Which differentiable functions R to R are bijective?

[murrayE]

I innocently gave my students a problem: Which differentiable functions f: R \rightarrow R are bijective? "Innocently", I say, because I'm finding it hard to come up with any simple set of conditions that are both necessary and sufficient. Here's what I can say so far:

(1) If f'(x) \neq 0 for all real x, then f is injective. (Easy)

(2) If f'(x) > 0 for all x and lim f(x) = +\infty as x \rightarrow +\infty and lim f(x) =- \infty as x \rightarrow -\infty, or if f'(x) < 0 for all x and lim f(x) =- \infty as x \rightarrow +\infty and lim f(x) = +\infty as x \rightarrow -\infty, then f is bijective.

(3) There are many bijective differentiable functions f: R \rightarrow R that are bijective but for which f'(x) = 0 at one or more x. For example, f(x) = x^3

Any ideas on a clean necessary & sufficient set of conditions?

 

[TylerH]

The rule is that f'(x) must equal 0 at countably many values of x.

To simplify, you could say that f'(x) must be more than or equal to 0 for all x or less than or egual to for all x. \mbox{(\forall x \in \Re f&#039;(x) \ge 0 \vee \forall x \in \Re f&#039;(x) \le 0) \wedge f&#039;(x)=0 at countably many x} \Leftrightarrow bijective

Tex is being annoying. I hope you can read that.

 

[I like Serena]

The rule is that f'(x) must equal 0 at countably many values of x.

To simplify, you could say that f'(x) must be more than or equal to 0 for all x or less than or egual to for all x. \mbox{(\forall x \in \Re f&#039;(x) \ge 0 \vee \forall x \in \Re f&#039;(x) \le 0) \wedge f&#039;(x)=0 at countably many x} \Leftrightarrow bijective

Tex is being annoying. I hope you can read that.

Why can f'(x) only be 0 at countably many values of x?

 

[TylerH]

Why can f'(x) only be 0 at countably many values of x?

I had it backwards. That would be the rule for f being injective. Also, ignore everything below that line...

Sorry.

 

[I like Serena]

I'd say:

For differentiable functions f from R to R we have:

(1) f \mbox{ is injective } \iff \forall x \in \mathbb{R}: f&#039;(x) \ge 0 \quad \vee \quad \forall x \in \mathbb{R}: f&#039;(x) \le 0

(2) f is bijective iff f is injective and f approaches plus resp. minus infinity in its limits.

 

[murrayE]

I'd say:

For differentiable functions f from R to R we have:

(1) f \mbox{ is injective } \iff \forall x \in \mathbb{R}: f&#039;(x) \ge 0 \quad \vee \quad \forall x \in \mathbb{R}: f&#039;(x) \le 0

(2) f is bijective iff f is injective and f approaches plus resp. minus infinity in its limits.

Statement (1) is wrong because a constant function would satisfy the condition yet not be injective! Certainly a strictly positive derivative everywhere or a strictly negative derivative everywhere suffices to be injective. As already mentioned, there is an issue about what kind of zeros the derivative can have.

Part of the difficulty of this problem is that a differentiable function need not have a continuous derivative, and that a differentiable bijection (e.g., x^3) need not have a differentiable inverse.

 

[I like Serena]

Statement (1) is wrong because a constant function would satisfy the condition yet not be injective! Certainly a strictly positive derivative everywhere or a strictly negative derivative everywhere suffices to be injective. As already mentioned, there is an issue about what kind of zeros the derivative can have.

Part of the difficulty of this problem is that a differentiable function need not have a continuous derivative, and that a differentiable bijection (e.g., x^3) need not have a differentiable inverse.

True.
Here's a new effort, although for now I'll have to assume the "countably many places" clause to be correct.

For differentiable functions f from R to R we have:

(1) f \mbox{ is injective} \iff (\forall x \in \mathbb{R}: f&#039;(x) \ge 0 \quad \vee \quad \forall x \in \mathbb{R}: f&#039;(x) \le 0) \quad \wedge \quad (f&#039;(x) = 0 \mbox{ in countably many places})

(2) f \mbox{ is bijective} \iff f \mbox{ is injective and } |f(x)| \to \infty \mbox{ for } x \to -\infty \mbox{ and for } x \to +\infty

 

[murrayE]

True.
Here's a new effort, although for now I'll have to assume the "countably many places" clause to be correct.

For differentiable functions f from R to R we have:

(1) f \mbox{ is injective} \iff (\forall x \in \mathbb{R}: f&#039;(x) \ge 0 \quad \vee \quad \forall x \in \mathbb{R}: f&#039;(x) \le 0) \quad \wedge \quad (f&#039;(x) = 0 \mbox{ in countably many places})...

No, that still cannot be correct, and for the same reason, namely, a constant function (and even a function that's nondecreasing and constant on some interval) would satisfy the condition yet not be injective.

 

[I like Serena]

No, that still cannot be correct, and for the same reason, namely, a constant function (and even a function that's nondecreasing and constant on some interval) would satisfy the condition yet not be injective.

A constant function would have f'(x)=0 in an uncountable number of places.
The same is true for a function that is constant on some interval.
But for instance f(x)=x + sin x is bijective although f'(x)=0 for an infinite (but countable) number of values.

More specifically, for each x₀ with f'(x₀)=0 there must be an interval <x₀-eps,x₀+eps> for a sufficiently small eps >0, where f'(x) not 0, if x not x₀.

 

[murrayE]

No, that still cannot be correct, and for the same reason, namely, a constant function (and even a function that's nondecreasing and constant on some interval) would satisfy the condition yet not be injective.

Sorry, I forgot you included the qualification about the derivative being zero at only countably many points.

It's not clear to me why, with that qualification, we have now a necessary and sufficient condition for a differentiable function on the line being injective.

I can certainly imagine a smooth step-like function with denumerably many steps that's bijective. Without continuity of the derivative (and that's NOT assumed), does injectivity prevent a limit point of zeros of the derivative? does injectivity prohibit an uncountable set of zeros of the derivative that, nonetheless, don't force the function to be constant on some interval?

 

[I like Serena]

I can certainly imagine a smooth step-like function with denumerably many steps that's bijective.

Yes, f(x)=x + sin x would be such a function.

Without continuity of the derivative (and that's NOT assumed), does injectivity prevent a limit point of zeros of the derivative? does injectivity prohibit an uncountable set of zeros of the derivative that, nonetheless, don't force the function to be constant on some interval?

For injectivity the function can not be constant on any interval.
In other words, there must be clear separations between the points where f' is zero (some epsilon > 0). I think that qualifies for the zeroes to be countable, but I have to admit that I'm not entirely sure yet.
Could there be a function with an uncountable number of inflection points that is injective nonetheless?

 

[Citan Uzuki]

The zeros of f' don't have to be countable. In fact, the zeros of f' don't even have to have measure zero! In fact, for any ε>0, we can find an injective function f:R→R such that f'(x) = 0 on the complement of a set of measure epsilon. Consider the following: let U be a dense open subset of R having measure ε, and let C be the complement of that set. Define g(x):=\mathrm{d}(x,C)=\min \{|x-y|:y \in C\}. Observe that g(x) is continuous and that g(x)=0 \Leftrightarrow x \in C. Now, let f(x) = \int_{0}^{x} g(t) \mathrm{d}t. By the fundamental theorem of calculus, f'(x) = g(x), and f is strictly increasing, since for any a<b, U∩(a,b) is a nonempty open set by the density of U, and hence \mu (\{x \in (a,b): g(x)&gt;0\}) = \mu (U \cap (a,b)) &gt; 0, so f(b)-f(a) = \int_{a}^{b} g(t) \mathrm{d}t &gt; 0. With additional extra work we can arrange for f to be surjective as well (essentially by multiplying g with a function that blows up fast enough).

I think the best we can do here is to say that f is bijective iff the following hold:

a) (∀x, f'(x) ≥ 0) or (∀x, f'(x) ≤ 0)
b) f'(x) ≠ 0 on a dense set of points, and
c) \int_{-\infty}^{0} |f&#039;(t)| \mathrm{d}t = \int_{0}^{\infty} |f&#039;(t)| \mathrm{d}t = \infty

Edit: in a previous version of this post, I asserted condition c was that the integral over the whole real line of |f'(t)| was infinite. However that only guarantees that the range of f will be at least a semi-infinite interval, not necessarily that it will be the whole real line. This version should be correct.

 

[I like Serena]

The zeros of f' don't have to be countable. In fact, the zeros of f' don't even have to have measure zero! In fact, for any ε>0, we can find an injective function f:R→R such that f'(x) = 0 on the complement of a set of measure epsilon. Consider the following: let U be a dense open subset of R having measure ε, and let C be the complement of that set. Define g(x):=\mathrm{d}(x,C)=\min \{|x-y|:y \in C\}. Observe that g(x) is continuous and that g(x)=0 \Leftrightarrow x \in C. Now, let f(x) = \int_{0}^{x} g(t) \mathrm{d}t. By the fundamental theorem of calculus, f'(x) = g(x), and f is strictly increasing, since for any a<b, U∩(a,b) is a nonempty open set by the density of U, and hence \mu (\{x \in (a,b): g(x)&gt;0\}) = \mu (U \cap (a,b)) &gt; 0, so f(b)-f(a) = \int_{a}^{b} g(t) \mathrm{d}t &gt; 0. With additional extra work we can arrange for f to be surjective as well (essentially by multiplying g with a function that blows up fast enough).

I think the best we can do here is to say that f is bijective iff the following hold:

a) (∀x, f'(x) ≥ 0) or (∀x, f'(x) ≤ 0)
b) f'(x) ≠ 0 on a dense set of points, and
c) \int_{-\infty}^{0} |f&#039;(t)| \mathrm{d}t = \int_{0}^{\infty} |f&#039;(t)| \mathrm{d}t = \infty

Edit: in a previous version of this post, I asserted condition c was that the integral over the whole real line of |f'(t)| was infinite. However that only guarantees that the range of f will be at least a semi-infinite interval, not necessarily that it will be the whole real line. This version should be correct.

I don't get it.
In your example any two a, b in C and to the left of U would have f(a)=f(b) which violates injectivity.

Btw, I like c).

 

[Citan Uzuki]

U is dense in R, so there are no a, b to the left of U.