By Bertsekas D.P., Nedic A., Ozdaglar A.E.

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B) Consider the function f (x1 , x2 ) = (x2 − px21 )(x2 − qx21 ), where 0 < p < q and let x∗ = (0, 0). We first show that g(α) = f (x∗ + αd) is minimized at α = 0 for all d ∈ 2 . We have g(α) = f (x∗ + αd) = (αd2 − pα2 d21 )(αd2 − qα2 d21 ) = α2 (d2 − pαd21 )(d2 − qαd21 ). Also, g (α) = 2α(d2 − pαd21 )(d2 − qαd21 ) + α2 (−pd21 )(d2 − qαd21 ) + α2 (d2 − pαd21 )(−qd21 ). Thus g (0) = 0. Furthermore, g (α) = 2(d2 − pαd21 )(d2 − qαd21 ) + 2α(−pd21 )(d2 − qαd21 ) + 2α(d2 − pαd21 )(−qd21 ) + 2α(−pd21 )(d2 − qαd21 ) + α2 (−pd21 )(−qd21 ) + 2α(d2 − pαd21 )(−qd21 ) + α2 (−pd21 )(−qd21 ).

10) 2c0 The choice of c0 ensures, through the definition of cf , the existence of some c1 > c0 , some x ∈ n , and some scalar β such that f (w) ≥ − 1 w−x 2c1 2 ∀ w. + β, Together with Eq. 10), this implies that − 1 wk − x 2c1 2 + 1 wk − xk 2c0 2 ≤ α − β, for all sufficiently large k. Dividing this relation by wk as k → ∞, we get 1 1 − + ≤ 0, 2c1 2c0 2 and taking the limit from which it follows that c1 ≤ c0 . This is a contradiction by our choice of c1 . 10. By assumption, we have that f (x) < ∞ for some x ∈ n .

Hence, the result follows. Finally, let us show that RX = RX1 × · · · × RXm . Let y = (y1 , . . , ym ) ∈ RX . By definition, this implies that for all x ∈ X and α ≥ 0, we have x + αy ∈ X. From this, it follows that for all xi ∈ Xi and α ≥ 0, xi + αyi ∈ Xi , so that yi ∈ RXi , implying that y ∈ RX1 × · · · × RXm . Conversely, let y = (y1 , . . , ym ) ∈ RX1 × · · · × RXm . By definition, for all xi ∈ Xi and α ≥ 0, we have xi + αyi ∈ Xi . From this, we get for all x ∈ X and α ≥ 0, x + αy ∈ X, thus showing that y ∈ RX .

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