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.

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# Proof Using Shortest Distance

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