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

[Pere Callahan]

Thank you for your reply. Oh, yes I forgot about the x_0 in the denominator.. stupid...anyway: so there exist \xi\in(0,1) with
<br /> f&#039;(\xi)=\frac{f(x_0)}{x_0}<br />
and so
<br /> |f(\xi)|\geq|f&#039;(\xi)|=\left|\frac{f(x_0)}{x_0}\right|\geq|f(x_0)|<br />
Luckily, |x_0|\leq 1, 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

 

[Lazerlike42]

Thank you for your reply. Oh, yes I forgot about the x_0 in the denominator.. stupid...anyway: so there exist \xi\in(0,1) with
<br /> f&#039;(\xi)=\frac{f(x_0)}{x_0}<br />
and so
<br /> |f(\xi)|\geq|f&#039;(\xi)|=\frac{f(x_0)}{x_0}\geq|f(x_0)|<br />
Luckily, |x_0|\leq 1[\itex], so this mistake is easily corrected. The proof then shows, you&#039;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.<br /> <br /> As for your question: Just differentiate log f(x) using the chain rule<img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f642.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":smile:" title="Smile :smile:" data-smilie="1"data-shortname=":smile:" />

<br /> <br /> Pere, I updated my post responding to yours about my epsilon-delta reasoning with proper epsilon-delta, if you happen to care all that much. :)

 

[gabbagabbahey]

Shouldn't there be some sort of simple proof by induction?

perhaps something along the lines of the following:

If you move an infinitesimally small distance \epsilon from the point x_0 (to the left or the right), then Taylor's theorem tells you that f(x_0 \pm \epsilon)=f(x_0)+f&#039;(x_0)\epsilon. Since |f&#039;(x)|\leq |f(x)| and f(0)=0, f'(0)=0 and therefor f(\pm \epsilon)=0+(0)\epsilon=0. And |f&#039;(x)|\leq |f(x)|\implies f&#039;(\pm \epsilon)=0 and using the same argument as before, f(\pm 2\epsilon)=0...repeat ad nauseam

 

[Dick]

Nah. Don't like that much either. That's actually sort of a Riemann sum version of Tiny Tim's original suggestion of just integrating f'(t)/f(t) and ignoring potential pathologies of the function f. I was really hoping there was a way of just stating this in a simple way. But maybe there's not. You can certainly reduce the problem to saying there is an x0 such that f(x0)=0 and an x1 such that f(x1)>0 and f(x)>0 for x between x0 and x1. Can anyone show f'(x)/f(x) is even integrable between x0 and x1? This is really more of a real analysis type question. If f(x) is a 'normal average everyday function' (i.e. has a convergent power series), as I've pointed out before, it's easy.

 

[lurflurf]

Isn't that really the same information I would get from:

|f(y)-f(x)|=|\int_x^y f&#039;(t) dt| \le \int_x^y |f&#039;(t)| dt \le \int_x^y |f(t)| dt

What does wlog x<=z have to do with anything?

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

 

[Lazerlike42]

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.

 

[Dick]

Thank you for your reply. Oh, yes I forgot about the x_0 in the denominator.. stupid...anyway: so there exist \xi\in(0,1) with
<br /> f&#039;(\xi)=\frac{f(x_0)}{x_0}<br />
and so
<br /> |f(\xi)|\geq|f&#039;(\xi)|=\left|\frac{f(x_0)}{x_0}\right|\geq|f(x_0)|<br />
Luckily, |x_0|\leq 1, 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

That's it. I can live with that.

 

[Dick]

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.

 

[Pere Callahan]

That's it. I can live with that.

Thanks, so I can now easily go for a walk to cool my mind.

 

[lurflurf]

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

 

[Dick]

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.

 

[Pere Callahan]

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...

 

[lurflurf]

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.

 

[Pere Callahan]

The function f=0 everywhere so f'/f is defined nowhere.

Okay, admitted.