Grönwall's inequality


In mathematics, Grönwall's inequality allows one to bound a function that is known to satisfy a certain 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.
Grönwall's inequality is an important tool to obtain various estimates in the theory of ordinary and stochastic differential equations. 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.
It is named for Thomas Hakon Grönwall. 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.
The differential form was proven by Grönwall in 1919.
The integral form was proven by Richard Bellman in 1943.
A nonlinear generalization of the Grönwall–Bellman inequality is known as Bihari–LaSalle inequality. Other variants and generalizations can be found in Pachpatte, B.G..

Differential form

Let denote an interval of the real line of the form or or with. Let and be real-valued continuous functions defined on. If is differentiable in the interior of and satisfies the differential inequality
then is bounded by the solution of the corresponding differential equation :
for all.
Remark: There are no assumptions on the signs of the functions and .

Proof

Define the function
Note that satisfies
with and for all. By the quotient rule
Thus the derivative of the function is non-positive and the function is bounded above by its value at the initial point of the interval :
which is Grönwall's inequality.

Integral form for continuous functions

Let denote an interval of the real line of the form or or with. Let, and be real-valued functions defined on . Assume that and are continuous and that the negative part of is integrable on every closed and bounded subinterval of .
Remarks:
Define
Using the product rule, the chain rule, the derivative of the exponential function and the fundamental theorem of calculus, we obtain for the derivative
where we used the assumed integral inequality for the upper estimate. Since and the exponential are non-negative, this gives an upper estimate for the derivative of . Since, integration of this inequality from to gives
Using the definition of for the first step, and then this inequality and the functional equation of the exponential function, we obtain
Substituting this result into the assumed integral inequality gives Grönwall's inequality.
If the function is non-decreasing, then part, the fact, and the fundamental theorem of calculus imply that

Integral form with locally finite measures

Let denote an interval of the real line of the form or or with. Let and be measurable functions defined on and let be a continuous non-negative measure on the Borel σ-algebra of satisfying for all . Assume that is integrable with respect to in the sense that
and that satisfies the integral inequality
If, in addition,
then satisfies Grönwall's inequality
for all, where denotes to open interval.

Remarks

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

Detailed proof

Claim 1: Iterating the inequality

For every natural number including zero,
with remainder
where
is an -dimensional simplex and

Proof of Claim 1

We use mathematical induction. For this is just the assumed integral inequality, because the empty sum is defined as zero.
Induction step from to :
Inserting the assumed integral inequality for the function into the remainder gives
with
Using the Fubini–Tonelli theorem to interchange the two integrals, we obtain
Hence is proved for.

Claim 2: Measure of the simplex

For every natural number including zero and all in
with equality in case is continuous for.

Proof of Claim 2

For, the claim is true by our definitions. Therefore, consider in the following.
Let denote the set of all permutations of the indices in. For every permutation define
These sets are disjoint for different permutations and
Therefore,
Since they all have the same measure with respect to the -fold product of, and since there are permutations in , the claimed inequality follows.
Assume now that is continuous for. Then, for different indices, the set
is contained in a hyperplane, hence by an application of Fubini's theorem its measure with respect to the -fold product of is zero. Since
the claimed equality follows.

Proof of Grönwall's inequality

For every natural number, implies for the remainder of that
By assumption we have. Hence, the integrability assumption on implies that
and the series representation of the exponential function imply the estimate
for all in . If the function is non-negative, then it suffices to insert these results into to derive the above variant of Grönwall's inequality for the function .
In case is continuous for, gives
and the integrability of the function permits to use the dominated convergence theorem to derive Grönwall's inequality.