Task:Unnamed inequality

Statement

Let [ilmath]\lambda\in(0,1)\subset\mathbb{R} [/ilmath] be given. Then for all [ilmath]t\in\mathbb{R}_{\ge 0} [/ilmath] we claim:

• [ilmath]t^\lambda \le 1 - \lambda + \lambda t[/ilmath]

Proof

Define:

Behaviour:	[ilmath]<1[/ilmath]	[ilmath]=1[/ilmath]	[ilmath]>1[/ilmath]
[ilmath]f'(t)[/ilmath]	[ilmath]<0[/ilmath]	[ilmath]0[/ilmath]	[ilmath]>0[/ilmath]
[ilmath]f(t)[/ilmath]	[ilmath]\searrow[/ilmath]	[ilmath]\longrightarrow[/ilmath]	[ilmath]\nearrow[/ilmath]
	[ilmath]>0[/ilmath]	[ilmath]=0[/ilmath]	[ilmath]>0[/ilmath]
So [ilmath]f(1)[/ilmath] is a minimum

Broughall table for [ilmath]f(1)[/ilmath]
Here [ilmath]d[/ilmath] is some arbitrarily small positive value.
Reasoning: [ilmath]t^{\lambda-1} [/ilmath] for [ilmath]\lambda\in (0,1)\subset\mathbb{R} [/ilmath] means [ilmath]\lambda-1\in(-1,0)\subset\mathbb{R} [/ilmath]. Thus [ilmath](1\pm d)^{\lambda-1}=\frac{1}{(1\pm d)^{1-\lambda} }[/ilmath] and now [ilmath]1-\lambda\in(0,1)[/ilmath].

1. Thus [ilmath](1-d)^{1-\lambda} < 1[/ilmath]
   • so [ilmath]\frac{1}{(1-d)^{1-\lambda} } > 1[/ilmath]
   • so [ilmath]1-\frac{1}{(1-d)^{1-\lambda} } <0[/ilmath]
   • and finally [ilmath]\lambda(1-\frac{1}{(1-d)^{1-\lambda} }) <0[/ilmath]
     • As [ilmath]\lambda\in(0,1)[/ilmath] (i.e. "positive")
   • Thus [ilmath]\lambda(1-(1-d)^{\lambda-1})<0[/ilmath]
2. Thus [ilmath](1+d)^{1-\lambda} > 1[/ilmath]
   • so [ilmath]\frac{1}{(1+d)^{1-\lambda} } < 1[/ilmath]
   • so [ilmath]1-\frac{1}{(1+d)^{1-\lambda} } >0[/ilmath]
   • and finally [ilmath]\lambda(1-\frac{1}{(1+d)^{1-\lambda} }) >0[/ilmath]
     • As [ilmath]\lambda\in(0,1)[/ilmath]
   • Thus [ilmath]\lambda(1-(1+d)^{\lambda-1})>0[/ilmath]

• [ilmath]f:\mathbb{R}_{\ge 0}\rightarrow\mathbb{R} [/ilmath] by [ilmath]f:t\mapsto 1-\lambda+\lambda t-t^\lambda [/ilmath], so [ilmath]f(t):=1-\lambda+\lambda t - t^\lambda[/ilmath]. Then:
  • [ilmath]f':t\mapsto \lambda - \lambda t^{\lambda-1}[/ilmath] by differentiation (for [ilmath]f':\mathbb{R}_{>0}\rightarrow\mathbb{R} [/ilmath] - we'd have to use a limit-from-the-right to find [ilmath]f'(0)[/ilmath] if it exists.
    • We wish to find a turning point of [ilmath]f'(t)=\lambda - \lambda t^{\lambda-1}=\lambda(1-t^{\lambda-1})[/ilmath], that is [ilmath]f'(a)=0[/ilmath] for some [ilmath]a[/ilmath]
      • Suppose [ilmath]\lambda - \lambda t^{\lambda-1} = 0[/ilmath], then [ilmath]\lambda - \lambda t^{\lambda-1}=\lambda(1-t^{\lambda-1})=0[/ilmath] and clearly:
        • If [ilmath]t=1[/ilmath] then [ilmath]1-t^{\lambda-1}=1-1=0[/ilmath] and we're done (there is no other root)
    • Thus [ilmath]1[/ilmath] is a turning point of [ilmath]f[/ilmath] (that is [ilmath]f'(1)=0[/ilmath])
    • Next we must find out what [ilmath]f[/ilmath] is doing. We can use a Broughall table (shown right)
      • Or we can use the second derivative:
        • [ilmath]f' '(t)=-\lambda(\lambda-1)t^{\lambda-2}[/ilmath]
          • The quadratic curve [ilmath]\lambda(\lambda-1)[/ilmath] has roots [ilmath]0[/ilmath] and [ilmath]1[/ilmath] and tends towards [ilmath]+\infty[/ilmath] as tends to [ilmath]\pm\infty[/ilmath], thus is standard [ilmath]\cup[/ilmath] shaped.
          • The quadratic curve [ilmath]-\lambda(\lambda-1)[/ilmath] is the same but flipped about the [ilmath]x[/ilmath]-axis. Thus a [ilmath]\cap[/ilmath] shape.
          • As [ilmath]\lambda\in(0,1)[/ilmath] and this is between the roots, we see [ilmath]-\lambda(\lambda-1)<0[/ilmath] over the relevant domain.
          • [ilmath]t > 0[/ilmath] so [ilmath]t^\text{anything real} > 0[/ilmath] (remember we have to do [ilmath]t=0[/ilmath] separately, with a directed limit)
        • So [ilmath]-\lambda(\lambda-1)t^{\lambda-2} [/ilmath] is a [ilmath]\cap[/ilmath]-shape with respect to [ilmath]\lambda[/ilmath] and [ilmath]>0[/ilmath] always on the relevant [ilmath]\lambda[/ilmath]-domain.
        • We conclude [ilmath]f' '(1) > 0[/ilmath] (we can use this in the Broughall table, knowing [ilmath]f'(1)=0[/ilmath] and is increasing at [ilmath]1[/ilmath] means [ilmath]f'(\text{slightly after }1)>0[/ilmath] and as it must increase to zero [ilmath]f'(\text{slightly before }1)<0[/ilmath], as shown
        • Thus [ilmath]f(1)[/ilmath] is a minimum
    • We have shown [ilmath]f[/ilmath] is at a global minimum for all relevant [ilmath]\lambda[/ilmath] and all [ilmath]t[/ilmath] except [ilmath]t=0[/ilmath]
      • We evaluate [ilmath]f(1)[/ilmath] to get [ilmath]f(1)=0[/ilmath]. Thus [ilmath]f(t)>0[/ilmath] for [ilmath]t\in\mathbb{R}_{>0} [/ilmath] (but not [ilmath]t=0[/ilmath] yet!)
    • We evaluate [ilmath]f(0)[/ilmath] to get [ilmath]1-\lambda[/ilmath] (as [ilmath]t^\lambda=0^\lambda=0[/ilmath] as [ilmath]\lambda\in(0,1)[/ilmath] and [ilmath]0^x[/ilmath] is only undefined for [ilmath]x=0[/ilmath])
      • We observe that as [ilmath]\lambda\in(0,1)[/ilmath] that [ilmath]1-\lambda\in(0,1)[/ilmath] also, thus [ilmath]f(0)=1-\lambda>0[/ilmath] as required.
  • We have seen that [ilmath]f(t)\ge 0[/ilmath] for all [ilmath]t\in\mathbb{R}_{\ge 0}[/ilmath]
    • Thus [ilmath]1-\lambda+\lambda t-t^\lambda\ge 0[/ilmath] (we will no longer explicitly mention the domain of [ilmath]t[/ilmath] or [ilmath]\lambda[/ilmath])
    • So [ilmath]1-\lambda+\lambda t \ge t^\lambda[/ilmath]
      • As required
• This completes the proof.