{"history":[],"article":{"id":"c400ee2d-408e-4556-81c4-4caafd2f3958","title":"Grönwall's inequality","slug":"gr-nwall-s-inequality","content":":::summary\n:::\n\n# \nIn [mathematics](https://bliptext.com/articles/mathematics), **Grönwall's inequality** (also called **Grönwall's lemma** or the **Grönwall–Bellman inequality**) allows one to bound a function that is known to satisfy a certain [differential](https://bliptext.com/articles/differential-inequality) or [integral inequality](https://bliptext.com/articles/integral-inequality) by the solution of the corresponding [differential](https://bliptext.com/articles/differential-inequality) or [integral equation](https://bliptext.com/articles/integral-equation).\nThere are two forms of the lemma, a differential form and an integral form.\nFor the latter there are several variants.\nGrönwall's inequality is an important tool to obtain various estimates in the theory of [ordinary](https://bliptext.com/articles/ordinary-differential-equation) and [stochastic differential equations](https://bliptext.com/articles/stochastic-differential-equation).\nIn particular, it provides a [comparison theorem](https://bliptext.com/articles/comparison-theorem) that can be used to prove [uniqueness](https://bliptext.com/articles/uniqueness-quantification) of a solution to the [initial value problem](https://bliptext.com/articles/initial-value-problem); see the [Picard–Lindelöf theorem](https://bliptext.com/articles/picard-lindel-f-theorem).\nIt is named for [Thomas Hakon Grönwall](https://bliptext.com/articles/thomas-hakon-gr-nwall) (1877–1932).\nGrönwall is the Swedish spelling of his name, but he spelled his name as Gronwall in his scientific publications after emigrating to the United States.\nThe inequality was first proven by Grönwall in 1919 (the integral form below with\nα\nand\nβ\nbeing constants). [Richard Bellman](https://bliptext.com/articles/richard-bellman) proved a slightly more general integral form in 1943.\nA nonlinear generalization of the Grönwall–Bellman inequality is known as [Bihari–LaSalle inequality](https://bliptext.com/articles/bihari-lasalle-inequality).\nOther variants and generalizations can be found in Pachpatte, B.G. (1998).\n\n# Differential form\nLet I denote an [interval](https://bliptext.com/articles/interval-mathematics) of the [real line](https://bliptext.com/articles/real-line) of the form [a, \\infty) or [a, b] or [a, b) with a < b. Let \\beta and u be real-valued [continuous functions](https://bliptext.com/articles/continuous-functions) defined on I. If u is [differentiable](https://bliptext.com/articles/derivative) in the [interior](https://bliptext.com/articles/interior-topology) I^\\circ of I (the [interval](https://bliptext.com/articles/interval-mathematics) I without the end points a and possibly b) and satisfies the differential inequality\nthen u is bounded by the solution of the corresponding differential equation :\nfor all t \\in I.\n**Remark:** There are no assumptions on the signs of the functions \\beta and u.\n\n## Proof\nDefine the function\nNote that v satisfies\nwith v(a) = 1 and v(t) > 0 for all t \\in I. By the [quotient rule](https://bliptext.com/articles/quotient-rule)\nThus the derivative of the function u(t)/v(t) is non-positive and the function is bounded above by its value at the initial point a of the interval I:\nwhich is Grönwall's inequality.\n\n# Integral form for continuous functions\nLet\nI\ndenote an [interval](https://bliptext.com/articles/interval-mathematics) of the [real line](https://bliptext.com/articles/real-line) of the form or or with\na < b\n. Let\nα\n,\nβ\nand\nu\nbe real-valued functions defined on\nI\n. Assume that\nβ\nand\nu\nare continuous and that the negative part of\nα\nis integrable on every closed and bounded subinterval of\nI\n.\nβ\nis non-negative and if\nu\nsatisfies the integral inequality\nα\nis non-decreasing, then\n**Remarks:**\nα\nand\nu\n.\nu\nis not needed for the integral form.\nβ\nand\nu\n, see the version in the next section.\n\n## Proof\n(a) Define\nUsing the [product rule](https://bliptext.com/articles/product-rule), the [chain rule](https://bliptext.com/articles/chain-rule), the derivative of the [exponential function](https://bliptext.com/articles/exponential-function) and the [fundamental theorem of calculus](https://bliptext.com/articles/fundamental-theorem-of-calculus), we obtain for the derivative\nwhere we used the assumed integral inequality for the upper estimate.\nSince\nβ\nand the exponential are non-negative, this gives an upper estimate for the derivative of v(s).\nSince v(a)=0, integration of this inequality from\na\nto\nt\ngives\nUsing the definition of v(t) from the first step, and then this inequality and the property, we obtain\nSubstituting this result into the assumed integral inequality gives Grönwall's inequality.\n(b) If the function\nα\nis non-decreasing, then part (a), the fact\nα(s) ≤ α(t)\n, and the fundamental theorem of calculus imply that\n\n# Integral form with locally finite measures\nLet\nI\ndenote an [interval](https://bliptext.com/articles/interval-mathematics) of the [real line](https://bliptext.com/articles/real-line) of the form or or with\na < b\n. Let\nα\nand\nu\nbe [measurable functions](https://bliptext.com/articles/measurable-function) defined on\nI\nand let\nμ\nbe a continuous non-negative measure on the [Borel σ-algebra](https://bliptext.com/articles/borel-algebra) of\nI\nsatisfying\nμ( [ a, t ] ) < ∞\nfor all\nt ∈ I\n(this is certainly satisfied when\nμ\nis a [locally finite measure](https://bliptext.com/articles/locally-finite-measure)). Assume that\nu\nis integrable with respect to\nμ\nin the sense that\nand that\nu\nsatisfies the integral inequality\nIf, in addition,\nα\nis non-negative or\nt μ( [ a, t ] )\nis continuous for\nt ∈ I\nand the function\nα\nis integrable with respect to\nμ\nin the sense that\nthen\nu\nsatisfies Grönwall's inequality\nfor all\nt ∈ I\n, where\nIs,t\ndenotes to open interval.\n\n## Remarks\nα\nand\nu\n.\nα\nis the zero function and\nu\nis non-negative, then Grönwall's inequality implies that\nu\nis the zero function.\nu\nwith respect to\nμ\nis essential for the result.\nFor a [counterexample](https://bliptext.com/articles/counterexample), let\nμ\ndenote [Lebesgue measure](https://bliptext.com/articles/lebesgue-measure) on the [unit interval](https://bliptext.com/articles/unit-interval), define\nand\nfor\nt ∈\n, and let\nα\nbe the zero function.\nα\nis a non-negative constant and\nu\nis bounded on bounded intervals, but doesn't assume that the measure\nμ\nis locally finite.\nCompared to the one given below, their proof does not discuss the behaviour of the remainder\nRn(t)\n.\n\n## Special cases\nμ\nhas a density\nβ\nwith respect to Lebesgue measure, then Grönwall's inequality can be rewritten as\nα\nis non-negative and the density\nβ\nof\nμ\nis bounded by a constant\nc\n, then\nα\nis non-decreasing, then\n\n## Outline of proof\nThe proof is divided into three steps.\nThe idea is to substitute the assumed integral inequality into itself\nn\ntimes.\nThis is done in Claim 1 using mathematical induction.\nIn Claim 2 we rewrite the measure of a simplex in a convenient form, using the permutation invariance of product measures.\nIn the third step we pass to the limit\nn\nto infinity to derive the desired variant of Grönwall's inequality.\n\n## Detailed proof\n\n### Claim 1: Iterating the inequality\nFor every natural number\nn\nincluding zero,\nwith remainder\nwhere\nis an\nn\n-dimensional [simplex](https://bliptext.com/articles/simplex) and\n\n### Proof of Claim 1\nWe use [mathematical induction](https://bliptext.com/articles/mathematical-induction).\nFor\nthis is just the assumed integral inequality, because the [empty sum](https://bliptext.com/articles/empty-sum) is defined as zero.\nInduction step from\nn\nto\nn + 1\nInserting the assumed integral inequality for the function\nu\ninto the remainder gives\nwith\nUsing the [Fubini–Tonelli theorem](https://bliptext.com/articles/fubini-s-theorem) to interchange the two integrals, we obtain\nHence [Claim 1](https://bliptext.com/articles/) is proved for\nn + 1\n.\n\n### Claim 2: Measure of the simplex\nFor every natural number\nn\nincluding zero and all\ns < t\nin\nI\nwith equality in case\nt μ( [ a, t ] )\nis continuous for\nt ∈ I\n.\n\n### Proof of Claim 2\nFor\n, the claim is true by our definitions.\nTherefore, consider\nn ≥ 1\nin the following.\nLet\nSn\ndenote the set of all [permutations](https://bliptext.com/articles/permutation) of the indices in\n{1, 2, . . ., n}\n. For every permutation\nσ ∈ Sn\ndefine\nThese sets are disjoint for different permutations and\nTherefore,\nSince they all have the same measure with respect to the\nn\n-fold product of\nμ\n, and since there are\nn!\npermutations in\nSn\n, the claimed inequality follows.\nAssume now that\nt μ( [ a, t ] )\nis continuous for\nt ∈ I\n. Then, for different indices\ni, j ∈ {1, 2,.\n. ., n}\n, the set\nis contained in a [hyperplane](https://bliptext.com/articles/hyperplane), hence by an application of [Fubini's theorem](https://bliptext.com/articles/fubini-s-theorem) its measure with respect to the\nn\n-fold product of\nμ\nis zero.\nSince\nthe claimed equality follows.\n\n### Proof of Grönwall's inequality\nFor every natural number\nn\n, [Claim 2](https://bliptext.com/articles/) implies for the remainder of [Claim 1](https://bliptext.com/articles/) that\nBy assumption we have\nμ(Ia,t) < ∞\n. Hence, the integrability assumption on\nu\nimplies that\n[Claim 2](https://bliptext.com/articles/) and the [series representation](https://bliptext.com/articles/characterizations-of-the-exponential-function) of the exponential function imply the estimate\nfor all\ns < t\nin\nI\n. If the function\nα\nis non-negative, then it suffices to insert these results into [Claim 1](https://bliptext.com/articles/) to derive the above variant of Grönwall's inequality for the function\nu\n.\nIn case\nt μ( [ a, t ] )\nis continuous for\nt ∈ I\n, [Claim 2](https://bliptext.com/articles/) gives\nand the integrability of the function\nα\npermits to use the [dominated convergence theorem](https://bliptext.com/articles/dominated-convergence-theorem) to derive Grönwall's inequality.","current_revision":null,"created_by":"kACnuQmH5k9Nr5VNVEpVeg7XpvUmMctD","created_at":"2025-01-25T21:37:02.159Z","updated_at":"2025-01-25T21:37:02.159Z","search_vector":"'grönwall':1 'inequ':3","search_content":" \n \n\n# \nIn [mathematics](https://bliptext.com/articles/mathematics), **Grönwall's inequality** (also called **Grönwall's lemma** or the **Grönwall–Bellman inequality**) allows one to bound a function that is known to satisfy a certain [differential](https://bliptext.com/articles/differential-inequality) or [integral inequality](https://bliptext.com/articles/integral-inequality) by the solution of the corresponding [differential](https://bliptext.com/articles/differential-inequality) or [integral equation](https://bliptext.com/articles/integral-equation).\nThere are two forms of the lemma, a differential form and an integral form.\nFor the latter there are several variants.\nGrönwall's inequality is an important tool to obtain various estimates in the theory of [ordinary](https://bliptext.com/articles/ordinary-differential-equation) and [stochastic differential equations](https://bliptext.com/articles/stochastic-differential-equation).\nIn particular, it provides a [comparison theorem](https://bliptext.com/articles/comparison-theorem) that can be used to prove [uniqueness](https://bliptext.com/articles/uniqueness-quantification) of a solution to the [initial value problem](https://bliptext.com/articles/initial-value-problem); see the [Picard–Lindelöf theorem](https://bliptext.com/articles/picard-lindel-f-theorem).\nIt is named for [Thomas Hakon Grönwall](https://bliptext.com/articles/thomas-hakon-gr-nwall) (1877–1932).\nGrönwall is the Swedish spelling of his name, but he spelled his name as Gronwall in his scientific publications after emigrating to the United States.\nThe inequality was first proven by Grönwall in 1919 (the integral form below with\nα\nand\nβ\nbeing constants). [Richard Bellman](https://bliptext.com/articles/richard-bellman) proved a slightly more general integral form in 1943.\nA nonlinear generalization of the Grönwall–Bellman inequality is known as [Bihari–LaSalle inequality](https://bliptext.com/articles/bihari-lasalle-inequality).\nOther variants and generalizations can be found in Pachpatte, B.G. (1998).\n\n# Differential form\nLet I denote an [interval](https://bliptext.com/articles/interval-mathematics) of the [real line](https://bliptext.com/articles/real-line) of the form [a, \\infty) or [a, b] or [a, b) with a 0 for all t \\in I. By the [quotient rule](https://bliptext.com/articles/quotient-rule)\nThus the derivative of the function u(t)/v(t) is non-positive and the function is bounded above by its value at the initial point a of the interval I:\nwhich is Grönwall's inequality.\n\n# Integral form for continuous functions\nLet\nI\ndenote an [interval](https://bliptext.com/articles/interval-mathematics) of the [real line](https://bliptext.com/articles/real-line) of the form or or with\na < b\n. Let\nα\n,\nβ\nand\nu\nbe real-valued functions defined on\nI\n. Assume that\nβ\nand\nu\nare continuous and that the negative part of\nα\nis integrable on every closed and bounded subinterval of\nI\n.\nβ\nis non-negative and if\nu\nsatisfies the integral inequality\nα\nis non-decreasing, then\n**Remarks:**\nα\nand\nu\n.\nu\nis not needed for the integral form.\nβ\nand\nu\n, see the version in the next section.\n\n## Proof\n(a) Define\nUsing the [product rule](https://bliptext.com/articles/product-rule), the [chain rule](https://bliptext.com/articles/chain-rule), the derivative of the [exponential function](https://bliptext.com/articles/exponential-function) and the [fundamental theorem of calculus](https://bliptext.com/articles/fundamental-theorem-of-calculus), we obtain for the derivative\nwhere we used the assumed integral inequality for the upper estimate.\nSince\nβ\nand the exponential are non-negative, this gives an upper estimate for the derivative of v(s).\nSince v(a)=0, integration of this inequality from\na\nto\nt\ngives\nUsing the definition of v(t) from the first step, and then this inequality and the property, we obtain\nSubstituting this result into the assumed integral inequality gives Grönwall's inequality.\n(b) If the function\nα\nis non-decreasing, then part (a), the fact\nα(s) ≤ α(t)\n, and the fundamental theorem of calculus imply that\n\n# Integral form with locally finite measures\nLet\nI\ndenote an [interval](https://bliptext.com/articles/interval-mathematics) of the [real line](https://bliptext.com/articles/real-line) of the form or or with\na < b\n. Let\nα\nand\nu\nbe [measurable functions](https://bliptext.com/articles/measurable-function) defined on\nI\nand let\nμ\nbe a continuous non-negative measure on the [Borel σ-algebra](https://bliptext.com/articles/borel-algebra) of\nI\nsatisfying\nμ( [ a, t ] ) < ∞\nfor all\nt ∈ I\n(this is certainly satisfied when\nμ\nis a [locally finite measure](https://bliptext.com/articles/locally-finite-measure)). Assume that\nu\nis integrable with respect to\nμ\nin the sense that\nand that\nu\nsatisfies the integral inequality\nIf, in addition,\nα\nis non-negative or\nt μ( [ a, t ] )\nis continuous for\nt ∈ I\nand the function\nα\nis integrable with respect to\nμ\nin the sense that\nthen\nu\nsatisfies Grönwall's inequality\nfor all\nt ∈ I\n, where\nIs,t\ndenotes to open interval.\n\n## Remarks\nα\nand\nu\n.\nα\nis the zero function and\nu\nis non-negative, then Grönwall's inequality implies that\nu\nis the zero function.\nu\nwith respect to\nμ\nis essential for the result.\nFor a [counterexample](https://bliptext.com/articles/counterexample), let\nμ\ndenote [Lebesgue measure](https://bliptext.com/articles/lebesgue-measure) on the [unit interval](https://bliptext.com/articles/unit-interval), define\nand\nfor\nt ∈\n, and let\nα\nbe the zero function.\nα\nis a non-negative constant and\nu\nis bounded on bounded intervals, but doesn't assume that the measure\nμ\nis locally finite.\nCompared to the one given below, their proof does not discuss the behaviour of the remainder\nRn(t)\n.\n\n## Special cases\nμ\nhas a density\nβ\nwith respect to Lebesgue measure, then Grönwall's inequality can be rewritten as\nα\nis non-negative and the density\nβ\nof\nμ\nis bounded by a constant\nc\n, then\nα\nis non-decreasing, then\n\n## Outline of proof\nThe proof is divided into three steps.\nThe idea is to substitute the assumed integral inequality into itself\nn\ntimes.\nThis is done in Claim 1 using mathematical induction.\nIn Claim 2 we rewrite the measure of a simplex in a convenient form, using the permutation invariance of product measures.\nIn the third step we pass to the limit\nn\nto infinity to derive the desired variant of Grönwall's inequality.\n\n## Detailed proof\n\n### Claim 1: Iterating the inequality\nFor every natural number\nn\nincluding zero,\nwith remainder\nwhere\nis an\nn\n-dimensional [simplex](https://bliptext.com/articles/simplex) and\n\n### Proof of Claim 1\nWe use [mathematical induction](https://bliptext.com/articles/mathematical-induction).\nFor\nthis is just the assumed integral inequality, because the [empty sum](https://bliptext.com/articles/empty-sum) is defined as zero.\nInduction step from\nn\nto\nn + 1\nInserting the assumed integral inequality for the function\nu\ninto the remainder gives\nwith\nUsing the [Fubini–Tonelli theorem](https://bliptext.com/articles/fubini-s-theorem) to interchange the two integrals, we obtain\nHence [Claim 1](https://bliptext.com/articles/) is proved for\nn + 1\n.\n\n### Claim 2: Measure of the simplex\nFor every natural number\nn\nincluding zero and all\ns < t\nin\nI\nwith equality in case\nt μ( [ a, t ] )\nis continuous for\nt ∈ I\n.\n\n### Proof of Claim 2\nFor\n, the claim is true by our definitions.\nTherefore, consider\nn ≥ 1\nin the following.\nLet\nSn\ndenote the set of all [permutations](https://bliptext.com/articles/permutation) of the indices in\n{1, 2, . . ., n}\n. For every permutation\nσ ∈ Sn\ndefine\nThese sets are disjoint for different permutations and\nTherefore,\nSince they all have the same measure with respect to the\nn\n-fold product of\nμ\n, and since there are\nn!\npermutations in\nSn\n, the claimed inequality follows.\nAssume now that\nt μ( [ a, t ] )\nis continuous for\nt ∈ I\n. Then, for different indices\ni, j ∈ {1, 2,.\n. ., n}\n, the set\nis contained in a [hyperplane](https://bliptext.com/articles/hyperplane), hence by an application of [Fubini's theorem](https://bliptext.com/articles/fubini-s-theorem) its measure with respect to the\nn\n-fold product of\nμ\nis zero.\nSince\nthe claimed equality follows.\n\n### Proof of Grönwall's inequality\nFor every natural number\nn\n, [Claim 2](https://bliptext.com/articles/) implies for the remainder of [Claim 1](https://bliptext.com/articles/) that\nBy assumption we have\nμ(Ia,t) < ∞\n. Hence, the integrability assumption on\nu\nimplies that\n[Claim 2](https://bliptext.com/articles/) and the [series representation](https://bliptext.com/articles/characterizations-of-the-exponential-function) of the exponential function imply the estimate\nfor all\ns < t\nin\nI\n. If the function\nα\nis non-negative, then it suffices to insert these results into [Claim 1](https://bliptext.com/articles/) to derive the above variant of Grönwall's inequality for the function\nu\n.\nIn case\nt μ( [ a, t ] )\nis continuous for\nt ∈ I\n, [Claim 2](https://bliptext.com/articles/) gives\nand the integrability of the function\nα\npermits to use the [dominated convergence theorem](https://bliptext.com/articles/dominated-convergence-theorem) to derive Grönwall's inequality.","revisionCount":0},"hasMore":false,"nextCursor":null}

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