I want to derive a Gronwall-type inequality from the inequality below. Here all the functions are nonnegative, continuous and if you need some assumptions you may use that. $$ f^2(t) \leqslant g^2(t) + \int_0^t (f(s) +c) f(s) ds \;\;\;\; (t \in [0,T]) $$ So please help!
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2 Answers
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Consider the function $$h(t):=\int_0^t f(s)e^{t-s}ds\,,$$ which solves the ODE $h'=h+f$ with $h(0)=0$, so $$h(t):=\int_0^t \Big(h(s)+f(s)\Big)ds\, .$$ Adding the term $-c\, h(t)$ to both sides, your inequality takes a more familiar form of a Gronwall inequality: $$f(t)^2-c\;h(t)\le g(t)^2 + \int_0^t \Big(f(s)^2 -c\;h(s) \Big)ds $$ relative to the function $f(t)^2-c\; h(t)$.
answered Nov 4, 2012 at 19:59
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Consider g=0, c=2, f(t)=t.
By the way, what did you mean by "if you need some assumptions?" Are there folks out there who prove things without them?
answered Nov 4, 2012 at 16:59