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Let $f:\mathbb{R}_+\rightarrow\mathbb{R}$ be a $C^1$ function such that $f(0)>0$ and there exist an increasing sequence of positive reals $a_1,...,a_k$ with $k\geq 2$ satisfying $f(a_k)=0$.

Define $F(s):=\int_0^s f(\sigma)d\sigma$.

I want to show that the following conditions are equivalent:

i) $F(a_k)>\max\{F(s):\ 0\leq s\leq a_{k-1}\}$

ii) $F(a_k)>F(a_{k-1})$

Also, it is true that $\max\{F(s):\ 0\leq s\leq a_{k-1}\}=F(a_{k-1})$? if i) or ii) are satisfied?

Edit: The values $a_k$ need not to be the only zeroes of $f$.

Edit 2: In this edit I will change some of the hypothesis, please verify if in this case, i) is equivalently to ii). The only hypothesis that changes, is the hypothesis about the $a_i's$.

There exist $0<a_1<b_1<a_2<b_2<...<b_{m-1}<a_m$ such that for all $k=1,...,m-1$, $f(x)\leq 0$ in $(a_k,b_k)$ and $f(x)\geq 0$ in $(b_k,a_{k+1})$. $f$ does not have zeroes on the interval $(0,a_1)$.

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are those the only zeros of the function? can there be zeros in between them? – cats Feb 18 '13 at 21:23
I Think that unless $f$ is non-negative, it is necessary to have zeroes between then @lyj – Tomás Feb 18 '13 at 23:08

The assertion seems false, even if the $a_j$ are the only zeros. Clearly (ii) implies (i), while (i) implies the first statement in (ii); but I don't see any reason why (i) should imply the second statement in (ii). (unless $f$ is assumed to be nonnegative, in which case both (i) and (ii) are trivially true)

For example, take $f(x) = (x+1)\cos x$ (so that $F(s) = s\sin s +\sin s + \cos s$) and $a_1=\pi/2$, $a_2=3\pi/2$, and $a_3=5\pi/2$. This satisfies (i) but not (ii).

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Greg, you can see that Halil Duru edited my question and he changed all it mean. If you take a look in the first post, you will see that the second statement of ii) is separated from it. I will revert it. – Tomás Feb 18 '13 at 22:01

With the above version : 1 implies 2 but 2 doesn't imply 1. And 3 (the last statement) is false.

(Are you sure about the range of f?Or probably you should replace $k$ with $i$ in the statements 1,2,3.)

For example ; f(x)= $cos$ $x$ and $a_1=3\pi/2$ and $a_2=5\pi/2$ .

And with the second hypothesis set (even with the positivity in the first interval), still 2 doesn't imply 1.

You need much stronger hypothesis for equivalence.

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I had already seen it. – Halil Duru Feb 19 '13 at 1:03
Ok, let me try again. I added a new hypothesis: I am asking now that in the interval $(0,a_1)$, $f$ does not have zeroes, or equivalently, $f$ is positive in this interval. – Tomás Feb 19 '13 at 12:02

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