Show that f(x) is constant

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  • #26
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I don't know how to write "lim" with "xn → x" below on Microsoft Word's equation editor. And now I understand why my book has 0<|x-y|<∂ in the definition of continuity; I always assumed the "0<" part was irrelevant.

Perhaps you could try lim from {n rightarrow infty}?


Yes, that is one reason for 0<|x-y|<δ, which is necessary for derivatives.

In the case of a regular limit there is another reason though.
It is possible that the limit of a function is not equal to the function value in that point.
However, in that case the limit still exists.
So the definition of a limit in general requires that 0<|x-y|<δ.
 
  • #27
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Perhaps you could try lim from {n rightarrow infty}?


Yes, that is one reason for 0<|x-y|<δ, which is necessary for derivatives.

In the case of a regular limit there is another reason though.
It is possible that the limit of a function is not equal to the function value in that point.
However, in that case the limit still exists.
So the definition of a limit in general requires that 0<|x-y|<δ.


Did you like how I used the world "thereupon"? I get sick of using "consequently", so I went to the Thesaurus to find a cool synonym.
 
  • #28
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Did you like how I used the world "thereupon"? I get sick of using "consequently", so I went to the Thesaurus to find a cool synonym.

I noticed the word and I had to smile. :)
What about "therefore" or "yields"?

I have to warn you though, English is not my native language, so I don't know what is proper.
I'd like to think that I'm better at math than at language. :wink:
 
  • #29
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Homework Statement



(I don't have my book with me, so this may not be the correct word-for-word representation of the exercise)

Suppose f(x) is differentiable on the whole real line. Show that f(x) is constant if for all real numbers x and y,

|f(x)-f(y)| ≤ (x-y)2.​

Homework Equations



Definition of a derivative

The Attempt at a Solution



f(x)-f(y) ≤ |f(x)-f(y)|, so by basic algebra we have [f(x)-f(y)]/(x-y) ≤ x - y. Letting x approach y on both sides of the inequality yields f '(x) ≤ 0.

........ Now I somehow need to show that f '(x) ≥ 0. Ideas?

I'm somehow not convinced that that represents f'(x) why not f'(y) since writing x as [tex] y+Δy [/tex] allows the Newton's quotient to be f'(y).

i.e. [tex] lim_ {\delta y->0}\frac {f(y+\triangle y)-f(y)}{y+\triangle y -y}≤ \triangle y[/tex]
 
  • #30
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Again the Mean value theorem can come to the rescue.
Can you apply it to f(x)/x?


Choose y such that 0 < y < ∞. The function f(y)/y is continuous on [0, y] and differentiable on (0, y), so there exists an x in (0, y) such that (f(x)/x)' = [f(y) - f(0)]/[y - 0] = f(y) / y. Am I close?
 
  • #31
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I'm somehow not convinced that that represents f'(x) why not f'(y) since writing x as [tex] y+Δy [/tex] allows the Newton's quotient to be f'(y).

i.e. [tex] lim_ {\delta y->0}\frac {f(y+\triangle y)-f(y)}{y+\triangle y -y}≤ \triangle y[/tex]

I worked out all the kinks. Don't worry.

screen-capture-59.png
 
  • #32
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Choose y such that 0 < y < ∞. The function f(y)/y is continuous on [0, y] and differentiable on (0, y), so there exists an x in (0, y) such that (f(x)/x)' = [f(y) - f(0)]/[y - 0] = f(y) / y. Am I close?

Close yes. :smile:
It should be: f'(x) = [f(y) - f(0)]/[y - 0] = f(y) / y
 
  • #33
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Close yes. :smile:
It should be: f'(x) = [f(y) - f(0)]/[y - 0] = f(y) / y

Got it.

Choose x with 0 < x < ∞. The function f is continuous on [0, x] and differential on (0, x). Thus, by the Mean Value Theorem, this exists an x0 with 0 < x0 < x and

f(x) = f(x) - f(0) = f '(x0) * (x - 0) = x * f '(x0).

Because f ' is monotonically increasing and x > 0, we have x * f '(x0) ≤ x * f '(x) and accordingly x * f '(x) ≥ f(x), as desired.

http://collegestudybreak.files.wordpress.com/2010/07/success.jpg [Broken]
 
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