Unraveling Inequality: Can We Bridge the Gap?

[Miike012]

The problem and my question and thoughts are all in the picture.

 

[Poopsilon]

No it is not that the graph is necessarily discontinuous at zero. What it means is that the graph is approaching zero at an increasingly steep angle so that even though we may have f(h)→0 as h→0 its angle of approach becomes so steep that it overwhelms h as h→0 and so the ratio f(h)/h blows up as h→0.

Consider the function:

f(x) = |x|^β + |xsin(1/x)| for x≠0
f(0) = 0
(modified from Rudin's PMA).

This function should satisfy all your conditions, graph it, notice its behavior near zero, the fact that it is continuous at zero. But nevertheless it is not differentiable at zero.

 

[phinds]

Do you think you could make the font size just a bit smaller? I mean I can actually READ this, although not easily enough that I'm willing to.

 

[Miike012]

No it is not that the graph is necessarily discontinuous at zero. What it means is that the graph is approaching zero at an increasingly steep angle so that even though we may have f(h)→0 as h→0 its angle of approach becomes so steep that it overwhelms h as h→0 and so the ratio f(h)/h blows up as h→0.

Consider the function:

f(x) = |x|^β + |xsin(1/x)| for x≠0
f(0) = 0
(modified from Rudin's PMA).

This function should satisfy all your conditions, graph it, notice its behavior near zero, the fact that it is continuous near zero. But nevertheless it is not differentiable at zero.

it would be diff at zero if β>1 right?

 

[Miike012]

what if it said

f(x) ≥ |x|^β
or
f(x) ≥ x^β
or
|f(x)|≤ |x|^β
and so on...

If you can see where I am going with this... I understand that the inequalities have a lot to do with solving the problem, but because I don't understand inequalities I can't solve it...

for instance I know that if, |f(x) - m|< z then this statement means the distacnce from f(x) to m is less than z. and knowing this I can proceed to solve the inequality because i know the meaning... but I have not Idea what |f(x)|≥ |x|^b means...

 

[Poopsilon]

Actually I think not because I have that xsin(1/x) term screwing things up for you. But now that I look at it, I think just using the function f(x) = |x|^β for 0<β<1 would probably be a much simpler function that exhibits the same 'steepining' behavior, which is continuous at zero but not differentiable at zero. And in this case it would be differentiable if β>1.

 

[Poopsilon]

Well as far as inequalities go, a couple good alternate representations to keep in mind when you're dealing with them is that.

if |a|<b than -b<a<b.

Second, if you have |x| and you want to get rid of the absolute value bars a tricky way is to write it in the form $\sqrt{x^2}$; assuming you're interpreting the square-root as a function and not the solution to an equation (bit of a subtlety).

Your first two: f(x) ≥ |x|^β and f(x) ≥ x^β are a bit problematic. The first one is actually equivalent to the the original case |f(x)| ≥ |x|^β, since the right side is always positive and so the left side must be as well for the inequality to hold. Your second will involve only the right side of the plane, since in real analysis it doesn't make sense to take even roots of negative numbers, (If you want to avoid this you need to restrict the values which beta may take).

As for interpreting |f(x)|≥|x|^β, well since the right side is always positive we obtain by my first rule the inequality:

-|f(x)| ≤ |x|^β ≤ |f(x)|. Thus we see that for each x there is some interval around zero which f(x) must be outside.

Edit: Actually forget my rule for this case, as it doesn't really apply well. But it's still the case that it means that for each x there is some interval around zero which f(x) must be outside.

 

[SammyS]

Do you think you could make the font size just a bit smaller? I mean I can actually READ this, although not easily enough that I'm willing to.

If you click on the thumbnail, then click on the resulting image, the image will open in another window. A zoom-in character will replace your mouse arrow on that image. If you zoom-in, the result is very readable, if not understandable. LOL

--- if you're willing to do all of that.

 

[phinds]

If you click on the thumbnail, then click on the resulting image, the image will open in another window. A zoom-in character will replace your mouse arrow on that image. If you zoom-in, the result is very readable, if not understandable. LOL

--- if you're willing to do all of that.

Ah to be young and have good eyes again ... the image you are talking about IS the image I was talking about.