# For nonnegative continuous $f$, if $f'(x)-f(x)\leq 0, \forall x\geq 0$ and $f(0)=0$, find the value of $f(1)$. [duplicate]

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 $f(1)$?

## marked as duplicate by Paramanand Singh, Winther, 6005, Claude Leibovici calculus StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Aug 2 '15 at 8:24

• The approach might be trying to prove $f(x)=0$ for all $x\geq 0$ – user negative one over twelve Aug 2 '15 at 7:03
• 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. – Paramanand Singh Aug 2 '15 at 7:38