Can a Function Be a Strict Contraction with a Derivative of 1 at a Point?

[mattmns]

Here is the question
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Give an example of a function $f:[a,b] \to \mathbb{R}$ which is continuously differentiable and which is a strict contraction, but such that $|f'(x)| = 1$ for at least one value of $x \in [a,b]$.
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Continuously differentiable means that the first derivative is continuous.

Strict contraction means that $d(f(x),f(y)) < d(x,y)$ for all $x,y \in [a,b]$ (in our case the metric would just be the standard metric on R (d(x,y) = |x-y|)).------

When I was looking at the problem I figured we need something that has slope less than 1, then hits 1 for exactly one point, and then the slope goes back down below 1 again.

My first guess was sin (and cos).

For example. $f:[0,\pi]\to \mathbb{R}$ defined by $f(x) = -cos(x)$. This gives us $f'(x) = sin(x)$. Which will give us that $f'(\pi/2) = 1$.

I think this example should work, it is obviously continuously differentiable, but my problem is proving that f is a STRICT contraction. We know that f is a contraction by a previous exercise (if f is diff and $|f'(x)| \leq 1$ then f is a contraction). But getting the strict part I am having trouble with.

First, is it even true that it is a strict contraction? If so, any hints on how to prove it. Also, are there any interesting, or simpler, examples for this problem? Thanks!

 

[siddharth]

I'm confused with the question.

If $$f:[a,b] \rightarrow \mathbb{R}$$ is continuously differentiable, then using the mean value theorem,

$$\frac{f(x)-f(y)}{x-y} = f'(c) , c \in [x,y]$$

So, won't f be a strict contraction only if $$f'(x)<1 \quad \forall x \in [a,b]$$?

 

[mattmns]

I am not 100% sure, but the problem is as I stated it, I just double checked a minute ago.

I had thought of the same thing, but I think there is something with taking the limit that allows for equals. The argument you gave is similar to the one I gave for the previous problem (which was that if f is differentiable and $|f'(x)| \leq 1$ then f is a contraction).

 

[AKG]

I'm confused with the question.

If $$f:[a,b] \rightarrow \mathbb{R}$$ is continuously differentiable, then using the mean value theorem,

$$\frac{f(x)-f(y)}{x-y} = f'(c) , c \in [x,y]$$

So, won't f be a strict contraction only if $$f'(x)<1 \quad \forall x \in [a,b]$$?

You're arguing backwards. What you say would only be true if for all c in [a,b], there exists x,y in [a,b] with [f(x)-f(y)]/[x-y] = f'(c).

 

[AKG]

Here is the question
--------
Give an example of a function $f:[a,b] \to \mathbb{R}$ which is continuously differentiable and which is a strict contraction, but such that $|f'(x)| = 1$ for at least one value of $x \in [a,b]$.
--------

Continuously differentiable means that the first derivative is continuous.

Strict contraction means that $d(f(x),f(y)) < d(x,y)$ for all $x,y \in [a,b]$ (in our case the metric would just be the standard metric on R (d(x,y) = |x-y|)).


------

When I was looking at the problem I figured we need something that has slope less than 1, then hits 1 for exactly one point, and then the slope goes back down below 1 again.

My first guess was sin (and cos).

For example. $f:[0,\pi]\to \mathbb{R}$ defined by $f(x) = -cos(x)$. This gives us $f'(x) = sin(x)$. Which will give us that $f'(\pi/2) = 1$.

I think this example should work, it is obviously continuously differentiable, but my problem is proving that f is a STRICT contraction. We know that f is a contraction by a previous exercise (if f is diff and $|f'(x)| \leq 1$ then f is a contraction). But getting the strict part I am having trouble with.

First, is it even true that it is a strict contraction? If so, any hints on how to prove it. Also, are there any interesting, or simpler, examples for this problem? Thanks!

-cos(x) should work. If it didn't, there would be a point (x,f(x)) on the graph such that either the line with slope 1 or the line with slope -1 through (x,f(x)) intersects the graph of -cos(x) at another point. See graphically why this cannot be, and use this understanding to motivate a proof using calculus as to why this cannot be.

 

[siddharth]

You're arguing backwards. What you say would only be true if for all c in [a,b], there exists x,y in [a,b] with [f(x)-f(y)]/[x-y] = f'(c).

Yeah, I understood my error, Thanks.