# Bootstrap proof of Gronwall's inequality (exercise 1.26 in Tao)

I am trying to do Exercise 1.26 in Tao's book on nonlinear dispersive equations. I wanted to check if my proof seemed correct/in the spirit of the exercise (proving Gronwall via a bootstrap), since it is more convoluted than I would have thought. I would be interested if there is a more direct (bootstrap) proof. (More detailed questions listed below.) (Also included at the end is a lemma's proof that is needed, but is unrelated to the bootstrap method, so is included at the end.)

Exercise: Use the bootstrap method to give another proof of Gronwall's integral inequality: for $$A, B, u \geq 0$$ continuous on an interval $$[t_1, t_2]$$, if $$u(t) \leq A + \int_{t_0}^t B(s) u(s)\, ds,$$ then $$u(t) \leq A\exp(\int_0^t B(s)\, ds).$$

Proof of exercise. The proof is by bootstrap argument. Let $$H(t)$$ be the statement that $$u(t') \leq (1 + 2\epsilon)A\exp(\int_{t_0}^{t'} B(s)\, ds) \hspace{5mm} \forall t' \in [t_0, t]$$ and $$C(t)$$ the statement that $$u(t') \leq (1 + \epsilon)A\exp(\int_{t_0}^{t'} B(s)\, ds) \hspace{5mm} \forall t' \in [t_0, t].$$ To complete the bootstrap argument one must show $$H(t) \implies C(t)$$. We do this as follows: \begin{align} u(t) &\leq A + \int_{t_0}^t B(s) u(s)\, ds \\ &\leq A + \int_{t_0}^t B(s)(1 + 2\epsilon)A\exp(\int_{t_0}^s B(\tau)\, d\tau)\, ds \\ &=: F(t). \end{align} We want to show that this is less than or equal to $$(1 + \epsilon)A\exp(\int_{t_0}^t B(s)\, ds)$$, which we call $$G(t)$$. To do this, we note that: $$F(t_0) = A < (1 + \epsilon)A = G(t_0). \tag{1}$$ Thus we can conclude there is some $$\delta > 0$$ such that on $$[t_0, t_0 + \delta]$$, $$F(t) < G(t)$$. Therefore $$H(t) \implies C(t)$$, so we have proven the desired result, but only on $$t\in [t_0, t_0 + \delta]$$. Since we are given so little hypotheses on $$u$$, I don't see how to conclude that $$t_0 + \delta$$ can be immediately replaced by $$t_1$$.

Therefore, we need to do the following. First, we note that the above proof can be repeated on $$[t_0 + \delta, t_0 + 2\delta]$$ to show that $$u(t) \leq A'\exp(\int_{t_0 + \delta}^t B(s)\, ds) \hspace{5mm} \forall t_0 + \delta \leq t \leq t_0 + 2\delta,$$ where $$A' = A + \int_{t_0}^{t_0 + \delta} B(s)u(s)\, ds. \tag{2}$$ Note that in order to make this $$\delta$$ uniform one needs in (1) to choose it based on $$B$$'s maximum, which is finite. Thus the above proof pushes through to show (2). Next, one needs the following lemma:

Lemma. Suppose $$t_0 \leq t_1 \leq t_2$$, and suppose we have: $$u(t) \leq A\exp(\int_{t_0}^t B(s)\, ds) \hspace{5mm} \forall t_0 \leq t \leq t_1 \tag{A}$$ and $$u(t) \leq A'\exp(\int_{t_1}^t B(s)\, ds) \hspace{5mm} \forall t_1 \leq t \leq t_2 \tag{B},$$ where $$A' = A + \int_{t_0}^{t_1} B(s) u(s)\, ds.$$ Then we have: $$u(t) \leq A\exp(\int_{t_0}^t B(s)\, ds) \hspace{5mm} \forall t_0 \leq t \leq t_2. \tag{C}$$

Essentially, this lemma says that we can "glue" the results above on $$[t_0, t_0 + \delta], [t_0 + \delta, t_0 + 2\delta]$$ to achieve the desired conslusion on $$[t_0, t_0 + 2\delta]$$. Since $$\delta$$ was uniform one can continue until $$t_1$$ is reached, completing the proof.

Questions: 1) Does the above proof seem correct? What I really want to know is if this seems like the intended "bootstrap method" proof, which is maybe harder to say (and more subjective). For instance, the bootstrap method came in only as one small portion of the argument; I could not immediately conclude the exercise after using the bootstrap method, more work had to be done. Is there a simpler/shorter proof where the bootstrap method is the main ingredient used?

For completeness, here is a proof of the lemma. Note that since (A) implies (C) for $$t_0 \leq t \leq t_1$$, we only need to show (C) for $$t_1 \leq t \leq t_2$$. By (B), we have, for all $$t_1 \leq t \leq t_2$$: \begin{align} u(t) &\leq (A + \int_{t_0}^{t_1}B(s)u(s)\, ds)\exp(\int_{t_1}^t B(s)\, ds) \\ &= (A + \int_{t_0}^{t_1}B(s)u(s)\, ds)\exp(\int_{t_0}^t B(s)\, ds)\exp(-\int_{t_0}^{t_1} B(s)\, ds). \end{align} The goal is to show that this is less or equal to $$A\exp(\int_{t_0}^t B(s)\, ds)$$. Therefore it suffices to show that $$F(t_1) := A + \int_{t_0}^{t_1}B(s)u(s)\, ds \leq A\exp(\int_{t_0}^{t_1} B(s)\, ds) =: G(t_1).$$ To show this, note that $$F(t_0) = A = G(t_0)$$. Also, $$F'(t) = B(t) u(t), \hspace{5mm} G'(t) = AB(t) \exp(\int_{t_0}^t B(s)\, ds).$$ But we have already, by assumption (A), that $$F'(t) \leq G'(t)$$ exactly, for all $$t_0 \leq t \leq t_1$$. Therefore integrating we find $$F(t_1) \leq G(t_1)$$, which was the desired conclusion.

I think your proof is correct, but it can be made simpler by first observing that $$\int_{t_0}^t B(s)e^{\int_{t_0}^sB(s')\,ds'}\,ds=e^{\int_{t_0}^tB(s)\,ds}-1$$ The inequality $$F(t)\leq G(t)$$ is then equivalent to $$e^{\int_{t_0}^{t}B(s)\,ds}\leq 2$$, which does not necessarily hold for all $$t$$. It looks like you fixed this by working on intervals of size $$\delta$$, but instead you can let $$H(t)$$ be $$u(s)\leq (1+M\epsilon)Ae^{\int_{t_0}^sB(t)\,dt}$$ for all $$s\in [t_0,t]$$ where $$M$$ is chosen so that $$(M-1)e^{\int_{t_0}^{t_1}B(t)\,dt}\leq M$$