I quote from this previous question:
Let $L$ be a differentiable function defined on $\mathbb{R}\times\Omega$ with $\Omega\subseteq\mathbb{R}^n$. I will say it has exponential growth if for all $O\subset\subset\Omega$ open there exists a constant $C(O)$ such that:
$$|\nabla_\xi L(x,\xi)|\leq C(O)L(x,\xi),$$
for all $\xi\in O$. A photocopy I was given from a book I can't identify uses this condition to deduce the following:
$$\frac{d}{dt}L(x,a+tb)=\langle\nabla_\xi L(x,a+tb),b\rangle\leq c|b|L(x,a),$$
and I understand that, and then:
$$L(x,a+b)\leq L(x,a)e^{c|b|}.$$
How is this last inequality deduced from the previous one?
Trouble is, the second displayed equation is mistyped. What I should have written, i.e. what I can derive from the exponential growth condition, is:
$$\frac{d}{dt}L(x,a+tb)=\langle\nabla_\xi L(x,a+tb),b\rangle\leq c|b|L(x,a+tb),$$
and if I integrate this from 0 to 1, as the answer to the previous question suggested, I get:
$$L(x,a+b)-L(x,a)\leq\int_0^1c|b|L(x,a+tb)dt,$$
which I cannot see how to turn into the last displayed equation in the quoted text. So I am asking:
Can I either derive the mistyped inequality from the exponential growth, or the final displayed equation from the correctly typed version of the mistyped inequality? And in either case, how do I do that?
The last inequality is a simple application of Gronwall's inequality. But we will prove it directly.
Let us use the notation \begin{align} f(t) = L(x, a+tb) \end{align} then we have \begin{align} f'(t) \leq c|b|f(t). \end{align} Observe by the method of integrating factors, we have \begin{align} f'(t)-c|b| f(t) \leq 0 \ \ \Rightarrow \ \ \frac{d}{dt}[e^{-b|c|t}f(t)] \leq 0 \end{align} which means \begin{align} e^{-b|c|t}f(t)-e^{-b|c|t_0}f(t_0)= \int^t_{t_0} \frac{d}{ds}[e^{-b|c|s}f(s)]\ ds \leq 0 \ \ \Rightarrow \ \ e^{-b|c|t}f(t) \leq e^{-c|b|t_0}f(t_0)= f(0). \end{align} Hence we have the desired inequality \begin{align} f(t) \leq f(0)e^{b|c|t} \ \ \Rightarrow \ \ L(x, a+tb) \leq L(x, a)e^{b|c|t}. \end{align} Now plug in $t=1$.