Throughout, β£ β
β£ |\cdot| β£ β
β£ absolute value on R \mathbb{R} R d R ( s , t ) = β£ s β t β£ d_{\mathbb{R}}(s,t)=|s-t| d R β ( s , t ) = β£ s β t β£ s , t s,t s , t The Absolute Value Metric on the Real Line .
Claim 1. Fix t β R t\in\mathbb{R} t β R
Case c = 0 c=0 c = 0 By claim 1 of Basic Properties of the Exponential Function , exp β‘ ( 0 ) = 1 \exp(0)=1 exp ( 0 ) = 1 E 0 ( u ) = 1 E_0(u)=1 E 0 β ( u ) = 1 u β R u\in\mathbb{R} u β R E 0 E_0 E 0 β 0 0 0 c exp β‘ ( c t ) = 0 c\exp(ct)=0 c exp ( c t ) = 0 Ξ΅ > 0 \varepsilon>0 Ξ΅ > 0 Ξ΄ = 1 \delta=1 Ξ΄ = 1 E 0 E_0 E 0 β differentiable at t t t 0 = c exp β‘ ( c t ) 0=c\exp(ct) 0 = c exp ( c t )
Case c β 0 c\ne0 c ξ = 0 Then β£ c β£ > 0 |c|>0 β£ c β£ > 0 Basic Properties of the Exponential Function , exp β‘ \exp exp 0 0 0 exp β‘ β² ( 0 ) = exp β‘ ( 0 ) = 1 \exp'(0)=\exp(0)=1 exp β² ( 0 ) = exp ( 0 ) = 1 Basic Properties of the Exponential Function , exp β‘ ( c t ) > 0 \exp(ct)>0 exp ( c t ) > 0
Let Ξ΅ β R \varepsilon\in\mathbb{R} Ξ΅ β R Ξ΅ > 0 \varepsilon>0 Ξ΅ > 0
Ξ΅ β² = Ξ΅ exp β‘ ( c t ) β β£ c β£ \varepsilon'=\frac{\varepsilon}{\exp(ct)\,|c|} Ξ΅ β² = exp ( c t ) β£ c β£ Ξ΅ β is a positive real, so by the differentiability of exp β‘ \exp exp 0 0 0 Ξ΄ β² > 0 \delta'>0 Ξ΄ β² > 0 k k k 0 < β£ k β£ < Ξ΄ β² 0<|k|<\delta' 0 < β£ k β£ < Ξ΄ β²
β£ exp β‘ ( k ) β 1 k β 1 β£ < Ξ΅ β² , \left|\frac{\exp(k)-1}{k}-1\right|<\varepsilon' , β k exp ( k ) β 1 β β 1 β < Ξ΅ β² , the difference quotient of exp β‘ \exp exp 0 0 0 ( exp β‘ ( k ) β exp β‘ ( 0 ) ) / k = ( exp β‘ ( k ) β 1 ) / k (\exp(k)-\exp(0))/k=(\exp(k)-1)/k ( exp ( k ) β exp ( 0 )) / k = ( exp ( k ) β 1 ) / k
Put Ξ΄ = Ξ΄ β² / β£ c β£ \delta=\delta'/|c| Ξ΄ = Ξ΄ β² /β£ c β£ h h h 0 < β£ h β£ < Ξ΄ 0<|h|<\delta 0 < β£ h β£ < Ξ΄ k = c h k=ch k = c h 0 < β£ k β£ = β£ c β£ β β£ h β£ < β£ c β£ β Ξ΄ = Ξ΄ β² 0<|k|=|c|\,|h|<|c|\,\delta=\delta' 0 < β£ k β£ = β£ c β£ β£ h β£ < β£ c β£ Ξ΄ = Ξ΄ β² Basic Properties of the Exponential Function ,
E c ( t + h ) β E c ( t ) = exp β‘ ( c t + c h ) β exp β‘ ( c t ) = exp β‘ ( c t ) ( exp β‘ ( k ) β 1 ) , E_c(t+h)-E_c(t)=\exp(ct+ch)-\exp(ct)=\exp(ct)\bigl(\exp(k)-1\bigr), E c β ( t + h ) β E c β ( t ) = exp ( c t + c h ) β exp ( c t ) = exp ( c t ) ( exp ( k ) β 1 ) , and since h = k / c h=k/c h = k / c
E c ( t + h ) β E c ( t ) h = exp β‘ ( c t ) β c β exp β‘ ( k ) β 1 k . \frac{E_c(t+h)-E_c(t)}{h}=\exp(ct)\,c\,\frac{\exp(k)-1}{k} . h E c β ( t + h ) β E c β ( t ) β = exp ( c t ) c k exp ( k ) β 1 β . Therefore
β£ E c ( t + h ) β E c ( t ) h β c β exp β‘ ( c t ) β£ = exp β‘ ( c t ) β β£ c β£ β β£ exp β‘ ( k ) β 1 k β 1 β£ < exp β‘ ( c t ) β β£ c β£ β Ξ΅ β² = Ξ΅ . \left|\frac{E_c(t+h)-E_c(t)}{h}-c\,\exp(ct)\right|=\exp(ct)\,|c|\,\left|\frac{\exp(k)-1}{k}-1\right|<\exp(ct)\,|c|\,\varepsilon'=\varepsilon . β h E c β ( t + h ) β E c β ( t ) β β c exp ( c t ) β = exp ( c t ) β£ c β£ β k exp ( k ) β 1 β β 1 β < exp ( c t ) β£ c β£ Ξ΅ β² = Ξ΅ . As Ξ΅ > 0 \varepsilon>0 Ξ΅ > 0 E c E_c E c β t t t E c β² ( t ) = c exp β‘ ( c t ) E_c'(t)=c\exp(ct) E c β² β ( t ) = c exp ( c t )
Claim 2. By claim 1, E c E_c E c β t β R t\in\mathbb{R} t β R t t t interior point of the interval R \mathbb{R} R Differentiability at an Interior Point Implies Continuity There , applied with I = R I=\mathbb{R} I = R t t t E c E_c E c β continuous at t t t R \mathbb{R} R for every t t t R \mathbb{R} R β \blacksquare β