Find a function less than a fraction of itself

[member 428835]

Homework Statement

Give an example of a function $u$ such that $u:I\to I=(−1,1)$ such that $|u(x)−u(y)|\leq 0.5|x−y|$ for all $x,y\in I$, and $u(x) \neq x$ for all $x\in I$.

Relevant Equations

Nothing comes to mind

Well doesn't $u(x) = 0.4 x$ work? Seems too easy, but the phrasing at the end "for all $x\in I$" makes me think since $0.4x = x$ only at $x=0$, and not all of $I$, that this is okay. But am I wrong?

 

[fresh_42]

$u(x) \neq x$ for all $x\in I$ means, that none of the $x$ should have this property. Thus linear functions are ruled out.

 

[member 428835]

$u(x) \neq x$ for all $x\in I$ means, that none of the $x$ should have this property. Thus linear functions are ruled out.

But then $\sin(x)$ is also ruled out? And so is $x^{2}$. Any ideas?

 

[Mark44]

But then $\sin(x)$ is also ruled out? And so is $x^{2}$. Any ideas?

How about fractional powers? You want a function that is bounded above by y = (1/2)x on [0, 1] and bounded below by the same function on [-1, 0].

 

[member 428835]

How about fractional powers? You want a function that is bounded above by y = (1/2)x on [0, 1] and bounded below by the same function on [-1, 0].

But at $x=0$ all fractional powers are zero. So it won't work, right?

 

[Mark44]

But at $x=0$ all fractional powers are zero. So it won't work, right?

You could just make the function undefined at x = 0.

 

[member 428835]

You could just make the function undefined at x = 0.

But the function requires $I\to I$.

 

[Mark44]

But the function requires $I\to I$.

I don't think that precludes omitting 0 in how the function is defined.

However, if you don't like that, one can interpret the requirement

and $u(x) \neq x$ for all $x\in I$.

to mean that u(x) is not identically equal to x. I read this as different from $\forall x \in (-1, 1), f(x) \ne x$; i.e., for each x in I, $f(x) \ne x$.

 

[member 428835]

I don't think that precludes omitting 0 in how the function is defined.

However, if you don't like that, one can interpret the requirement to mean that u(x) is not identically equal to x. I read this as different from $\forall x \in (-1, 1), f(x) \ne x$; i.e., for each x in I, $f(x) \ne x$.

So is there even such a function that takes $I\to I$, satisfying the inequality, where it does not cross the line $y=x$ anywhere in $I$? I also read it that way at first, but it seemed way too easy.

 

[PeroK]

So is there even such a function that takes $I\to I$, satisfying the inequality, where it does not cross the line $y=x$ anywhere in $I$? I also read it that way at first, but it seemed way too easy.

Can you show that $u$ must be continuous?

Can you draw such a function? Maybe forget formulas to begin with. What would it look like?

 

[PeroK]

PS if $u$ must also be onto, then it is impossible. You might want to try proving this.

Hint: assume first that $u(0) > 0$ and reach a contradiction to the required inequality.

 

[fresh_42]

As I read it, the function doesn't need to be surjective, but defined everywhere on $I=(-1,1)$ as the condition involves all points of $I$. This means you have three tasks:
a) let $u(x)$ be a composite function, e.g. two linear ones with a 'jump' at $x=0$
b) prove what @PeroK said: there is no such function if it has to be surjective on the closed interval $[-1,1]$
c) this leads to what is probably meant by the exercise: $u(x)=x$ has to be at the points $x=\pm 1$ for otherwise it is not possible, which leads back to @Mark44 's parabola - just lift it a bit above the origin

 

[PeroK]

There is actually a simple solution. A simple family of solutions, in fact.

 

[member 428835]

What about the function $u(x) = 2$? Then $|u(x)-u(y)| = |2-2| = 0 \leq 0.5|x-y|$. Additionally, in $I$ we note that $u(x) \neq x$.

 

[PeroK]

What about the function $u(x) = 2$? Then $|u(x)-u(y)| = |2-2| = 0 \leq 0.5|x-y|$. Additionally, in $I$ we note that $u(x) \neq x$.

$2 \notin I$

 

[member 428835]

$2 \notin I$

Ahhhhh bummer! Ok ok, let me think about what you posted in your previous comment.

 

[member 428835]

As I read it, the function doesn't need to be surjective, but defined everywhere on $I=(-1,1)$ as the condition involves all points of $I$. This means you have three tasks:
a) let $u(x)$ be a composite function, e.g. two linear ones with a 'jump' at $x=0$
b) prove what @PeroK said: there is no such function if it has to be surjective on the closed interval $[-1,1]$
c) this leads to what is probably meant by the exercise: $u(x)=x$ has to be at the points $x=\pm 1$ for otherwise it is not possible, which leads back to @Mark44 's parabola - just lift it a bit above the origin

A composite linear function with a jump at $x=0$ seems impossible to force $|u(x)-u(y)| \leq 0.5|x-y|$ since the $\lim_{x \to 0^-}u \neq \lim_{x \to 0^+}u$, yet $\lim_{x\to 0} 0.5x = 0$. Am I missing something?

 

[PeroK]

Given a composite linear function with a jump at $x=0$ seems impossible to force $|u(x)-u(y)| \leq 0.5|x-y|$ since the $\lim_{x \to 0^-}u \neq \lim_{x \to 0^+}u$, yet $\lim_{x\to 0} 0.5x = 0$. Am I missing something?

The function must be continuous. The only real constraint is that $u$ must not cross the line $y=x$.

Why not keep $u$ above that line?

When you tried $u(x) = 2$, you would have been closer with $u(x) = 1$.

 

[member 428835]

The function must be continuous. The only real constraint is that $u$ must not cross the line $y=x$.

Why not keep $u$ above that line?

When you tried $u(x) = 2$, you would have been closer with $u(x) = 1$.

Yea, I was about to respond with that but $1\notin I$.

If we agree the function must be continuous, then are you saying make like a $V$ but not cross $y=x$? Seems like a good solution!

 

[PeroK]

Yea, I was about to respond with that but $1\notin I$.

If we agree the function must be continuous, then are you saying make like a $V$ but not cross $y=x$? Seems like a good solution!

You don't need a V.

 

[PeroK]

Yea, I was about to respond with that but $1\notin I$.

If we agree the function must be continuous, then are you saying make like a $V$ but not cross $y=x$? Seems like a good solution!

Okay. Go for a V. I keep looking at the OP to see whether I'm missing something!

 

[WWGD]

But then $\sin(x)$ is also ruled out? And so is $x^{2}$. Any ideas?

How about rescaling something that is increasing , with |f(-1)|=|f(1)|? Just rescale by the value at either endpoint.

 

[member 428835]

Wait, I totally spaced the function is $y=0.5x$ and not $y=x$. In this case ,why not $u(x) = 0.7$? Doesn't this work?

 

[PeroK]

Wait, I totally spaced the function is $y=0.5x$ and not $y=x$. In this case ,why not $u(x) = 0.7$? Doesn't this work?

I can't understand the issue here. It's ptobably time to wrap this up and move on. Any linear function for which $u(1) = 1$ and whose gradient is $\le 1/2$ would do.

And, indeed, any differentiable function for which $u(1) = 1$ and whose gradient is $\le 1/2$ would do.

The same for functions that lie below the line $y = x$ and intersect it at $x = -1$.

Maybe there was something about the working of the problem that confuses everyone. Maybe it was too much focus on formulas rather than a focus on some simple geometric constraints.

The only two constraints were

a) $|u(x) - u(y)| \le 0.5 |x-y|$

This enforces continuity and means the function can't be too steep.

b) $u(x) \ne x$

This means that the function must lie above or below the line $y=x$.

For example, the straight line joining $(-1, 0)$ with $(1, 1)$ would do. It has a constant gradient of $1/2$ and intersects the line $y=x$ outside the open interval $I$.

 

[member 428835]

I can't understand the issue here. It's ptobably time to wrap this up and move on. Any linear function for which $u(1) = 1$ and whose gradient is $\le 1/2$ would do.

And, indeed, any differentiable function for which $u(1) = 1$ and whose gradient is $\le 1/2$ would do.

The same for functions that lie below the line $y = x$ and intersect it at $x = -1$.

Maybe there was something about the working of the problem that confuses everyone. Maybe it was too much focus on formulas rather than a focus on some simple geometric constraints.

The only two constraints were

a) $|u(x) - u(y)| \le 0.5 |x-y|$

This enforces continuity and means the function can't be too steep.

b) $u(x) \ne x$

This means that the function must lie above or below the line $y=x$.

For example, the straight line joining $(-1, 0)$ with $(1, 1)$ would do. It has a constant gradient of $1/2$ and intersects the line $y=x$ outside the open interval $I$.

Yea sorry, I was really spacing out. But thanks! Kinda too late for the HW anyways, buuuuuuut I appreciate you sticking with it!

So if the interval $I = \mathbb R$ and we require the inequality be strictly less than, then would there be any such $u(x)$?

 

[PeroK]

Yea sorry, I was really spacing out. But thanks! Kinda too late for the HW anyways, buuuuuuut I appreciate you sticking with it!

So if the interval $I = \mathbb R$ and we require the inequality be strictly less than, then would there be any such $u(x)$?

It's fairly easy to see that there is no function. Just look at $u(0)$ and apply the MVT using the inequality given.

 

[member 428835]

It's fairly easy to see that there is no function. Just look at $u(0)$ and apply the MVT using the inequality given.

Got it, thanks. Didn't think about MVT, but it makes sense.

 

[PeroK]

Got it, thanks. Didn't think about MVT, but it makes sense.

Although, in fact I meant the intermediate value theorem.