I encountered a problem in a book with a proof given. But I am a bit skeptic about it. I hope someone can help shed some light.(adsbygoogle = window.adsbygoogle || []).push({});

Let [tex]\{g_{i}\}[/tex] be a set of vectors and imagine a cone defined as [tex]K = \left\{v \,\bigg|\, v =-\sum_{i}\lambda_{i}g_{i}, \textup{ where }\lambda_{i}\geq 0 \ , \forall i \right\}[/tex].

Let [tex]f \notin K[/tex] and let [tex]u \in K[/tex] be the closest point to [tex]f[/tex]. Obviously [tex]u[/tex] is the projected point of [tex]f[/tex] onto [tex]K[/tex]. The objective is to prove that if [tex] d = u - f [/tex], then [tex]g_{i}^\top d \leq 0, \, \forall i[/tex]. (Note that [tex]d \neq 0[/tex].)

The proof given is by contradiction: Suppose that is not true, that is, [tex]\hat{g}_{i}^\top d = s_{i}[/tex] for some scalar [tex]s_{i} > 0, \, \forall i[/tex], where [tex]\hat{g}_{i}= g_{i}/\|g_{i}\|[/tex]. It is not difficult to see that [tex](u-s_{i}\hat{g}_{i}) \in K, \, \forall i[/tex], i.e., it remains in the cone even by small or large perturbation on the vector [tex]u[/tex]. Now, we shall show the perturbed point has smaller distance. Indeed this is the case since for any [tex]i[/tex],

[tex]

\begin{align*}

\|(u-s_{i}\hat{g}_{i})-f \|^{2} &= \|(u-f)-s_{i}\hat{g}_{i}\|^{2}= \|(u-f)\|^{2}-2 s_{i}\hat{g}_{i}^\top (u-f)+s_{i}^{2}\|\hat{g}_{i}\|^{2} \\

&= \|d\|-2s_{i}\hat{g}_{i}^\top d+s_{i}^{2} \\

&= \|d\|-2s_{i}^{2}+s_{i}^{2} \\

&= \|d\|-s_{i}^{2}\leq \|d\|,

\end{align*}

[/tex]

which contradicts with the assumption that [tex]u[/tex] is the nearest point in [tex]K[/tex] to [tex]f[/tex] -- done!!!.

All looks good, however if I let [tex]\hat{g}_{i}^\top d = t_{i}[/tex] for which the scalar [tex]t_{i}< 0,\, \forall i[/tex] but sufficiently close to 0 such that [tex](u-t_{i}\hat{g}_{i}) \in K[/tex] for any [tex]i[/tex], then using the same derivation I arrive at [tex]\|(u-t_{i}\hat{g}_{i})-f \|^{2}= \|d\|-t_{i}^{2}\leq \|d\|[/tex] too! This means it can contradict even for the case [tex]g_{i}^\top d < 0[/tex]. I now question the validity of this proof. I welcome your comment.

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Proof Using Shortest Distance

**Physics Forums | Science Articles, Homework Help, Discussion**