# Bound the Euclidean distance between two points on a cone

We have a cone $$y^\top D y=0$$ ($$[y]_1 \geq 0$$), where $$y \in \mathbb R^{n+1}$$, $$[y]_1$$ is the first element of $$y$$, and $$$$D=\left[ {\begin{array}{*{20}{c}} 1&{{0_{1 \times n}}}\\ {{0_{n \times 1}}}&{ - {I_n}} \end{array}} \right] .$$$$ $$y^*$$ and $$\tilde y^*$$ are two points on the cone and we can decompose $$\tilde y^*-y^*$$ into a weighted sum of s set of unit vectors $$\tilde y^*-y^*=\beta_1 z_1+\beta_2 z_2+\dots+\beta_l z_l+\beta^\bot z^\bot,$$ where $$z_i,i=1,\ldots,l$$ ($$l \in \{1,\ldots,n\}$$, i.e., $$l$$ can vary from $$1$$ to $$n$$ ) are orthogonal to each other and we can calculate them beforehand, and $$z^\bot$$ is orthogonal to $$z_i,i=1,\ldots,l$$.

Now we can already bound $$\beta^\bot z^\bot$$ by a constant $$\theta$$ $$\|\beta^\bot z^\bot\| \leq \theta.$$ My question is can we bound $$\|\tilde y^*-y^*\|$$ in this problem?

• How are your unit vectors related to the cone? (Your decomposition is possible for any choice of an orthonormal basis $\{z_1,\dots z_n\}$)? Dec 12, 2019 at 1:54
• @GReyes $\{z_1,\dots,z_l\}$ are the eigenvectors of singular matrix $(A^\top A+\lambda D)$ associated with eigenvalue $0$, where $A$ can be any full column rank matrix. Dec 12, 2019 at 2:40
The vector $$t=\bar y^*-y^*$$ can be unbounded even when $$z^\perp=0$$, $$n=1$$, and $$l=1$$. Indeed, if $$n=1$$ then the cone (which we denote by $$C$$) is a union of two rays $${(x,x):x\ge 0}$$ and $${(x,-x):x\ge 0}$$. Graphically it is easy to guess and then to check that $$C-C=\{(x,y)\in\Bbb R^2: |x|\le |y|\}$$. It follows that for each vector $$z^\perp\in\Bbb R^2$$ a set $$\{t\in C-C: (t,z^\perp)=0\}$$ is unbounded.