How Can We Prove That a Function is Constant Using Derivatives?

In summary, the function f is differentiable, but does not have a constant function f' at x=0. To solve for x, one must use the Mean Value Theorem to find a point c such that |f '(c0)|>0.
  • #36
lurflurf said:
without loss of generality assume x<=y to avoid considering the cases y<x

The integral was a wrong turn.
I noticed this is Baby Rudin exercise 5.26.

Show f=0 on [0,1] and f=0 on R will follow.
0<x<1
f(x)=f(x)-f(0)=(x-0)f'(t)=x f'(t) (mean value theorem with 0<t<x<1)
|f(x)|=x|f'(t)|<=x|f(t)| (by given inequality)
M:=sup(t|0<t<x||f(x)|)
|f(x)|<=x M
M<=x M
(1-x)M<=0
M=0
|f|=0 on [0,1]
(f(1)=f(1-)=0

Looks like or similar to a more concise version of what I did, but what are you saying when you say M:=sup(t|0<t<x||f(x)|) ? I'm unsure what t|0<t<x||f(x)| means.
 
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  • #37
Pere Callahan said:
Thank you for your reply. Oh, yes I forgot about the [itex]x_0[/itex] in the denominator.. stupid...anyway: so there exist [itex]\xi\in(0,1)[/itex] with
[tex]
f'(\xi)=\frac{f(x_0)}{x_0}
[/tex]
and so
[tex]
|f(\xi)|\geq|f'(\xi)|=\left|\frac{f(x_0)}{x_0}\right|\geq|f(x_0)|
[/tex]
Luckily, [itex]|x_0|\leq 1[/itex], so this mistake is easily corrected. The proof then shows, you're right, that f=0 on [0,1] in particular that f(1)=0. But then you can do the same type of thing to show that f=0 on [1,2]... and inductively on the whole of R.

As for your question: Just differentiate log f(x) using the chain rule:smile:

That's it. I can live with that.
 
  • #38
lurflurf said:
without loss of generality assume x<=y to avoid considering the cases y<x

The integral was a wrong turn.
I noticed this is Baby Rudin exercise 5.26.

Show f=0 on [0,1] and f=0 on R will follow.
0<x<1
f(x)=f(x)-f(0)=(x-0)f'(t)=x f'(t) (mean value theorem with 0<t<x<1)
|f(x)|=x|f'(t)|<=x|f(t)| (by given inequality)
M:=sup(t|0<t<x||f(x)|)
|f(x)|<=x M
M<=x M
(1-x)M<=0
M=0
|f|=0 on [0,1]
(f(1)=f(1-)=0

After you get to |f(x)|<=x|f(t)| you appear to be selectively taking a sup of f over x and t independently (though I am guessing about that - but it's all I could figure out that you could possibly mean). You can't do that.
 
  • #39
Dick said:
That's it. I can live with that.
Thanks, so I can now easily go for a walk to cool my mind.:smile:
 
  • #40
Dick said:
After you get to |f(x)|<=x|f(t)| you appear to be selectively taking a sup of f over x and t independently (though I am guessing about that - but it's all I could figure out that you could possibly mean). You can't do that.

That is not what I mean, but I do see I was needlessly confusing. How about this?

|f(x)|<=x|f(t)|
On [0,x] f (being a continuous function on a closed interval) has extreme points in the interval, so too does |f|.
M=|f(x')| where x' is a point in [0,x] such that for any y in [0,x], |f(y)|<=|f(x')|
The cases of x'=0 and x'=x are easily handled, otherwise chose t' such that 0<t'<x'<x<1
and
|f(x')|<=x'|f(t')|
|f(x')|<=x'|f(x')|
(1-x')|f'(x')|<=0
|f'(x')|=0
f=0 on [0,x]
f=0 on R
 
  • #41
Sure. Now it's basically the same thing other people are saying. I don't know why this problem gives me such a hard time every time I see it. Probably just forgetting to restrict to [0,1] first. I don't know. But, then I also seem to spend a lot of my time trying to figure out what's wrong with bad proofs other people post.
 
  • #42
Dick said:
Can anyone show f'(x)/f(x) is even integrable between x0 and x1? .
What's the problem here. f is differentiable, hence continuous. f'(x)/f(x) has certianly no singularities, because it is dominated by one, so is continuous as well. But continuous functions are integrable...
 
  • #43
Pere Callahan said:
What's the problem here. f is differentiable, hence continuous. f'(x)/f(x) has certianly no singularities, because it is dominated by one, so is continuous as well. But continuous functions are integrable...

The function f=0 everywhere so f'/f is defined nowhere.
 
  • #44
lurflurf said:
The function f=0 everywhere so f'/f is defined nowhere.
Okay, admitted.
 

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