Suppose that a continuous and integrable function $g$,$g\ge 0$ on the interval $[e,f]$ and $A\geq 0$ is a constant such that $g(x)\le A \int_{e}^{x} g(s) ds,$ then $g(x)\equiv 0$ for all $x\in [e,f]$.
How to deduce $g(x)\equiv 0$?
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0Do we know that $g$ is differentiable? – 2012-01-28
2 Answers
Put $F(x):=\int_e^xg(t)dt$. Then $F$ is differentiable and F'(x)\leq AF(x), so F'(x)e^{-Ax}-Ae^{-Ax}F(x)\leq 0, hence $\frac d{dx}\left(F(x)e^{-Ax}\right)\leq 0$. We deduce that $F(x)e^{-Ax}\leq F(0)e^{-A0}$ but $F(0)=0$ so $F(x)\leq 0$. On the other hand $F(x)\geq 0$ since $g\geq 0$, so $F(x)=0$ for all $x$ and F'(x)=g(x)=0 for all $x\in ]e,f[$ and by continuity for all $x\in [e,f]$.
Hint: If $g(x)\leqslant h(x)$ for every $x$ in $[e,f]$, then $g(x)\leqslant (Th)(x)$ for every $x$ in $[e,f]$, where $ (Th)(x)=A\int_e^xh(s)\mathrm ds. $ It follows that $g\leqslant h_n$ for every $n\geqslant0$, where the sequence $(h_n)_{n\geqslant0}$ is defined by $h_0=h$ and $h_{n+1}=Th_n$ for every $n\geqslant0$.
Appplication: Show that there exists a constant $C$ such that $g(x)\leqslant AC$ for every $x$ in $[e,f]$. Compute each $h_n$ when $h_0$ is the constant function $h=AC$. Show that for every fixed $x$ in $[e,f]$, $h_n(x)\to0$ when $n\to\infty$. Conclude.
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0@Mathematics: [Gronwall](http://en.wikipedia.org/wiki/Gronwall%27s_inequality). – 2012-02-06