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The exercise is:

Let $K \geq 0$, $f,g \geq 0$ continuous functions from $[a,b]$ to $\Bbb R$ and $x_0 \in ]a,b[$. Suppose that $f(x) \leq K + \left|\int_{x_0}^x f(t)g(t) \ \mathrm{d}t\right|,$ for $a \leq x \leq b$. Show that: $$f(x) \leq Ke^{\left|\int_{x_0}^x f(t)g(t) \ \mathrm{d}t\right|}.$$

I feel like I'm just going in circles here. Since it's a problem working with absolute values, I broke it into cases, but I'm getting something a bit weird.. and not exactly what the exercise asks. There goes:

Case 1: If $x \geq x_0$, we have as hypothesis $f(x) \leq K + \int_{x_0}^x f(t)g(t) \ \mathrm{d}t$. Then, we define: $$u(x) = K + \int_{x_0}^x f(t)g(t) \ \mathrm{d}t,$$ and differentiating, we have: $$u'(x) = f(x)g(x) \leq u(x)g(x) \implies u'(x) - g(x)u(x) \leq 0 \implies \left(e^{-\int_{x_0}^x g(t) \ \mathrm{d}t}u(x)\right)' \leq 0.$$ Applying $\int_{x_0}^x$, we obtain: $$e^{-\int_{x_0}^x g(t) \ \mathrm{d}t}u(x) - K \leq 0 \implies f(x) \leq u(x) \leq Ke^{\int_{x_0}^x g(t) \ \mathrm{d}t}.$$

Case 2: If $x < x_0$, we have as hypothesis $f(x) \leq K - \int_{x_0}^x f(t)g(t) \ \mathrm{d}t$. Then, we define: $$u(x) = K - \int_{x_0}^x f(t)g(t) \ \mathrm{d}t,$$ and differentiating, we have: $$u'(x) = -f(x)g(x) \geq -u(x)g(x) \implies u'(x) + g(x)u(x) \geq 0 \implies \left(e^{\int_{x_0}^x g(t) \ \mathrm{d}t}u(x)\right)' \geq 0.$$ Applying $\int_{x_0}^x$, we obtain: $$e^{\int_{x_0}^x g(t) \ \mathrm{d}t}u(x) - K \leq 0 \implies f(x) \leq u(x) \leq Ke^{-\int_{x_0}^x g(t) \ \mathrm{d}t}.$$

I'm uncomfortable with this swapped inequality, and there's no $f$ in the exponential. Can someone take a look at this? Thanks.

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  • $\begingroup$ Your proofs look fine to me, so unless I'm overlooking the same glitch as you did, your estimates hold. Note that your estimates are weaker if $f \leqslant 1$, however. I trust you have looked at wikipedia or something similar already to see how the standard proof looks. $\endgroup$ – Daniel Fischer Oct 27 '14 at 20:19
  • $\begingroup$ Yes, I based myself on my lecture notes (which have the standard proof).. but the problem is, although what I've done may be right, it is not close from what was asked.. $\endgroup$ – Ivo Terek Oct 27 '14 at 20:23

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