Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $f:[a,\infty)\rightarrow \mathbb{R}, \ \ f\in C^2[a,\infty)$ such that $$ \\ f(a)>0 , \ \ f'(a)<0, \ \ f''(x)\leq 0 \ \ \forall x\in [a,\infty)$$ Prove that

$$ \exists !~t\in (a,\infty):f(t)=0$$

share|improve this question
    
Source? Motivation? Effort on part of poster? –  Gerry Myerson Jul 14 '13 at 12:47
1  
Words?${}{}{}{}{}{}$ –  Javier Badia Jul 14 '13 at 12:51
    
Eyes pain looking at math symbols. –  Torsten Hĕrculĕ Cärlemän Jul 14 '13 at 12:54
1  
Is it not preferred to state a mathematical question using only mathematical symbols(if it is possible)? If not, I will use words onwards. –  EricAm Jul 14 '13 at 13:06
    
@Babak S: Can you tell me why $t$ is preferred over $x'$? I don't see why. –  EricAm Jul 14 '13 at 13:07

2 Answers 2

up vote 3 down vote accepted

From the mean value theorem, $f'(x)-f'(a)=(x-a)f''(\xi)$ for some $\xi\in(a,x)$, hence $f'(x)\le f'(a)<0$ for all $x\in(a,\infty)$ because $x-a\ge0$ and $f''(\xi)\le 0$. As a consequence $\frac{f'(x)}{f'(a)}\ge1$ for all $x\ge a$.

Let $x=a-\frac{f(a)}{f'(a)}$. Then $x>a$ and $f(x)-f(a)=(x-a)f'(\xi)=-f(a)\frac{f'(\xi)}{f'(a)}\le -f(a)$ because $f(a)>0$ and $\frac{f'(\xi)}{f'(a)}\ge1$. In other words, $f(x)\le 0<f(a)$ and by the intermediate value theorem, there exists $t\in[a,x]$ with $f(t)=0$.

If there were two solutions $f(t_1)=f(t_2)=0$ with $t_1<t_2$, then by Rolle $f'(x)=0$ for some $x\in(t_1,t_2)$, contradicting $f'(x)<0$. Hence the solution is unique.

share|improve this answer

I will only give the idea which you can formalise yourself. It isn't too difficult, and should be a useful exercise.

The value of $f(a)$ is greater than zero, it's differential lower than zero. This means that, for $\epsilon$ sufficiently small, values $f(x')$ for $x'\in (a, \epsilon)$ will be lower than $f(a)$.

If you take such an $x'$, you are in the same situation again, since all second derivatives negative or zero means that $f'(x')\le f'(a)$ for all $x'\in(a,\infty)$. So you know that, about $f(x')$, you can do the same trick as before.

The only difficulty then is to show that you can chose an epsilon small enough s.t. the trick works for any $x'\in (a,\infty)$ (i.e. chose an epsilon uniformly).

You can then iterate this, and since $\epsilon\gt 0$, you will reach $f(x')\le 0$ after finite iterations. The best part: you can continue doing this iteration to infininty, and your function will only decrease in value.

This means you will have captures that one unique x', since it would otherwise be a contradiction to the intermediate value theorem.

Edit: Corrected terminology. I used mean-value instead of intermediate.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.