,{\mathcal {L}}_{\theta })\) for
\(\theta \in(\frac{\pi }{4},\frac{\pi}{2})\). Moreover, we note that
$$\begin{aligned} \tan \theta \cdot \operatorname {dist}(x,{\mathcal {L}}_{\theta }) =& \sqrt{\tan^{2} \theta \bigl(x_{1}^{2}+\|x_{2}\|^{2}\bigr)}\\ \geq& \sqrt{x_{1}^{2}\tan^{2}\theta + \|x_{2}\|^{2}}=\operatorname {dist}\bigl(Ax,{\mathcal {K}}^{n}\bigr). \end{aligned}$$
Thus, it follows that
$$\operatorname{dist}(x,{\mathcal {L}}_{\theta }) \leq \operatorname{dist}\bigl(Ax,{\mathcal {K}}^{n}\bigr) \leq\tan \theta \cdot \operatorname{dist}(x, {\mathcal {L}}_{\theta }), $$
and
\(\operatorname{dist}(Ax,{\mathcal {K}}^{n}) = \tan \theta \cdot \operatorname{dist}(x,{\mathcal {L}}_{\theta })\) holds only in the case of
\(x_{2}=0\).
Case 3. If
\(x\notin {\mathcal {L}}_{\theta }\cup(-{\mathcal {L}}_{\theta }^{\ast})\), then we have
\(-\|x_{2}\|\tan \theta < x_{1}<\|x_{2}\|\cot \theta \) and
$$\begin{aligned} \operatorname{dist}(x,{\mathcal {L}}_{\theta }) =& \bigl\Vert x-x^{\theta }_{+} \bigr\Vert \\ =& \sqrt{\frac{\cot^{2}\theta }{1+\cot^{2}\theta }\bigl(x_{1}\tan \theta -\|x_{2} \|\bigr)_{-}^{2} + \frac{\tan^{2}\theta }{1+\tan^{2}\theta }\bigl(x_{1}\cot \theta +\|x_{2} \|\bigr)_{-}^{2}} \\ =& \sqrt{\frac{\cot^{2}\theta }{1+\cot^{2}\theta }\bigl(x_{1}\tan \theta -\|x_{2} \|\bigr)^{2}}. \end{aligned}$$
Since
\(-\|x_{2}\|\tan \theta < x_{1}<\|x_{2}\|\cot \theta \), it follows immediately that
\(-\|x_{2}\|\tan^{2} \theta < x_{1} \tan \theta <\|x_{2}\|\). Again, there are two subcases for the element
Ax. If
\(-\|x_{2}\|\tan^{2} \theta <-\|x_{2}\| <x_{1} \tan \theta <\|x_{2}\|\), then we have
\(Ax \notin {\mathcal {K}}^{n} \cup(-{\mathcal {K}}^{n})\). Thus, it follows that
$$\begin{aligned} \operatorname{dist}\bigl(Ax,{\mathcal {K}}^{n}\bigr) =& \bigl\Vert Ax-(Ax)_{+} \bigr\Vert \\ =& \sqrt{\frac{1}{2}\bigl(x_{1}\tan \theta -\|x_{2} \|\bigr)_{-}^{2}+\frac{1}{2}\bigl(x_{1}\tan \theta +\|x_{2} \|\bigr)_{-}^{2}} \\ =& \sqrt{\frac{1}{2}\bigl(x_{1}\tan \theta -\|x_{2} \|\bigr)^{2}}. \end{aligned}$$
This together with
\(\theta \in(\frac{\pi}{4},\frac{\pi}{2})\) yields
$$\operatorname{dist}\bigl(Ax,{\mathcal {K}}^{n}\bigr) > \operatorname{dist}(x,{\mathcal {L}}_{\theta }). $$
Moreover, by the expressions of
\(\operatorname{dist}(Ax,{\mathcal {K}}^{n})\) and
\(\operatorname{dist}(x,{\mathcal {L}}_{\theta })\), it is easy to verify
$$\operatorname{dist}\bigl(Ax,{\mathcal {K}}^{n}\bigr) = \sqrt{\frac{2}{1+\tan^{2}\theta }} \operatorname{dist}(x,{\mathcal {L}}_{\theta }). $$
On the other hand, if
\(-\|x_{2}\|\tan^{2} \theta < x_{1} \tan \theta \leq-\|x_{2}\|<\|x_{2}\|\), then we have
\(Ax\in-{\mathcal {K}}^{n}\), which implies
$$\operatorname{dist}\bigl(Ax,{\mathcal {K}}^{n}\bigr) = \|Ax\|= \sqrt{x_{1}^{2} \tan^{2}\theta +\|x_{2}\|^{2}}. $$
Therefore, it follows that
$$\begin{aligned} \operatorname{dist}(x,{\mathcal {L}}_{\theta }) =& \sqrt{\frac{\cot^{2}\theta }{1+\cot^{2}\theta }\bigl(x_{1}\tan \theta -\|x_{2} \|\bigr)^{2}} \\ < & \sqrt{\frac{1}{2}\bigl(x_{1}\tan \theta -\|x_{2} \|\bigr)^{2}}\\ \leq&\sqrt{x_{1}^{2}\tan ^{2}\theta +\|x_{2}\|^{2}} = \operatorname{dist}\bigl(Ax,{\mathcal {K}}^{n}\bigr). \end{aligned}$$
Since
$$\begin{aligned}& \tan^{2}\theta \bigl(x_{1}\tan \theta -\|x_{2} \|\bigr)^{2}-\bigl(1+\tan^{2}\theta \bigr) \bigl(x_{1}^{2} \tan^{2}\theta +\| x_{2}\|^{2}\bigr) \\& \quad= -2x_{1}\|x_{2}\|\tan^{3}\theta -x_{1}^{2} \tan^{2}\theta -\|x_{2}\|^{2} \\& \quad= -x_{1}\tan \theta \cdot\bigl(\|x_{2}\|\tan^{2} \theta +x_{1}\tan \theta \bigr)+\|x_{2}\|\bigl(-x_{1}\tan ^{3}\theta -\|x_{2}\|\bigr) \\& \quad\geq \|x_{2}\|\bigl(-x_{1}\tan^{3}\theta - \|x_{2}\|\bigr) \\& \quad\geq 0, \end{aligned}$$
where the first inequality holds due to
\(-\|x_{2}\|\tan^{2}\theta < x_{1}\tan \theta \), and the second inequality holds due to
\(x_{1}\tan \theta <-\|x_{2}\|\) and
\(\theta \in(\frac{\pi}{4},\frac{\pi}{2})\), we have
$$\operatorname{dist}\bigl(Ax,{\mathcal {K}}^{n}\bigr) \leq\tan \theta \cdot \operatorname{dist}(x,{\mathcal {L}}_{\theta }). $$
From all the above analyses and the fact that
\(\max\{\tan \theta , \sqrt{\frac{2}{1+\tan^{2}\theta }} \}=\tan \theta \) for
\(\theta \in(\frac{\pi}{4},\frac{\pi}{2})\), we can conclude that
$$\operatorname{dist}(x,{\mathcal {L}}_{\theta }) \leq \operatorname{dist}\bigl(Ax,{\mathcal {K}}^{n}\bigr) \leq\tan \theta \cdot \operatorname{dist}(x, {\mathcal {L}}_{\theta }). $$
Thus, the proof is complete. □
The following example elaborates more why \(\operatorname{dist}(Ax,{\mathcal {K}}^{n}) = \tan \theta \cdot \operatorname{dist}(x,{\mathcal {L}}_{\theta })\) holds only in the cases of \(x=(x_{1},x_{2})\in {\mathcal {L}}_{\theta }\) or \(x_{2}=0\).