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i have that $b:[0,+\infty)\rightarrow [0,+\infty)$ a $C^1$ function and $B:\mathbb{R}\rightarrow[0,+\infty)$ such that $B(t)=\int_0^{|t|}b(s)s\,ds$

I want to prove that if $B(t)=0$ then $t=0$ and i have this condition on $b$:

*) b is increasing

*) there exist a constant c>0 such that $|b'(t)t|\leq c b(t), \forall t\geq0$

*) thereexist positive constants $b_1,b_2>1$ such that $$b_1 \leq \frac{b(t)t^2}{B(t)}\leq b_2 , \forall t\neq0$$

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    Of the three conditions, the last one is about $B$, not $b$. Are you sure it's a given condition? Or maybe it's part of the question, something to be proven? Please clarify your question by stating more clearly what's given and what needs to be shown. – 2017-01-03
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    $B$ has a relation with $b$ , i want to prove that if $B(t)=0$ we have that $t=0$ – 2017-01-03
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    By $b$ increasing, do you mean "strictly" increasing? – 2017-01-03

2 Answers 2

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The conditions on $b$ indicate to start with a by part integration with $u'(s)=s$ and $v(s)=b(s)$. You get $B(t)= \frac{1}{2}(b(|t|)t^2-\int_0^{|t|}b'(s)s^2ds)$ If $B(t)=0$ and $t>0$ then ou have $b(t)t^2= \int_0^{t}b'(s)s^2ds$ Using the first and second conditions on $b$ you have the inequality $b(t)t^2\leq cB(t) =0$. Or we supposed that $t>0$ so $b(t)=0$. However $b$ is positive and increasing so it equals to 0 on [0,t] which is impossible and then $t=0$

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    Ok I didn't know... I'll update my answer. – 2017-01-03
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(Note: we can see from the given formula for $B(t)$ that it's an even function, but I'm not gonna use that.) Let's prove that $B(t)\neq0$ for any $t\neq0$. For this proof, we only need to use the fact that $b:[0,+\infty)\to[0,+\infty)$ is increasing (which is why I asked for clarification on whether all conditions that you listed as given are actually given).

Let $t\neq0$. Clearly, $b(s)s\ge0$ for all $s\in[0,+\infty]$. Therefore,

$$B(t)=\int_0^{|t|}b(s)s\,ds=\int_0^{|t|/2}b(s)s\,ds+\int_{|t|/2}^{|t|}b(s)s\,ds\ge\int_{|t|/2}^{|t|}b(s)s\,ds.$$

Let $K=b(|t|/2)$; since $b:[0,+\infty)\to[0,+\infty)$ is increasing, $K=b(|t|/2)>b(0)\ge0$, i.e. $K>0$. Then for any $s\in[|t|/2,|t|]$, we have

$$b(s)\ge b(|t|/2)=K \quad\text{and}\quad b(s)s\ge b(|t|/2)\cdot\frac{|t|}{2}=\frac{K|t|}{2}>0.$$

Finally, we can go back to the integral to estimate that

$$B(t)\ge\int_{|t|/2}^{|t|}b(s)s\,ds\ge\frac{K|t|}{2}\cdot\frac{|t|}{2}=\frac{K|t|^2}{4}>0.$$