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 $f : (0, \infty) \rightarrow \mathbb{R}$ be differentiable and suppose that $|f(x)| \leq \frac{C}{x^k}$ ($k$ is non-negative) and this inequality would not hold for a smaller $k$ (even if you change $C$). Suppose this also holds for $|f'(x)|$ but with a possibly different $C$, but same $k$. Show that $\lim_{x \rightarrow 0^+} f(x)$ exists.

share|improve this question
2  
Could you be more specific about your statement : "this inequality would not hold for a smaller $k$" ? Does this mean $$\forall C'>0, \ \forall k'<k, \ \exists x>0, \ |f(x)| > C'/x^{k'}$$ ? –  vanna Sep 24 '12 at 13:29
    
Yes, that's what I mean, vanna. –  Homer Sep 24 '12 at 14:16
2  
The whole elaborated setup is just a terribly fancy way to say that $k=0$. Where did the problem come from? –  fedja Oct 31 '12 at 12:29
    
Is $k$ assumed to be integer? And is the claim that the limit is finite, or would $+\infty$ or $-\infty$ also be included in "the limit exists"? –  Lukas Geyer Oct 31 '12 at 23:05
add comment

1 Answer

Taking the maximum of the constants for $f$ and $f'$ we can assume that $$|f(x)| \le C x^{-k} \text{ and } |f'(x)| \le Cx^{-k} \text{ for all } x>0.$$ Integrating the inequality for $f'$ gives $$|f(x)| \le \begin{cases} C_1 + C_2 x^{-(k-1)} & \text{for } k \ne 1 \\ C_1 + C_2 |\ln x| & \text{for } k=1\end{cases}$$ with some constants $C_1$ and $C_2$.

It is still not absolutely clear to me whether $k$ is supposed to be integer, and whether an infinite limit is allowed or not, but at least this estimate shows that the assumption is never satisfied for $k>1$, as follows. For $k>1$ the estimates imply that there exists $C_3$ with $$ |f(x)| \le C_3 x^{-(k-1)} \text{ for } 0<x<1 $$ and $$ |f(x)| \le C x^{-k} \le C x^{-(k-1)} \text{ for } x \ge 1.$$ Setting $C_4 = \max (C,C_3)$ we get $$ |f(x)| \le C_4 x^{-(k-1)} \text{ for } x>0. $$ This contradicts the assumption that this inequality does not hold for exponents $k'<k$.

Obviously the estimate can be satisfied for $k=0$, and in that case boundedness of the derivative implies that $f$ is uniformly Lipschitz continuous, hence it extends continuously to $[0,\infty)$.

For the case $0 < k \le 1$ I am waiting for clarification of the exact statement of the problem.

share|improve this answer
    
You don't says what is $C_3$ anywhere –  leo Nov 1 '12 at 4:34
1  
It is just some constant, you can choose $C_3=C_1+C_2$, if you want an explicit expression. –  Lukas Geyer Nov 1 '12 at 4:43
add comment

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.