Extra Maths: Analysis, Darboux’s Theorem

As before, these questions are aimed at first-year or second-year undergraduates studying a course in Real Analysis.

Question 1. Let f:[a,b]\longrightarrow\mathbb{R} be differentiable, with derivative f'. Suppose that f'(a)>0>f'(b). Show there exists c\in(a,b) such that f'(c)=0.

Compare Question 1 with Rolle’s Theorem.

Question 2 (Darboux’s Theorem). Let f:\mathbb{R}\longrightarrow\mathbb{R} be differentiable, with derivative f'. Show that f' has the intermediate value property.

Compare Question 2 with the Mean Value Theorem.

Question 3. Let f:[a,b]\longrightarrow\mathbb{R} be differentiable, with derivative f'. Show that there exists c\in[a,b] such that f' is continuous at c.

This is very difficult! See this article. Nevertheless, I would like to expand on this argument in a subsequent post. Watch this space.

Question 4. Suppose that g:\mathbb{R}\longrightarrow\mathbb{R} satisfies the intermediate value property. Need g admit an anti-derivative?

Actually, this one is straightforward, so I’ll give the argument right away.

Proof. Certainly not. In fact we have just seen that every derivative (so, a function that admits an anti-derivative) is continuous somewhere. However, there are plenty of functions that satisfy the intermediate value property which are continuous nowhere. For example, there are functions whose graphs are dense in the plane. \square

Remark. For an in-depth discussions of the continuity of derivatives, see this article.


Extra Maths: Analysis, Accumulation

Here are three fun Analysis questions, really aimed at my students who are studying their first/second course on Real Analysis, but hopefully it’s of wider interest as well.

We begin at the beginning, with a couple of definitions.

Definition. A function f:\mathbb{R}\longrightarrow\mathbb{R} has the IVT Property if, for all real numbers a,b,x with a<b and \min\{f(a),f(b)\}<x<\max\{f(a),f(b)\} there exists \xi\in[a,b] such that f(\xi)=x.

So, using this terminology, the Intermediate Value Theorem is the statement that a continuous function \mathbb{R}\longrightarrow\mathbb{R} has the IVT Property.

Definition. Let f:\mathbb{R}\longrightarrow\mathbb{R} and let a,l\in\mathbb{R}. We say that l is a weak limit of f at i if
for all \varepsilon>0 there exists b\in(a-\varepsilon,a+\varepsilon)\setminus\{a\} such that f(b)\in(l-\varepsilon,l+\varepsilon). Naturally, we say that f has a weak limit somewhere if there exists a,l\in\mathbb{R} such that l is a weak-limit of f at a.

We say that l\in\mathbb{R} is an accumulation point of a real sequence (a_{n}) if for all \varepsilon>0 there is an infinite set N\subseteq\mathbb{N} such that for all n\geq N we have |a_{n}-l|<\varepsilon.

The Questions

Question 1.
Show the existence or non-existence of a function f:\mathbb{R}\longrightarrow\mathbb{R} which has the IVT Property but which is nowhere continuous.

Question 2.
Prove or disprove the claim that every function f:\mathbb{R}\longrightarrow\mathbb{R} has a weak limit somewhere.

Question 3.
Find a real sequence (a_{n}) such that every real number l\in\mathbb{R} is an accumulation point of (a_{n}).

Where should we start? Perhaps my favourite of the three is the first one: to show the existence or non-existence of a function with the IVT property but which is nowhere continuous. There are several approaches we might take, beginning with the following lemma.

Lemma 1. Suppose that a function f:\mathbb{R}\longrightarrow\mathbb{R} has the property that, for all a,b with a<b, the restriction of f to the interval (a,b) is surjective onto \mathbb{R}. Then f has the IVT property.

Hopefully the proof of this lemma is quite clear: the IVT property asks, of such a function restricted to such an interval, that it attain certain values. So if f attains all real values, then the property is evidently satisfied.

Okay…. So what?

Lemma 2. Suppose that a function f:\mathbb{R}\longrightarrow\mathbb{R} has the property that, for all a,b with a<b, the restriction of f to the interval (a,b) is surjective onto \mathbb{R}. Then f is continuous nowhere.

Proof. Let a\in\mathbb{R} be given, and consider, for example, \varepsilon=1. For any \delta>0, by assumption f restricted to the open interval (a-\delta,a+\delta) is surjective onto \mathbb{R}. In particular, the image under f of (a-\delta,a+\delta) is not a subset of (f(a)-1,f(a)+1). \square

Ah-ha! Now it’s clear why we’re interested in these sorts of function. Such a `locally surjective function‘ would be a nowhere continuous function which nevertheless has the IVT property. The big question now is: does any locally surjective function exist?

Well, YES. Such functions do exist, and there are several ways to find them. First we give a slightly abstract construction, which might seem a little unsatisfactory.

Method 1. Recall that the cardinality of \mathbb{R} is 2^{\aleph_{0}}, and the cardinality of \mathbb{R}/\mathbb{Q} is also 2^{\aleph_{0}}. We may choose a bijection


Also, let \phi:\mathbb{R}\longrightarrow\mathbb{R}/\mathbb{Q} be the natural map sending each real number r to its coset modulo \mathbb{Q}, namely r+\mathbb{Q}. Note that \phi is surjective. Thus, the composition F\circ\phi:\mathbb{R}\longrightarrow\mathbb{R} is also a surjection. Indeed, by the density of the rationals in the reals, each proper open interval (a,b) contains a representative of each coset of \mathbb{Q} in \mathbb{R}. That is, for each r\in\mathbb{R}, there exists q\in\mathbb{Q} such that r+q\in(a,b). A consequence of this is that the restriction to (a,b) of \phi is still surjective onto \mathbb{R}/\mathbb{Q}. Finally, the restriction of F\circ\phi to (a,b) is still surjective onto \mathbb{R}, which shows that F\circ\phi is locally surjective, as required.\square

Wonderful, that was easy. But it wasn’t particularly constructive. Perhaps it felt a little unsatisfactory to rely on the cardinalities of these sets. Is there a better way?

That’s where I’ll leave it for the moment. To be continued!

In the meantime, try to construct a locally surjective function by using the decimal expansion of each real number.