Cauchy’s functional equation I

Next semester, I’m going to write my undergraduate thesis about the P\neq NP problem, and right now I’m trying to decide if I should write it in Danish or in English. So I decided to translate a few pages about Cauchy’s functional equation that I wrote in Danish last year. Today I’ll post the first half of this, and I will post the rest in a few days (update: It’s here. Here you can also find the notes in pdf.). If you have anything to say about the mathematics in this post or about my English, please post a comment.

Cauchy’s functional equation

Cauchy’s functional equation, f(x+y)=f(x)+f(y), f:\mathbb{R}\to\mathbb{R} looks very simple, and it has a class of simple solutions, f(x)=\lambda x, but there are many other and more interesting solutions. In these notes, I will show you what some of these “wild” solutions look like, and I will use them to prove that there exist a set A\subset \mathbb{R}, such that neither A nor \mathbb{R}\setminus A contains a measurable subset with positive measure. Section 1 is about Cauchy’s functional equation on the rational numbers, in section 2 I show that there some wild solutions on \mathbb{R}, and in section 3 I will show that their graphs are dense in \mathbb{R}^2. In section 4 I’ll show that these functions are ugly from a measure theoretical point of view, and in section 5, I’ll show that some of these functions are wilder than others. E.g., I will prove that there is a solution to Cauchy’s functional equation, that intersects any continuous function from \mathbb{R} to \mathbb{R}.

1: The simple solutions

First, we consider the equation over the rational numbers. That is, f(x+y)=f(x)+f(y),f:\mathbb{Q}\to\mathbb{Q} By setting x=y=0 we get f(0)=2f(0) and thus f(0)=0. Let’s set \lambda = f(1). If f(n)=\lambda n we get: f(n+1)=f(n)+f(1)=\lambda n+\lambda=\lambda (n+1) By definition of \lambda, we have f(n)=\lambda n for n=1, so by induction, f(n)=\lambda n for all n\in\mathbb{N}. More generally, we can prove that for x\in \mathbb{Q} and n\in\mathbb{N} we have f(nx)=nf(x): It is clearly true for n=1 and if it is true for n we get: f((n+1)x)=f(nx+x)= f(nx)+f(x)=nf(x)+f(x)=(n+1)f(x) Let x be a positive rational number, and write it as x=\frac{n}{m}, where n,m\in \mathbb{N}. Now, mf(x)=mf(n/m)=f(n)=nf(1)=\lambda n Dividing by m we get f(x)=f(n/m)=\lambda(n/m)=\lambda x. Furthermore, 0=f(0)=f(x+(-x))=f(x)+f(-x) so f(x)=-f(-x). Putting it all together we have f(x)=\lambda x for all x\in \mathbb{Q}. It is easy to verify that f(x)=\lambda x is a solution for the general equation on \mathbb{R}.

2: Existence of wild solutions

Now consider Cauchy’s functional equation on the real numbers, f(x+y)=f(x)+f(y), f:\mathbb{R}\to\mathbb{R} The proof from last section, tells us that f(q)=qf(1) for all rational numbers q, and using the same idea, we can prove that f(qx)=qf(x) for all q\in \mathbb{Q} and x\in \mathbb{R}. But this does not imply that f(x)=xf(1) for all the real numbers. However, if we assume that f is continuous, we can show that f(x)=\lambda x for all x\in \mathbb{R}: We simply choose a sequence (x_i)_{i\in \mathbb{N}} of rational numbers that converge to x. By continuity we get f(x)=f\left(\lim_{i\to\infty}x_i\right)=\lim_{i\to\infty}f(x_i)=\lim_{i\to\infty}\lambda x_i=\lambda x But it is much more fun if we do not have any assumptions on f! Using axiom of choice we can find non-continuous solutions. The idea is: A priori we only know that f(0)=0. Now we choose some value for f, e.g. f(1)=3. This determines f on all the rational numbers, f(q)=3q for q\in \mathbb{Q}, but the value of f is not determined on any irrational number. So we make another choice, let’s say f(\sqrt{2})=\pi. Now the functional equation tells us that f(q_1+q_2\sqrt{2})=3q_1+\pi q_2 for all q_1,q_2\in \mathbb{Q}. But for numbers x not on this from, we cannot determine the value of f(x). So we simply continue by choosing more and more values of the function. Unfortunately, we have to make infinitely many choices, so we need axiom of choice. In the rest of these notes, I will assume axiom of choice. To formalize the above, we consider the set of real numbers as a vector space over \mathbb{Q}, in much that same way as you can consider \mathbb{C} to be a two dimensional vector space over \mathbb{R}. An important difference is, that when we consider \mathbb{R} to be a vector space over \mathbb{Q} it is infinite dimensional: it even has uncountably many dimensions. We now use the axiom of choice to choose a basis (x_i)_{i\in I}, x_i\in \mathbb{R} (a so-called Hamel basis) and we choose some coefficients (\lambda_i)_{i\in I},\lambda_i\in \mathbb{R}. This defines a linear map from this vector space to itself: f\left(\sum_{i\in I}q_ix_i\right)=\sum_{i\in I}q_i\lambda_i, where the q_is are rational numbers, and only finitely many of them are non-zero. I called this function ‘linear’, so it sounds like it is a nice function. But it is not! It is only linear when we consider \mathbb{R} as a vector space over \mathbb{Q} and forget about the rest of the structure on \mathbb{R}. This function is only linear in the usual sense on \mathbb{R} if \frac{\lambda_i}{x_i} is the same for all i. All functions on this form are solutions to the Cauchy’s functional equation, and conversely all solutions to Cauchy’s functional equation are on this form.

3: What do I mean by “wild”?

A function can be more or less wild/ugly/pathological. Here is a list of possible definitions of what makes a function f wild. The list is ordered, such that any of the properties imply the one above.

  • f is discontinuous.
  • f is discontinuous in every point in \mathbb{R}.
  • f is unbounded on any open interval.
  • f is unbounded above on any open interval.
  • the graph of f, defined by \{(x,f(x))|x\in\mathbb{R}\}\subset \mathbb{R}^2, is dense in \mathbb{R}^2, that is, any point in \mathbb{R}^2 is the limit of a sequence of points in the graph of f.

All of these statements are true for any non-linear solution to Cauchy’s functional equation. I will show that the last one of these is true. Proof: Let f be a non-linear solution. If (x_1,y_1) and (x_2,y_2) are points in the graph of f, and q is a rational number, we see that the points (x_1,y_1)+(x_2,y_2)=(x_1+x_2,y_1+y_2) and q(x_1,y_1)=(qx_1,qy_1) are both in the graph too. In words, any linear combination over \mathbb{Q} of points in the graph are also in the graph. Since f is non-linear we can find real numbers x_1 and x_2, both non-zero, such that \frac{f(x_1)}{x_1}\neq \frac{f(x_2)}{x_2}. Now the two vectors (x_1,f(x_1)) and (x_2,f(x_2)) are linearly independent (over \mathbb{R}), so they span the plane. That is, any point (x,y) can be written as a(x_1,f(x_1))+b(x_2,f(x_2)) for some a,b\in \mathbb{R}. Let q_i and r_i be sequences of rational numbers with \lim_{i\to \infty}q_i=a and \lim_{i\to\infty}r_i=b. Now q_i(x_1,f(x_1))+r_i(x_2,f(x_2)) is a sequence of points in the plan converging to (x,y), so the graph is dense in \mathbb{R}^2.

About these ads
Explore posts in the same categories: Uncategorized

Tags: ,

You can comment below, or link to this permanent URL from your own site.

6 Comments on “Cauchy’s functional equation I”

  1. qubit1 Says:

    I like the style of writing – most expository mathematics tends to be very formal, bordering on daunting, and this isn’t!

  2. [...] this post I will finish what I began in my previous post. I have collected the two posts in a pdf-file, which might be a bit easier to read. Again you are [...]

  3. Johan Richter Says:

    There is a LaTex error in the second bulleted point of section 3.

  4. Very nice post. I just stumbled upon your blog and
    wanted to say that I have really enjoyed browsing your blog posts.
    After all I’ll be subscribing to your rss feed and I hope you write again soon!

  5. I like reading through an article that can make men and women think.

    Also, many thanks for permitting me to comment!

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s


Get every new post delivered to your Inbox.

Join 37 other followers

%d bloggers like this: