Grönwall's inequality

Define the function

${\displaystyle v(t)=\exp {\biggl (}\int _{a}^{t}\beta (s)\,\mathrm {d} s{\biggr )},\qquad t\in I.}$

Note that v satisfies

${\displaystyle v'(t)=\beta (t)\,v(t),\qquad t\in I^{\circ },}$

with v(a) = 1 and v(t) > 0 for all t ∈ I. By the quotient rule

${\displaystyle {\frac {d}{dt}}{\frac {u(t)}{v(t)}}={\frac {u'(t)\,v(t)-v'(t)\,u(t)}{v^{2}(t)}}={\frac {u'(t)\,v(t)-\beta (t)\,v(t)\,u(t)}{v^{2}(t)}}\leq 0,\qquad t\in I^{\circ },}$

Thus the derivative of the function ${\displaystyle u(t)/v(t)}$ is non-positive and the function is bounded above by its value at the initial point ${\displaystyle a}$ of the interval ${\displaystyle I}$:

${\displaystyle {\frac {u(t)}{v(t)}}\leq {\frac {u(a)}{v(a)}}=u(a),\qquad t\in I,}$

which is Grönwall's inequality.

(a) Define

${\displaystyle v(s)=\exp {\biggl (}{-}\int _{a}^{s}\beta (r)\,\mathrm {d} r{\biggr )}\int _{a}^{s}\beta (r)u(r)\,\mathrm {d} r,\qquad s\in I.}$

Using the product rule, the chain rule, the derivative of the exponential function and the fundamental theorem of calculus, we obtain for the derivative

${\displaystyle v'(s)={\biggl (}\underbrace {u(s)-\int _{a}^{s}\beta (r)u(r)\,\mathrm {d} r} _{\leq \,\alpha (s)}{\biggr )}\beta (s)\exp {\biggl (}{-}\int _{a}^{s}\beta (r)\mathrm {d} r{\biggr )},\qquad s\in I,}$

where we used the assumed integral inequality for the upper estimate. Since β and the exponential are non-negative, this gives an upper estimate for the derivative of v. Since v(a) = 0, integration of this inequality from a to t gives

${\displaystyle v(t)\leq \int _{a}^{t}\alpha (s)\beta (s)\exp {\biggl (}{-}\int _{a}^{s}\beta (r)\,\mathrm {d} r{\biggr )}\mathrm {d} s.}$

Using the definition of v(t) for the first step, and then this inequality and the functional equation of the exponential function, we obtain

${\displaystyle {\begin{aligned}\int _{a}^{t}\beta (s)u(s)\,\mathrm {d} s&=\exp {\biggl (}\int _{a}^{t}\beta (r)\,\mathrm {d} r{\biggr )}v(t)\\&\leq \int _{a}^{t}\alpha (s)\beta (s)\exp {\biggl (}\underbrace {\int _{a}^{t}\beta (r)\,\mathrm {d} r-\int _{a}^{s}\beta (r)\,\mathrm {d} r} _{=\,\int _{s}^{t}\beta (r)\,\mathrm {d} r}{\biggr )}\mathrm {d} s.\end{aligned}}}$

Substituting this result into the assumed integral inequality gives Grönwall's inequality.

(b) If the function α is non-decreasing, then part (a), the fact α(s) ≤ α(t), and the fundamental theorem of calculus imply that

${\displaystyle {\begin{aligned}u(t)&\leq \alpha (t)+{\biggl (}{-}\alpha (t)\exp {\biggl (}\int _{s}^{t}\beta (r)\,\mathrm {d} r{\biggr )}{\biggr )}{\biggr |}_{s=a}^{s=t}\\&=\alpha (t)\exp {\biggl (}\int _{a}^{t}\beta (r)\,\mathrm {d} r{\biggr )},\qquad t\in I.\end{aligned}}}$

• There are no continuity assumptions on the functions α and u.
• The integral in Grönwall's inequality is allowed to give the value infinity.
• If α is the zero function and u is non-negative, then Grönwall's inequality implies that u is the zero function.
• The integrability of u with respect to μ is essential for the result. For a counterexample, let μ denote Lebesgue measure on the unit interval [0, 1], define u(0) = 0 and u(t) = 1/t for t ∈ (0, 1], and let α be the zero function.
• The version given in the textbook by S. Ethier and T. Kurtz.[4] makes the stronger assumptions that α is a non-negative constant and u is bounded on bounded intervals, but doesn't assume that the measure μ is locally finite. Compared to the one given below, their proof does not discuss the behaviour of the remainder R_n(t).

• If the measure μ has a density β with respect to Lebesgue measure, then Grönwall's inequality can be rewritten as

${\displaystyle u(t)\leq \alpha (t)+\int _{a}^{t}\alpha (s)\beta (s)\exp {\biggl (}\int _{s}^{t}\beta (r)\,\mathrm {d} r{\biggr )}\,\mathrm {d} s,\qquad t\in I.}$

• If the function α is non-negative and the density β of μ is bounded by a constant c, then

${\displaystyle u(t)\leq \alpha (t)+c\int _{a}^{t}\alpha (s)\exp {\bigl (}c(t-s){\bigr )}\,\mathrm {d} s,\qquad t\in I.}$

• If, in addition, the non-negative function α is non-decreasing, then

${\displaystyle u(t)\leq \alpha (t)+c\alpha (t)\int _{a}^{t}\exp {\bigl (}c(t-s){\bigr )}\,\mathrm {d} s=\alpha (t)\exp(c(t-a)),\qquad t\in I.}$

The proof is divided into three steps. The idea is to substitute the assumed integral inequality into itself 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 n to infinity to derive the desired variant of Grönwall's inequality.

For every natural number n including zero,

${\displaystyle u(t)\leq \alpha (t)+\int _{[a,t)}\alpha (s)\sum _{k=0}^{n-1}\mu ^{\otimes k}(A_{k}(s,t))\,\mu (\mathrm {d} s)+R_{n}(t)}$

with remainder

${\displaystyle R_{n}(t):=\int _{[a,t)}u(s)\mu ^{\otimes n}(A_{n}(s,t))\,\mu (\mathrm {d} s),\qquad t\in I,}$

where

${\displaystyle A_{n}(s,t)=\{(s_{1},\ldots ,s_{n})\in I_{s,t}^{n}\mid s_{1}<s_{2}<\cdots <s_{n}\},\qquad n\geq 1,}$

is an n-dimensional simplex and

${\displaystyle \mu ^{\otimes 0}(A_{0}(s,t)):=1.}$

We use mathematical induction. For n = 0 this is just the assumed integral inequality, because the empty sum is defined as zero.

Induction step from n to n + 1: Inserting the assumed integral inequality for the function u into the remainder gives

${\displaystyle R_{n}(t)\leq \int _{[a,t)}\alpha (s)\mu ^{\otimes n}(A_{n}(s,t))\,\mu (\mathrm {d} s)+{\tilde {R}}_{n}(t)}$

with

${\displaystyle {\tilde {R}}_{n}(t):=\int _{[a,t)}{\biggl (}\int _{[a,q)}u(s)\,\mu (\mathrm {d} s){\biggr )}\mu ^{\otimes n}(A_{n}(q,t))\,\mu (\mathrm {d} q),\qquad t\in I.}$

Using the Fubini–Tonelli theorem to interchange the two integrals, we obtain

${\displaystyle {\tilde {R}}_{n}(t)=\int _{[a,t)}u(s)\underbrace {\int _{(s,t)}\mu ^{\otimes n}(A_{n}(q,t))\,\mu (\mathrm {d} q)} _{=\,\mu ^{\otimes n+1}(A_{n+1}(s,t))}\,\mu (\mathrm {d} s)=R_{n+1}(t),\qquad t\in I.}$

Hence Claim 1 is proved for n + 1.

For every natural number n including zero and all s < t in I

${\displaystyle \mu ^{\otimes n}(A_{n}(s,t))\leq {\frac {{\bigl (}\mu (I_{s,t}){\bigr )}^{n}}{n!}}}$

with equality in case t ↦ μ([a, t]) is continuous for t ∈ I.

For n = 0, the claim is true by our definitions. Therefore, consider n ≥ 1 in the following.

Let S_n denote the set of all permutations of the indices in {1, 2, . . . , n}. For every permutation σ ∈ S_n define

${\displaystyle A_{n,\sigma }(s,t)=\{(s_{1},\ldots ,s_{n})\in I_{s,t}^{n}\mid s_{\sigma (1)}<s_{\sigma (2)}<\cdots <s_{\sigma (n)}\}.}$

These sets are disjoint for different permutations and

${\displaystyle \bigcup _{\sigma \in S_{n}}A_{n,\sigma }(s,t)\subset I_{s,t}^{n}.}$

Therefore,

${\displaystyle \sum _{\sigma \in S_{n}}\mu ^{\otimes n}(A_{n,\sigma }(s,t))\leq \mu ^{\otimes n}{\bigl (}I_{s,t}^{n}{\bigr )}={\bigl (}\mu (I_{s,t}){\bigr )}^{n}.}$

Since they all have the same measure with respect to the n-fold product of μ, and since there are n! permutations in S_n, the claimed inequality follows.

Assume now that t ↦ μ([a, t]) is continuous for t ∈ I. Then, for different indices i, j ∈ {1, 2, . . . , n}, the set

${\displaystyle \{(s_{1},\ldots ,s_{n})\in I_{s,t}^{n}\mid s_{i}=s_{j}\}}$

is contained in a hyperplane, hence by an application of Fubini's theorem its measure with respect to the n-fold product of μ is zero. Since

${\displaystyle I_{s,t}^{n}\subset \bigcup _{\sigma \in S_{n}}A_{n,\sigma }(s,t)\cup \bigcup _{1\leq i<j\leq n}\{(s_{1},\ldots ,s_{n})\in I_{s,t}^{n}\mid s_{i}=s_{j}\},}$

the claimed equality follows.

For every natural number n, Claim 2 implies for the remainder of Claim 1 that

${\displaystyle |R_{n}(t)|\leq {\frac {{\bigl (}\mu (I_{a,t}){\bigr )}^{n}}{n!}}\int _{[a,t)}|u(s)|\,\mu (\mathrm {d} s),\qquad t\in I.}$

By assumption we have μ(I_a,t) < ∞. Hence, the integrability assumption on u implies that

${\displaystyle \lim _{n\to \infty }R_{n}(t)=0,\qquad t\in I.}$

Claim 2 and the series representation of the exponential function imply the estimate

${\displaystyle \sum _{k=0}^{n-1}\mu ^{\otimes k}(A_{k}(s,t))\leq \sum _{k=0}^{n-1}{\frac {{\bigl (}\mu (I_{s,t}){\bigr )}^{k}}{k!}}\leq \exp {\bigl (}\mu (I_{s,t}){\bigr )}}$

for all s < t in I. If the function α is non-negative, then it suffices to insert these results into Claim 1 to derive the above variant of Grönwall's inequality for the function u.

In case t ↦ μ([a, t]) is continuous for t ∈ I, Claim 2 gives

${\displaystyle \sum _{k=0}^{n-1}\mu ^{\otimes k}(A_{k}(s,t))=\sum _{k=0}^{n-1}{\frac {{\bigl (}\mu (I_{s,t}){\bigr )}^{k}}{k!}}\to \exp {\bigl (}\mu (I_{s,t}){\bigr )}\qquad {\text{as }}n\to \infty }$

and the integrability of the function α permits to use the dominated convergence theorem to derive Grönwall's inequality.