- #1

- 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

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter maximus101
- Start date

- #1

- 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

- 273

- 0

Which inequality?

- #3

- 22

- 0

sorry just fixed it

- #4

- 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

- 273

- 0

- #6

- 273

- 0

- #7

- 218

- 0

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

automatically imply the function is differentiable?

- #8

- 726

- 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

- 724

- 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

- 726

- 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.

Share: