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Let $f(x)$ be a non-negative continuous function such that $f'(x)-f(x)\leq 0, \forall x\geq 0$ and $f(0)=0,$find the value of $f(1)$.

$f'(x)-f(x)\leq 0$$\Rightarrow f'(x)\leq f(x)$$\Rightarrow \frac{f'(x)}{f(x)}\leq 1$$\Rightarrow \int\frac{f'(x)}{f(x)}dx\leq \int1 dx$$\Rightarrow \log f(x) \leq x$.Then could not solve further.Can someone assist me find $f(1)$?

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marked as duplicate by Paramanand Singh, Winther, 6005, Claude Leibovici calculus Aug 2 '15 at 8:24

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    $\begingroup$ The approach might be trying to prove $f(x)=0$ for all $x\geq 0$ $\endgroup$ – gamma Aug 2 '15 at 7:03
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    $\begingroup$ Hint: Let $g(x) = e^{-x}f(x) \geq 0$ so that $g'(x) = e^{-x}\{f'(x) - f(x)\} \leq 0$ so $g(x)$ is decreasing. $\endgroup$ – Paramanand Singh Aug 2 '15 at 7:38
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\begin{align*} f(1) &= e\left[ e^{-1} f(1) \right] \\ &= e\left[ e^0 f(0) + \int_0^1 \frac{d}{dx} (e^{-x} f(x)) \; dx \right] \\ &= e\left[ 0 + \int_0^1 e^{-x} \left[ f'(x) - f(x) \right] \right] \\ &\le e\left[ \int_0^1 e^{-x} \left[ 0 \right] \right] \\ &= 0. \end{align*}

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