Let $f:\mathbb R \to \mathbb R$ be a differentiable function such that $f(0)=0$ and $f'(x)>f(x),\forall x \in \mathbb R$ ; then is it true that $f(x)>0,\forall x>0$ ?
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perhaps looking at average value on $[0,t]$ might be relevant – Mirko Dec 15 '15 at 15:17
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@Mirko: For every $x>0$ , $\exists c_x \in (0,x)$ such that $f'(x)/x > f(x)/x=f'(c_x)>f(c_x)$ ; but I cannot proceed further – Dec 15 '15 at 15:21
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I posted a closely-related MSE question that is based on yours. – Mirko Dec 15 '15 at 20:58
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Because $f'(x)>f(x)$ we have that $ e^{-x} f(x)$ is strictly increasing.That is the conclusion.
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Your solution works even if we replace the requirement $f'(x)>f(x),\forall x \in \mathbb R$ with the weaker one $f'(x)>f(x),\forall x>0$. Even if $f'(0)<0$, though the latter only shows that no such $f$ exists. One may ask: For which $t\ge0$ does an $f$ exist with $f'(0)=t$ (and $f(0)=0$, $f'(x)>f(x)$ for all $x>0$)? Easily $t\ge1$ works, e.g. $f(x)=e^{tx}-1$. I wonder if any $t<1$ might work, in particular if $t=0$ works. (I may consider posting a separate question if I can't figure it.) – Mirko Dec 15 '15 at 19:04
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Yes. Hint: Consider $x=0$. Since $f'(0)>f(0)=0$, we know $$ \lim_{h\rightarrow 0}\frac{f(h)-f(0)}{h}=\lim_{h\rightarrow 0}\frac{f(h)}{h}>0. $$ Therefore, for $h$ positive and sufficiently close to $0$, the numerator is positive. Continue with this idea.
Michael Burr
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this makes the solution obvious. I didn't read the problem carefully, and was thinking of $f(0)=f'(0)=0$ and $f'(x)>f(x)$ for $x>0$. I wonder if one could derive the conclusion under these different assumptions (though this is a different question, and you did answer the original one). – Mirko Dec 15 '15 at 15:30
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@Michael Burr : Umm , I'm not getting the idea ; could you please elaborate a bit more ? Thanks in advance – Dec 15 '15 at 15:31
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1@SaunDev Inspired by the answer by Michael Burr here, and by your request for elaboration, I posted this answer to another question.. My answer there first discusses a general idea of "induction" over the reals (a well-known technique in disguise) and then illustrates it specifically with this answer here. (There are already several answers posted there, please locate mine, as the click on the above link only approximately locates it, thank you.) – Mirko Dec 15 '15 at 17:49
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@SaunDev I just noticed that my answer to which I refer in the previous comment is not the only one there that discusses so-called real induction, you may wish to see this answer by Jakub Konieczny, and also see the link to a paper by Pete L. Clark provided in the comments to this answer. – Mirko Dec 15 '15 at 18:16
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Let $y(x)=e^{-x}f(x)$. Then $ f$ (strictly) positive $ \iff y$ (striclty) positive.
$\forall x $, $ y'(x)=e^{-x}(f'(x)-f(x)) \ge 0$ and if $x> 0, y'(x)>0$.
Therefore $y$ is positive for $x \ge 0$.
Now lets suppose it exists $t>0$ such that $y(t)=0$. Then for $\epsilon >0 $ small enough, because $y'(t) >0$, for $x \in ]t-\epsilon,t[$ , $y(x) < y(t) $, which is in contradiction with the previous point.
yultan
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I think it should work like this:
By definition of limit
$\lim_{h \to 0^+}\frac{f(0+h)-f(0)}{h}=f^\prime(0)$
Since $f^\prime(0)>f(0)$ and $f(0)=0$, $\lim_{h \to 0^+}\frac{f(h)}{h}>0$ then $f(x)>0$ for $x$ in $]0,\delta]$.
$\lim_{h \to \delta^+}\frac{f(\delta+h)-f(\delta)}{h}=f^\prime(\delta)>f(\delta)>0\rightarrow\lim_{h \to \delta^+}{f(\delta+h)-f(\delta)}>0\rightarrow f(x)>0\ for\ x\in(\delta,\delta_1) $ And so on.. So yes it is true
user47354
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