Proof of Gronwall's Inequality
📂LemmasProof of Gronwall's Inequality
Theorem
Let’s assume there are two continuous functions f,w:I→R defined in an interval I⊂R that contains the minimum value a∈R. If w is ∀t∈I in w(t)≥0 and for some constant C∈R,
f(t)≤C+∫atw(s)f(s)ds,∀t∈I
then the following holds.
f(t)≤Cexp(∫atw(s)ds),∀t∈I
Explanation
The fact that interval I has a minimum value a means that I looks like the following with respect to a<b.
[a,b] or [a,b) or [a,∞)
Proof
α(t)=β(t)=C+∫atw(s)f(s)dsCexp(∫atw(s)ds)
When we define two functions α,β as shown above and suppose γ(t):=β(t)α(t), then γ(a)=1 and by the quotient rule of differentiation,
===γ′(t)[β(t)]2α′(t)β(t)−α(t)β′(t)(Cexp(∫atw(s)ds))2w(t)f(t)⋅Cexp(∫atw(s)ds)−(C+∫atw(s)f(s)ds)⋅w(t)Cexp(∫atw(s)ds)w(t)Cexp(∫atw(s)ds)f(t)−C−∫atw(s)f(s)ds
it follows. Given that f(t)≤C+∫atw(s)f(s)ds and w(t)≥0, then γ′(t)≤0. Therefore, γ(t) is a non-increasing function, and since γ(t)≤1, the following is established.
⟹⟹⟹⟹γ(t)=β(t)α(t)≤1α(t)≤β(t)C+∫atw(s)f(s)ds≤Cexp(∫atw(s)ds)f(t)≤C+∫atw(s)f(s)ds≤Cexp(∫atw(s)ds)f(t)≤Cexp(∫atw(s)ds)
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