- #1

maximus101

- 22

- 0

[tex]\forall[/tex] [tex]x,y[/tex][tex]\in[/tex] [tex]R[/tex]

does

| [tex]f(y)[/tex] - [tex]f(x)[/tex] | [tex]\mid[/tex] [tex]\leq[/tex] [tex](y-x)^2[/tex]

hold

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- Thread starter maximus101
- Start date

- #1

maximus101

- 22

- 0

[tex]\forall[/tex] [tex]x,y[/tex][tex]\in[/tex] [tex]R[/tex]

does

| [tex]f(y)[/tex] - [tex]f(x)[/tex] | [tex]\mid[/tex] [tex]\leq[/tex] [tex](y-x)^2[/tex]

hold

- #2

Buri

- 273

- 0

Which inequality?

- #3

maximus101

- 22

- 0

sorry just fixed it

- #4

maximus101

- 22

- 0

[tex]\dfrac{|f(y) - f(x)|}{|y-x|}\leqslant |y-x|[/tex]

was obtained by [tex](y-x)^2[/tex]=[tex]|y-x||y-x|[/tex] and dividing both sides by [tex]|y-x|[/tex].

then trying to use the fact that

[tex]\displaystyle\lim_{y\to x}\frac{f(y)-f(x)}{y-x}[/tex] is the derivative at x

but not sure what's next how to combine the two?

- #5

Buri

- 273

- 0

- #6

Buri

- 273

- 0

- #7

Yuqing

- 218

- 0

[tex]\dfrac{|f(y) - f(x)|}{|y-x|}\leqslant |y-x|[/tex]

automatically imply the function is differentiable?

- #8

JG89

- 728

- 1

[tex]\dfrac{|f(y) - f(x)|}{|y-x|}\leqslant |y-x|[/tex]

automatically imply the function is differentiable?

Yes. Using the squeeze theorem you can see that the derivative is actually 0 everywhere, implying the only functions satisfying the inequality are constant functions.

- #9

TylerH

- 729

- 0

If you indented for it to be the difference quotient, then a counter example would be [tex]r \mapsto r^3[/tex]. For all |x-y| >= 1, [tex]\left| x^3 - y^3 \right| > \left( x - y \right)^2[/tex]

- #10

JG89

- 728

- 1

Let y approach x. By the squeeze theorem, we see that the limit exists, is equal to 0, and is in fact the derivative of f evaluated at x. So f'(x) = 0 for all x in its domain, which means that f must be a constant function.

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