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1840 V. ANDRIEU, L. PRALY, AND A. ASTOLFIAppendix B. Pro<strong>of</strong> <strong>of</strong> Proposition 2.11. We give the pro<strong>of</strong> only in the 0-limit case since the ∞-limit case <strong>is</strong> similar. The function φ being a bijection, we canassume without loss <strong>of</strong> generality that it <strong>is</strong> a strictly increasing function (otherw<strong>is</strong>e wetake −φ). Th<strong>is</strong>, together with homogeneity in the 0-limit, implies that ϕ 0 <strong>is</strong> strictlypositive. Moreover, for each δ> 0, there ex<strong>is</strong>ts t 0 (δ) > 0 such that∣ φ(t) ∣∣∣∣ − ϕ 0 ≤ δ ∀ t ∈ (0,t 0 (δ)] .t d0By letting λ = φ(t), <strong>th<strong>is</strong></strong> givesϕ 0 − δ ≤λφ −1 (λ) d0 ≤ ϕ 0 + δ ∀ λ ∈ (0,φ(t 0 (δ))] , ∀ δ> 0 .Since for δ < ϕ 0 the term on the left <strong>is</strong> strictly positive, these inequalities give( 1) 1d 0ϕ 0 + δ≤ φ−1 (λ)λ 1d 01Then since the function δ ↦→ (ϕ ) 1d 00−δex<strong>is</strong>ts δ 1 (ɛ 1 ) > 0 sat<strong>is</strong>fying( 1ϕ 0) 1d 0− ɛ 1 ≤Th<strong>is</strong> yields(φ −1 (λ)∣λ 1d 0≤1ϕ 0 + δ 1 (ɛ 1 )( ) 11d 0∀ λ ∈ (0,φ −1 (t 0 (δ))], ∀ δ ∈ (0,ϕ 0 ) .ϕ 0 − δ) 1d 0<strong>is</strong> continuous at zero, for every ɛ 1 > 0 there≤(1ϕ 0 − δ 1 (ɛ 1 )) 1d 0( ) ∣11d 0 ∣∣∣∣− ≤ ɛ 1 ∀ λ ∈ (0,λ − (ɛ 1 )] ,ϕ 0with λ − (ɛ 1 )=φ(t 0 (δ 1 (ɛ 1 ))). With a similar argument, we getφ −1 ( ) ∣1(−λ) 1d 0 ∣∣∣∣ + ≤ ɛ 1 ∀ λ ∈ (0,λ + (ε 1 )]∣ d 0 ϕ 0λ 1for some λ + > 0. Let λ 0 = min{λ − ,λ + }.Now, for x ≠ 0 and λ> 0, we have∣φ −1 (λx)∣λ 1d 0( ) ∣1xd 0 ∣∣∣∣− = |x| 1d 0ϕ 0∣ ∣∣∣∣φ −1 (λx)(xλ) 1d 0≤( ) ∣11d 0 ∣∣∣∣− .ϕ 0Therefore, for any compact set C <strong>of</strong> R\{0} and any ɛ> 0, <strong>by</strong> letting ɛ 1 =we have|x| 1d 0 ɛ 1 ≤ ɛ, 0 < |λx| ≤ λ 0 (ɛ 1 ) ∀ λ ∈and thereforeφ −1 (λx)∣λ 1d 0( ) ∣1xd 0 ∣∣∣∣− ≤ ɛϕ 0∀ λ ∈(0,( 1ϕ 0) 1d 0+ ɛ 1 .ɛmax x∈C |x| 1d 0,]λ 0 (ɛ 1 ), ∀ x ∈ C,max x∈C |x|(]λ 0 (ɛ 1 )0,, ∀ x ∈ C.max x∈C |x|Th<strong>is</strong> establ<strong>is</strong>hes homogeneity in the 0-limit <strong>of</strong> the function φ −1 .<strong>Copyright</strong> © <strong>by</strong> <strong>SIAM</strong>. <strong>Unauthorized</strong> <strong>reproduction</strong> <strong>of</strong> <strong>th<strong>is</strong></strong> <strong>article</strong> <strong>is</strong> prohibited.

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