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Proof of Gronwall's Inequality 📂Lemmas

Proof of Gronwall's Inequality

Theorem

Let’s assume there are two continuous functions $f,w : I \to \mathbb{R}$ defined in an interval $I \subset \mathbb{R}$ that contains the minimum value $a \in \mathbb{R}$. If $w$ is $\forall t \in I$ in $w(t) \ge 0$ and for some constant $C \in \mathbb{R}$, $$ f(t) \le C + \int_{a}^{t} w(s) f(s) ds \qquad , \forall t \in I $$ then the following holds. $$ f(t) \le C \exp \left( \int_{a}^{t} w(s) ds \right) \qquad , \forall t \in I $$

Explanation

The fact that interval $I$ has a minimum value $a$ means that $I$ looks like the following with respect to $a < b$. $$ [a,b] \text{ or } [a,b) \text{ or } [a,\infty) $$

Proof 1

$$ \begin{align*} \alpha (t) =& C + \int_{a}^{t} w(s) f(s) ds \\ \beta (t) =& C \exp \left( \int_{a}^{t} w(s) ds \right) \end{align*} $$

When we define two functions $\alpha, \beta$ as shown above and suppose $\displaystyle \gamma (t) := {{ \alpha (t) } \over { \beta (t) }}$, then $\gamma (a) = 1$ and by the quotient rule of differentiation, $$ \begin{align*} & \gamma ' (t) \\ =& {{ \alpha ' (t) \beta (t) - \alpha (t) \beta ' (t) } \over { \left[ \beta (t) \right]^{2} }} \\ =& {{ w(t) f(t) \cdot C \exp \left( \int_{a}^{t} w(s) ds \right) - \left( C + \int_{a}^{t} w(s) f(s) ds \right) \cdot w (t) C \exp \left( \int_{a}^{t} w(s) ds \right) } \over { \left( C \exp \left( \int_{a}^{t} w(s) ds \right) \right)^{2} }} \\ =& w(t) {{ f(t) - C - \int_{a}^{t} w(s) f(s) ds } \over { C \exp \left( \int_{a}^{t} w(s) ds \right) }} \end{align*} $$ it follows. Given that $\displaystyle f(t) \le C + \int_{a}^{t} w(s) f(s) ds$ and $w(t) \ge 0$, then $\gamma ' (t) \le 0$. Therefore, $\gamma (t)$ is a non-increasing function, and since $\gamma (t) \le 1$, the following is established. $$ \begin{align*} & \gamma (t) = {{ \alpha (t) } \over { \beta (t) }} \le 1 \\ \implies& \alpha (t) \le \beta (t) \\ \implies& C + \int_{a}^{t} w(s) f(s) ds \le C \exp \left( \int_{a}^{t} w(s) ds \right) \\ \implies& f(t) \le C + \int_{a}^{t} w(s) f(s) ds \le C \exp \left( \int_{a}^{t} w(s) ds \right) \\ \implies& f(t) \le C \exp \left( \int_{a}^{t} w(s) ds \right) \end{align*} $$