Prove Uniqueness Theorem: |\phi(t) - \psi(t)| ≤ ∫0t

[thedude36]

I am having to justify the steps in a proof of the uniqueness theorem. I am supposed to show why the inequality follows from the initial equation.

http://i.imgur.com/AxApogj.png

[itex]\phi[/itex](t) - [itex]\psi[/itex](t) =∫₀^t 2s[[itex]\phi[/itex](t) - [itex]\psi[/itex](t)] ds


|[itex]\phi[/itex](t) - [itex]\psi[/itex](t)| =|∫₀^t 2s[[itex]\phi[/itex](t) - [itex]\psi[/itex](t)] ds| [itex]\leq[/itex] ∫₀^t 2s|[[itex]\phi[/itex](t) - [itex]\psi[/itex](t)]| ds , with t>0

I have no idea where to start. Mostly, I am unsure as to why pulling the 2s out will make it larger than the initial absolute value. Could anyone help?

 

[haruspex]

In the interval, s is always positive, but the other term may vary in sign. That would lead to some cancellation during the integration process. If we take the absolute value of the integrand, it can only increase the result:
|∫₀^t 2s[ϕ(t) - ψ(t)] ds| ≤ ∫₀^t |2s[ϕ(t) - ψ(t)]| ds = ∫₀^t 2s|[ϕ(t) - ψ(t)]| ds