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It seems that when the logarithmic derivative of a function exists, the function itself cannot be zero. But I failed to construct a convincent proof. Is this true?

$\dfrac{f'(x)}{f(x)}$ is bounded implies $f(x)$ cannot have zero as a limiting value.

Is this true?

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For $f'(x)/f(x)$ to even make sense, you need $f(x)\neq 0$. –  1015 Mar 4 '13 at 18:51
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Also, for $\ln f(x)$ to even make sense, you need $f(x)\ne 0$. –  Hagen von Eitzen Mar 4 '13 at 18:59
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@HagenvonEitzen I think you meant $f(x)>0$. –  1015 Mar 4 '13 at 19:04
    
Thanks, I was not precise enough. Of course f(x)≠0. But can f(x) have zero as a limiting value? –  will Mar 4 '13 at 20:23
    
I think this is kind of 0/0 limit... –  will Mar 4 '13 at 20:39

2 Answers 2

True when $x$ approaches a finite $x_0$, not when $x$ approaches $\pm \infty$, as explained below.

Assume $f$ is differentiable on $\mathbb{R}$.

Then the function $$ g(x):=\ln|f(x)| $$ is defined and differentiable on the open set $$ U=\{x\in\mathbb{R}\;;\; f(x)\neq 0\}. $$ Its derivative is $$ g'(x)=\frac{f'(x)}{f(x)}\qquad \forall\;x\in U. $$

Write $U=\bigcup_{n\geq 1}(a_n,b_n)$ as a countable disjoint union of open intervals. See this thread to see why this is possible.

Consider for instance a finite point $x_0=a_n$.

Note that $\lim_{x\rightarrow x_0^+}\ln |f(x)|=\lim_{x\rightarrow x_0^+}g(x)=-\infty$.

Now if $g'=(\ln |f|)'$ was bounded by, say, $M$ on $(x_0,x_0+\delta]$, the mean value theorem, and then the triangular inequality, would yield $$ |g(x)-g(x_0+\delta)|\leq M \delta\quad \Rightarrow\quad |g(x)|\leq |g(x_0+\delta)|+M\delta $$ for all $x\in(x_0,x_0+\delta]$.

Contradiction.

So the logarithmic derivative is unbounded when it approaches every finite boundary point of $U$.

Now regarding the behavior at $\pm\infty$, you can easily find examples where $f'/f$ is bounded while $f$ tends to $0$.

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I'm not sure I understood the $\Rightarrow$ properly. I get how the left-hand side is true but I'm not sure I understand the implication. Could you please add some more details? –  xavierm02 Mar 4 '13 at 21:12
    
@xavierm02 Sorry but what lhs are you talking about? And what implication? –  1015 Mar 4 '13 at 21:21
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@xavierm02 Ah, ok. This is the triangular inequality: $|g(x)|=|g(x)-g(y)+g(y)|\leq |g(x)-g(y)|+|g(y)|$. –  1015 Mar 4 '13 at 21:32
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@xavierm02 No, I use the fact that every open set in $\mathbb{R}$ is the disjoint union of countably many open intervals. –  1015 Mar 4 '13 at 21:42
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@xavierm02 See this thread: math.stackexchange.com/questions/318299/… –  1015 Mar 4 '13 at 21:46

Let $f:x\to \frac{1}{x}$ on $[1,+\infty)$, then we have $$\frac{f'(x)}{f(x)}=-\frac{1}{x},$$ so $\displaystyle\frac{f'}{f}$ is bounded on $[1,+\infty)$ Nevertheless $\displaystyle \lim_{+\infty}f=0.$

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The statement is true at finite points. –  1015 Mar 4 '13 at 20:50
    
Okay, thanks +1 –  Sami Ben Romdhane Mar 4 '13 at 21:05

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