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| In [[mathematics]], '''Gronwall's inequality''' (also called '''Grönwall's lemma''', '''Gronwall's lemma''' or '''Gronwall–Bellman inequality''') allows one to bound a function that is known to satisfy a certain [[differential inequality|differential]] or [[integral inequality]] by the solution of the corresponding differential or integral equation. There are two forms of the lemma, a differential form and an integral form. For the latter there are several variants.
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| Grönwall's inequality is an important tool to obtain various estimates in the theory of [[ordinary differential equation|ordinary]] and [[stochastic differential equation]]s. In particular, it provides a [[Comparison theorem]] that can be used to prove [[uniqueness]] of a solution to the [[initial value problem]]; see the [[Picard–Lindelöf theorem]].
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| It is named for [[Thomas Hakon Grönwall]] (1877–1932). Grönwall is the Swedish spelling of his name, but he spelled his name as Gronwall in his scientific publications after emigrating to the United States.
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| The differential form was proven by Grönwall in 1919.<ref name="gronwall">{{Citation | last = Gronwall | first = Thomas H. | author-link = Thomas Hakon Grönwall | title = Note on the derivatives with respect to a parameter of the solutions of a system of differential equations | journal = [[Annals of Mathematics|Ann. of Math.]] | volume = 20 | issue = 2 | pages = 292–296 | year = 1919 | jstor = 1967124 | mr = 1502565 | jfm = 47.0399.02}}</ref>
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| The integral form was proven by [[Richard Bellman]] in 1943.<ref>{{Citation | last = Bellman | first = Richard | author-link = Richard Bellman | title = The stability of solutions of linear differential equations | journal = [[Duke Mathematical Journal|Duke Math. J.]] | volume = 10 | issue = 4 | pages = 643–647 | year = 1943 | url = http://projecteuclid.org/euclid.dmj/1077472225 | mr = 0009408 | zbl = 0061.18502}}</ref>
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| A nonlinear generalization of the Gronwall–Bellman inequality is known as [[Bihari's inequality]].
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| == Differential form ==
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| Let {{math|''I''}} denote an [[Interval (mathematics)|interval]] of the [[real line]] of the form {{closed-open|''a'', ∞}} or {{closed-closed|''a'', ''b''}} or {{closed-open|''a'', ''b''}} with {{math|''a'' < ''b''}}. Let {{math|''β''}} and {{math|''u''}} be real-valued [[continuous functions]] defined on {{math|''I''}}. If {{math|''u''}} is [[derivative|differentiable]] in the [[Interior (topology)|interior]] {{math|''I''<sup>o</sup>}} of {{math|''I''}} (the interval {{math|''I''}} without the end points {{math|''a''}} and possibly {{math|''b''}}) and satisfies the differential inequality
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| :<math>u'(t) \le \beta(t)\,u(t),\qquad t\in I^\circ,</math>
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| then {{math|''u''}} is bounded by the solution of the corresponding differential ''equation'' {{math|''y'' ′(''t'') {{=}} ''β''(''t'') ''y''(''t'')}}:
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| :<math>u(t) \le u(a) \exp\biggl(\int_a^t \beta(s)\, \mathrm{d} s\biggr)</math>
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| for all {{math|''t'' ∈ ''I''}}.
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| '''Remark:''' There are no assumptions on the signs of the functions {{math|''β''}} and {{math|''u''}}.
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| === Proof ===
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| Define the function
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| :<math>v(t) = \exp\biggl(\int_a^t \beta(s)\, \mathrm{d} s\biggr),\qquad t\in I.</math>
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| Note that {{math|''v''}} satisfies
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| :<math>v'(t) = \beta(t)\,v(t),\qquad t\in I^\circ,</math>
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| with {{math|''v''(''a'') {{=}} 1}} and {{math|''v''(''t'') > 0}} for all {{math|''t'' ∈ ''I''}}. By the [[quotient rule]]
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| :<math>\frac{d}{dt}\frac{u(t)}{v(t)} = \frac{u'(t)\,v(t)-v'(t)\,u(t)}{v^2(t)} \le \frac{\beta(t)\,u(t)\,v(t) - \beta(t)\,v(t)\,u(t)}{v^2(t)} = 0,\qquad t\in I^\circ,</math>
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| Thus the derivative of the function <math>u(t)/v(t)</math> is non-positive and (by the [[mean value theorem]]) the function is bounded by its value at the initial point <math>a</math> of the interval <math>I</math>:
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| :<math>\frac{u(t)}{v(t)}\le \frac{u(a)}{v(a)}=u(a),\qquad t\in I,</math>
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| which is Gronwall's inequality.
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| == Integral form for continuous functions ==
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| Let {{math|''I''}} denote an [[Interval (mathematics)|interval]] of the [[real line]] of the form {{closed-open|''a'', ∞}} or {{closed-closed|''a'', ''b''}} or {{closed-open|''a'', ''b''}} with {{math|''a'' < ''b''}}. Let {{math|''α''}}, {{math|''β''}} and {{math|''u''}} be real-valued functions defined on {{math|''I''}}. Assume that {{math|''β''}} and {{math|''u''}} are continuous and that the negative part of {{math|''α''}} is integrable on every closed and bounded subinterval of {{math|''I''}}.
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| *(a) If {{math|''β''}} is non-negative and if {{math|''u''}} satisfies the integral inequality
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| ::<math>u(t) \le \alpha(t) + \int_a^t \beta(s) u(s)\,\mathrm{d}s,\qquad \forall t\in I,</math>
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| :then
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| ::<math> u(t) \le \alpha(t) + \int_a^t\alpha(s)\beta(s)\exp\biggl(\int_s^t\beta(r)\,\mathrm{d}r\biggr)\mathrm{d}s,\qquad t\in I.</math>
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| *(b) If, in addition, the function {{math|''α''}} is non-decreasing, then
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| ::<math>u(t) \le \alpha(t)\exp\biggl(\int_a^t\beta(s)\,\mathrm{d}s\biggr),\qquad t\in I.</math>
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| '''Remarks:'''
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| * There are no assumptions on the signs of the functions {{math|''α''}} and {{math|''u''}}.
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| * Compared to the differential form, differentiability of {{math|''u''}} is not needed for the integral form.
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| * For a version of Grönwall's inequality which doesn't need continuity of {{math|''β''}} and {{math|''u''}}, see the version in the next section.
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| === Proof ===
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| (a) Define
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| :<math>v(s) = \exp\biggl({-}\int_a^s\beta(r)\,\mathrm{d}r\biggr)\int_a^s\beta(r)u(r)\,\mathrm{d}r,\qquad s\in I.</math>
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| Using the [[product rule]], the [[chain rule]], the derivative of the [[exponential function]] and the [[fundamental theorem of calculus]], we obtain for the derivative
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| :<math>v'(s) = \biggl(\underbrace{u(s)-\int_a^s\beta(r)u(r)\,\mathrm{d}r}_{\le\,\alpha(s)}\biggr)\beta(s)\exp\biggl({-}\int_a^s\beta(r)\mathrm{d}r\biggr),
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| \qquad s\in I,</math>
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| where we used the assumed integral inequality for the upper estimate. Since {{math|''β''}} and the exponential are non-negative, this gives an upper estimate for the derivative of {{math|''v''}}. Since {{math|''v''(''a'') {{=}} 0}}, integration of this inequality from {{math|''a''}} to {{math|''t''}} gives
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| :<math>v(t) \le\int_a^t\alpha(s)\beta(s)\exp\biggl({-}\int_a^s\beta(r)\,\mathrm{d}r\biggr)\mathrm{d}s.</math>
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| Using the definition of {{math|''v''(''t'')}} for the first step, and then this inequality and the [[functional equation]] of the exponential function, we obtain
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| :<math>\begin{align}\int_a^t\beta(s)u(s)\,\mathrm{d}s
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| &=\exp\biggl(\int_a^t\beta(r)\,\mathrm{d}r\biggr)v(t)\\
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| &\le\int_a^t\alpha(s)\beta(s)\exp\biggl(\underbrace{\int_a^t\beta(r)\,\mathrm{d}r-\int_a^s\beta(r)\,\mathrm{d}r}_{=\,\int_s^t\beta(r)\,\mathrm{d}r}\biggr)\mathrm{d}s.
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| \end{align}</math>
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| Substituting this result into the assumed integral inequality gives Grönwall's inequality.
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| (b) If the function {{math|''α''}} is non-decreasing, then part (a), the fact {{math|''α''(''s'') ≤ ''α''(''t'')}}, and the fundamental theorem of calculus imply that
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| :<math>\begin{align}u(t)&\le\alpha(t)+\biggl({-}\alpha(t)\exp\biggl(\int_s^t\beta(r)\,\mathrm{d}r\biggr)\biggr)\biggr|^{s=t}_{s=a}\\
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| &=\alpha(t)\exp\biggl(\int_a^t\beta(r)\,\mathrm{d}r\biggr),\qquad t\in I.\end{align}</math>
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| == Integral form with locally finite measures ==
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| Let {{math|''I''}} denote an [[Interval (mathematics)|interval]] of the [[real line]] of the form {{closed-open|''a'', ∞}} or {{closed-closed|''a'', ''b''}} or {{closed-open|''a'', ''b''}} with {{math|''a'' < ''b''}}. Let {{math|''α''}} and {{math|''u''}} be [[measurable function]]s defined on {{math|''I''}} and let {{math|''μ''}} be a non-negative measure on the [[Borel σ-algebra]] of {{math|''I''}} satisfying {{math|''μ''(<nowiki>[</nowiki>''a'', ''t''<nowiki>]</nowiki>) < ∞}} for all {{math|''t'' ∈ ''I''}} (this is certainly satisfied when {{math|''μ''}} is a [[locally finite measure]]). Assume that {{math|''u''}} is integrable with respect to {{math|''μ''}} in the sense that
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| :<math>\int_{[a,t)}|u(s)|\,\mu(\mathrm{d}s)<\infty,\qquad t\in I,</math>
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| and that {{math|''u''}} satisfies the integral inequality
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| :<math>u(t) \le \alpha(t) + \int_{[a,t)} u(s)\,\mu(\mathrm{d}s),\qquad t\in I.</math>
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| If, in addition,
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| * the function {{math|''α''}} is non-negative or
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| * the function {{math|''t'' {{mapsto}} ''μ''(<nowiki>[</nowiki>''a'', ''t''<nowiki>]</nowiki>)}} is continuous for {{math|''t'' ∈ ''I''}} and the function {{math|''α''}} is integrable with respect to {{math|''μ''}} in the sense that
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| :: <math>\int_{[a,t)}|\alpha(s)|\,\mu(\mathrm{d}s)<\infty,\qquad t\in I,</math>
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| then {{math|''u''}} satisfies Grönwall's inequality
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| :<math>u(t) \le \alpha(t) + \int_{[a,t)}\alpha(s)\exp\bigl(\mu(I_{s,t})\bigr)\,\mu(\mathrm{d}s)</math>
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| for all {{math|''t'' ∈ ''I''}}, where {{math|''I<sub>s,t</sub>''}} denotes to open interval {{open-open|''s'', ''t''}}.
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| ===Remarks===
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| * There are no continuity assumptions on the functions {{math|''α''}} and {{math|''u''}}.
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| * The integral in Grönwall's inequality is allowed to give the value infinity.
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| * If {{math|''α''}} is the zero function and {{math|''u''}} is non-negative, then Grönwall's inequality implies that {{math|''u''}} is the zero function.
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| * The integrability of {{math|''u''}} with respect to {{math|''μ''}} is essential for the result. For a [[counterexample]], let {{math|''μ''}} denote [[Lebesgue measure]] on the [[unit interval]] {{closed-closed|0, 1}}, define {{math|''u''(0) {{=}} 0}} and {{math|''u''(''t'') {{=}} 1/''t''}} for {{math|''t'' ∈ }}{{open-closed|0, 1}}, and let {{math|''α''}} be the zero function.
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| * The version given in the textbook by S. Ethier and T. Kurtz.<ref>{{Citation | last = Ethier | first = Steward N. | last2 = Kurtz | first2 = Thomas G. | title = Markov Processes, Characterization and Convergence | place = New York | publisher = [[John Wiley & Sons]] | year = 1986 | page = 498 | isbn = 0-471-08186-8 | mr = 0838085 | zbl = 0592.60049}}</ref> makes the stronger assumptions that {{math|''α''}} is a non-negative constant and {{math|''u''}} is bounded on bounded intervals, but doesn't assume that the measure {{math|''μ''}} is locally finite. Compared to the one given below, their proof does not discuss the behaviour of the remainder {{math|''R<sub>n</sub>''(''t'')}}.
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| ===Special cases===
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| * If the measure {{math|''μ''}} has a density {{math|''β''}} with respect to Lebesgue measure, then Grönwall's inequality can be rewritten as
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| :: <math>u(t) \le \alpha(t) + \int_a^t \alpha(s)\beta(s)\exp\biggl(\int_s^t\beta(r)\,\mathrm{d}r\biggr)\,\mathrm{d}s,\qquad t\in I.</math>
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| * If the function {{math|''α''}} is non-negative and the density {{math|''β''}} of {{math|''μ''}} is bounded by a constant {{math|''c''}}, then
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| :: <math>u(t) \le \alpha(t) + c\int_a^t \alpha(s)\exp\bigl(c(t-s)\bigr)\,\mathrm{d}s,\qquad t\in I.</math>
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| * If, in addition, the non-negative function {{math|''α''}} is non-decreasing, then
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| :: <math>u(t) \le \alpha(t) + c\alpha(t)\int_a^t \exp\bigl(c(t-s)\bigr)\,\mathrm{d}s
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| =\alpha(t)\exp(c(t-a)),\qquad t\in I.</math>
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| === Outline of proof ===
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| The proof is divided into three steps. In idea is to substitute the assumed integral inequality into itself {{math|''n''}} times. This is done in Claim 1 using mathematical induction. In Claim 2 we rewrite the measure of a simplex in a convenient form, using the permutation invariance of product measures. In the third step we pass to the limit {{math|''n''}} to infinity to derive the desired variant of Grönwall's inequality.
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| === Detailed proof ===
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| ====Claim 1: Iterating the inequality====
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| For every natural number {{math|''n''}} including zero,
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| :<math>u(t) \le \alpha(t) + \int_{[a,t)} \alpha(s) \sum_{k=0}^{n-1} \mu^{\otimes k}(A_k(s,t))\,\mu(\mathrm{d}s) + R_n(t)</math>
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| with remainder
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| :<math>R_n(t) :=\int_{[a,t)}u(s)\mu^{\otimes n}(A_n(s,t))\,\mu(\mathrm{d}s),\qquad t\in I,</math> | |
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| where
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| :<math>A_n(s,t)=\{(s_1,\ldots,s_n)\in I_{s,t}^n\mid s_1<s_2<\cdots<s_n\},\qquad n\ge1,</math>
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| is an {{math|''n''}}-dimensional [[simplex]] and
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| :<math>\mu^{\otimes 0}(A_0(s,t)):=1.</math>
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| ====Proof of Claim 1====
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| We use [[mathematical induction]]. For {{math|''n'' {{=}} 0}} this is just the assumed integral inequality, because the [[empty sum]] is defined as zero.
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| Induction step from {{math|''n''}} to {{math|''n'' + 1}}:
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| Inserting the assumed integral inequality for the function {{math|''u''}} into the remainder gives
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| :<math>R_n(t)\le\int_{[a,t)} \alpha(s) \mu^{\otimes n}(A_n(s,t))\,\mu(\mathrm{d}s) +\tilde R_n(t)</math>
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| with
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| :<math>\tilde R_n(t):=\int_{[a,t)} \biggl(\int_{[a,q)} u(s)\,\mu(\mathrm{d}s)\biggr)\mu^{\otimes n}(A_n(q,t))\,\mu(\mathrm{d}q),\qquad t\in I.</math>
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| Using the [[Fubini's theorem|Fubini-Tonelli theorem]] to interchange the two integrals, we obtain
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| :<math>\tilde R_n(t)
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| =\int_{[a,t)} u(s)\underbrace{\int_{(s,t)} \mu^{\otimes n}(A_n(q,t))\,\mu(\mathrm{d}q)}_{=\,\mu^{\otimes n+1}(A_{n+1}(s,t))}\,\mu(\mathrm{d}s)
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| =R_{n+1}(t),\qquad t\in I.</math>
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| Hence [[#Claim 1: Iterating the inequality|Claim 1]] is proved for {{math|''n'' + 1}}.
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| ====Claim 2: Measure of the simplex====
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| For every natural number {{math|''n''}} including zero and all {{math|''s'' < ''t''}} in {{math|''I''}}
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| :<math>\mu^{\otimes n}(A_n(s,t))\le\frac{\bigl(\mu(I_{s,t})\bigr)^n}{n!}</math>
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| with equality in case {{math|''t'' {{mapsto}} ''μ''(<nowiki>[</nowiki>''a'', ''t''<nowiki>]</nowiki>)}} is continuous for {{math|''t'' ∈ ''I''}}.
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| ====Proof of Claim 2====
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| For {{math|''n'' {{=}} 0}}, the claim is true by our definitions. Therefore, consider {{math|''n'' ≥ 1}} in the following.
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| Let {{math|''S<sub>n</sub>''}} denote the set of all [[permutation]]s of the indices in {{math|{1, 2, . . . , ''n''}}}. For every permutation {{math|''σ'' ∈ ''S<sub>n</sub>''}} define
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| :<math>A_{n,\sigma}(s,t)=\{(s_1,\ldots,s_n)\in I_{s,t}^n\mid s_{\sigma(1)}<s_{\sigma(2)}<\cdots<s_{\sigma(n)}\}.</math>
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| These sets are disjoint for different permutations and
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| :<math>\bigcup_{\sigma\in S_n}A_{n,\sigma}(s,t)\subset I_{s,t}^n.</math>
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| Therefore,
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| :<math>\sum_{\sigma\in S_n} \mu^{\otimes n}(A_{n,\sigma}(s,t))
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| \le\mu^{\otimes n}\bigl(I_{s,t}^n\bigr)=\bigl(\mu(I_{s,t})\bigr)^n.</math>
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| Since they all have the same measure with respect to the {{math|''n''}}-fold product of {{math|''μ''}}, and since there are {{math|''n''!}} permutations in {{math|''S<sub>n</sub>''}}, the claimed inequality follows.
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| Assume now that {{math|''t'' {{mapsto}} ''μ''(<nowiki>[</nowiki>''a'', ''t''<nowiki>]</nowiki>)}} is continuous for {{math|''t'' ∈ ''I''}}. Then, for different indices {{math|''i'', ''j'' ∈ {1, 2, . . . , ''n''}}}, the set
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| :<math>\{(s_1,\ldots,s_n)\in I_{s,t}^n\mid s_i=s_j\}</math>
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| is contained in a [[hyperplane]], hence by an application of [[Fubini's theorem]] its measure with respect to the {{math|''n''}}-fold product of {{math|''μ''}} is zero. Since
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| :<math>I_{s,t}^n\subset\bigcup_{\sigma\in S_n}A_{n,\sigma}(s,t) \cup \bigcup_{1\le i<j\le n}\{(s_1,\ldots,s_n)\in I_{s,t}^n\mid s_i=s_j\},</math>
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| the claimed equality follows.
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| ====Proof of Grönwall's inequality====
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| For every natural number {{math|''n''}}, [[#Claim 2: Measure of the simplex|Claim 2]] implies for the remainder of [[#Claim 1: Iterating the inequality|Claim 1]] that
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| :<math>|R_n(t)| \le \frac{\bigl(\mu(I_{a,t})\bigr)^n}{n!} \int_{[a,t)} |u(s)|\,\mu(\mathrm{d}s),\qquad t\in I.</math>
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| By assumption we have {{math|''μ''(''I''<sub>''a'',''t''</sub>) < ∞}}. Hence, the integrability assumption on {{math|''u''}} implies that
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| :<math>\lim_{n\to\infty}R_n(t)=0,\qquad t\in I.</math>
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| [[#Claim 2: Measure of the simplex|Claim 2]] and the [[Characterizations of the exponential function|series representation]] of the exponential function imply the estimate
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| :<math>\sum_{k=0}^{n-1} \mu^{\otimes k}(A_k(s,t))
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| \le\sum_{k=0}^{n-1} \frac{\bigl(\mu(I_{s,t})\bigr)^k}{k!}
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| \le\exp\bigl(\mu(I_{s,t})\bigr)</math>
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| for all {{math|''s'' < ''t''}} in {{math|''I''}}. If the function {{math|''α''}} is non-negative, then it suffices to insert these results into [[#Claim 1: Iterating the inequality|Claim 1]] to derive the above variant of Grönwall's inequality for the function {{math|''u''}}.
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| In case {{math|''t'' {{mapsto}} ''μ''(<nowiki>[</nowiki>''a'', ''t''<nowiki>]</nowiki>)}} is continuous for {{math|''t'' ∈ ''I''}}, [[#Claim 2: Measure of the simplex|Claim 2]] gives
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| :<math>\sum_{k=0}^{n-1} \mu^{\otimes k}(A_k(s,t))
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| =\sum_{k=0}^{n-1} \frac{\bigl(\mu(I_{s,t})\bigr)^k}{k!}
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| \to\exp\bigl(\mu(I_{s,t})\bigr)\qquad\text{as }n\to\infty</math>
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| and the integrability of the function {{math|''α''}} permits to use the [[dominated convergence theorem]] to derive Grönwall's inequality.
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| ==References==
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| <references />
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| {{PlanetMath attribution|id=3901|title=Gronwall's lemma}}
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| {{DEFAULTSORT:Gronwall's inequality}}
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| [[Category:Inequalities]]
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| [[Category:Lemmas]]
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| [[Category:Ordinary differential equations]]
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| [[Category:Stochastic differential equations]]
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| [[Category:Articles containing proofs]]
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