## December 10, 2010

### Pri la Funkcia Ekvacio f(x + y) = f(x) + f(y)

#### Posted by Tom Leinster

Jam de longe, Cauchy pruvis ke kontinua funkcio $f(x)$ kiu verigas la funkcian ekvacion $f(x + y) = f(x) + f(y)$ kiuj ajn estu la nombroj $x, y$, necese estas homogena, unuagrada funkcio $f(x) \equiv A x$.

So begins Maurice Fréchet’s 1913 paper in L’Enseignement Mathématique. I came across it when I was trying to find the right reference for the solution of this functional equation. Apparently Cauchy was the first to prove that when $f$ is continuous, the only solutions are $f(x) = A x$ for some constant $A$. Mark Meckes had told me that Lebesgue measurability of $f$ was sufficient, and found a nice explanation at the Tricki.

But I needed to find the original reference. And when I tracked it down, I was surprised and intrigued to find that it was in Esperanto.

It turns out that Fréchet, apart from making numerous contributions to analysis, was a keen Esperantist, publishing many papers in the subject and (as Mark pointed out to me) serving as president of the Internacia Scienca Asocio Esperantista. I’m looking forward to citing it.

Incidentally, Fréchet’s result can be improved further. Mark Kormes, in 1926, showed that one only needs to assume $f$ to be bounded on some set of positive measure. (This is a weaker condition than measurability.) But, boringly, his paper is in English.

Posted at December 10, 2010 2:17 AM UTC

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### Re: Pri la Funkcia Ekvacio f(x + y) = f(x) + f(y)

Wow! Cool! But you should have posted this blog entry using rot13.

A cute fact, sort of obvious from what you said, is that starting with Zermelo-Fraenkel set theory, it’s perfectly consistent to assume either that $f(x + y) = f(x) + f(y)$ does have solutions other than the obvious $f(x) = A x$ ones, or that it doesn’t. The Axiom of Choice implies it does have other solutions… which unfortunately we are unable to write down. The Axiom of Determinancy implies it doesn’t.

Guess which axiom most mathematicians prefer.

And another, slightly harder puzzle: guess which famous mathematician wrote sentences like this:

He Inq Hd : Utr $\cdot$ $\not$ Pot . PAN Hd Inq Etc .

Posted by: John Baez on December 10, 2010 9:43 AM | Permalink | Reply to this

### Re: Pri la Funkcia Ekvacio f(x + y) = f(x) + f(y)

It’s funny that you mention the axiom of choice, because Mark and I were discussing that at the same time for wholly unrelated reasons.

I’d noticed two apparently conflicting attitudes to the axiom of choice in two different analysis books, both expressed rather… snarkily? Nik Weaver, in his book Lipschitz algebras, writes about something-or-other:

It even avoids using the axiom of choice, as if that mattered

(followed by a footnote about “mainstream mathematicians”). But Robert Strichartz, in his book A guide to distribution theory and Fourier transforms, writes apropos of something else:

wise-guys who like using the axiom of choice will have to worry about [continuity], along with wolves under the bed, etc.

Mark observed that although non-measurability may be nasty, so is the prospect of a nontrivial Banach space with no nonzero continuous linear functionals.

Posted by: Tom Leinster on December 10, 2010 10:02 AM | Permalink | Reply to this

### Re: Pri la Funkcia Ekvacio f(x + y) = f(x) + f(y)

Tom wrote:

Mark observed that although non-measurability may be nasty, so is the prospect of a nontrivial Banach space with no nonzero continuous linear functionals.

I’m not an expert on constructive mathematics, so even if we drop the axiom of choice, I don’t know if one can construct a nontrivial Banach space with a trivial dual. I’ll only worry about these beasts if someone hands me one.

Posted by: John Baez on December 10, 2010 10:09 AM | Permalink | Reply to this

### Re: Pri la Funkcia Ekvacio f(x + y) = f(x) + f(y)

Well, I carefully reproduced Mark’s phrasing—he said the prospect of such a space. But yeah, I see your point.

Posted by: Tom Leinster on December 10, 2010 10:13 AM | Permalink | Reply to this

### Re: Pri la Funkcia Ekvacio f(x + y) = f(x) + f(y)

The prospect of a wolf under the bed is also nasty, but I can live with the prospect as long as nobody actually exhibits such a wolf!

Posted by: John Baez on December 10, 2010 11:03 AM | Permalink | Reply to this

### Re: Pri la Funkcia Ekvacio f(x + y) = f(x) + f(y)

The problem with the prospect of a wolf under the bed is not that it ever actually bites you, but that every time you reach under the bed to pull out a stuffed animal you have to prove that it isn’t a wolf!

Posted by: Mike Shulman on December 11, 2010 6:38 AM | Permalink | Reply to this

### Re: Pri la Funkcia Ekvacio f(x + y) = f(x) + f(y)

There was some Polymath work on ‘definable’ Banach spaces described here. The proposal was to show that pretty much any Banach space you define (as some sort of function space on a measure space) contains one of the ‘standard ones’. Of course, I have no idea if the proofs would hold up constructively, but it seems that it is hard to find ‘definable’ Banach spaces (whatever these are) that are too crazy.

As far as the question that John posed, I would guess Frege, but looking up his Begriffsschrift, I realise that this isn’t him.

Aha - found it. I guess it wouldn’t give the game away if I said this person was well-known for their contributions to algebraic topology.

Posted by: David Roberts on December 10, 2010 10:34 AM | Permalink | Reply to this

### Re: Pri la Funkcia Ekvacio f(x + y) = f(x) + f(y)

If anyone gives up and wants to know the answer to my puzzle, they can click here.

Posted by: John Baez on December 14, 2010 10:27 AM | Permalink | Reply to this

### Re: Pri la Funkcia Ekvacio f(x + y) = f(x) + f(y)

I’m a bit confused about this Banach space business. The axiom of choice is equivalent, in the presence of the ZF axioms, to the statement that every vector space has a basis. Is it also equivalent to the statement that every vector space has a nontrivial dual?

Posted by: Tom Leinster on December 10, 2010 10:50 AM | Permalink | Reply to this

### Re: Pri la Funkcia Ekvacio f(x + y) = f(x) + f(y)

The always infallible Wikipedia has a footnote claiming, without proof or reference, that the axiom of choice is needed to show that the dual of an infinite-dimensional vector space is nonzero.

However, let it be known that all my previous comments on this blog entry concerned Banach spaces, and when I said ‘dual’ I meant the space of continuous linear functionals. Is the statement that every infinite-dimensional Banach space has a nontrivial dual in this sense equivalent to the axiom of choice? This is a different question than your question about vector spaces.

I’d bet a doughnut the answer is ‘yes’. But I don’t know a reference, not even a reference to a footnote in a Wikipedia article.

Posted by: John Baez on December 10, 2010 11:18 AM | Permalink | Reply to this

### Re: Pri la Funkcia Ekvacio f(x + y) = f(x) + f(y)

Whoops — I take back that bet. The Hahn-Banach theorem is not equivalent to axiom of choice: it’s implied by it, but doesn’t imply it. So, the statement that every infinite-dimensional Banach space has a nontrivial dual is weaker than the axiom of choice.

Sorry, no easy doughnut for you today.

However, the Hahn-Banach theorem does imply the existence of a Lebesgue nonmeasureable subset of the line!

(However, my computer warns me that the website that Wikipedia links to, that has a paper with the proof of this fact, also carries computer viruses. So watch out.)

Posted by: John Baez on December 10, 2010 11:25 AM | Permalink | Reply to this

### Re: Pri la Funkcia Ekvacio f(x + y) = f(x) + f(y)

For a nice Banach space like L^p, you need the axiom of choice to show the existence of a non-continuous linear functional.

Posted by: walt on December 10, 2010 2:46 PM | Permalink | Reply to this

### Re: Pri la Funkcia Ekvacio f(x + y) = f(x) + f(y)

Right, I know you were talking about the continuous duals of Banach spaces, whereas I (for one message) was talking about the algebraic dual of vector spaces, but I was just warming up.

Thanks for all the theorems. Why don’t I win a doughnut? Actually I don’t really like doughnuts, so perhaps I’d prefer not to win one. I’d rather have a coffee.

Posted by: Tom Leinster on December 10, 2010 12:15 PM | Permalink | Reply to this

### Re: Pri la Funkcia Ekvacio f(x + y) = f(x) + f(y)

Tom wrote:

Why don’t I win a doughnut? Actually I don’t really like doughnuts, so perhaps I’d prefer not to win one. I’d rather have a coffee.

Actually that ‘you’ was addressed at all the ravenous mathematicians whom I feared would pounce on my offer, win that bet and demand a doughnut. I’ll gladly buy you a coffee anytime, Tom!

Posted by: John Baez on December 10, 2010 2:37 PM | Permalink | Reply to this

### Re: Pri la Funkcia Ekvacio f(x + y) = f(x) + f(y)

Maybe here’s a Banach space for which it’s hard to ‘construct’ a nonzero continuous linear functional: the space of bounded operators on a countable-dimensional Hilbert space, mod the space of compact operators. In fact this Banach space is a Banach algebra, the Calkin algebra. I seem to recall that it requires the axiom of choice to construct a nonzero continuous linear functional on the Calkin algebra.

Maybe here’s another Banach space for which it’s hard to ‘construct’ a nonzero continuous linear functional: $\ell^\infty$ mod the closure of the space of sequences with just finitely many nonzero terms. I know we can get a nonzero continuous linear functional on this space starting from a Banach limit. But I also know it takes the axiom of choice to prove the existence of a Banach limit; we can’t ‘construct’ a Banach limit.

The idea in both cases is to take a Banach space and mod out by a closed subspace that’s quite huge, so the quotient space is rather wispy and elusive, and its dual even more so.

I’m putting scare quotes around ‘construct’ because I don’t really know for sure what it would mean to construct a continuous linear functional on these quotient spaces, though I would probably know a construction if I saw one.

Posted by: John Baez on December 10, 2010 11:01 AM | Permalink | Reply to this

### Re: Pri la Funkcia Ekvacio f(x + y) = f(x) + f(y)

Incidentally, in an appropriate sense, your two suggestions are almost the same. In (some parts of) Banach space theory, people would say that your first example is just a noncommutative version of the second, in the same sense that the Schatten classes are noncommutative versions of $\ell^p$ spaces. This is because the space of bounded operators on a countable-dimensional Hilbert space is the space of operators whose sequence of singular values lies in $\ell^\infty$, and the compact operators are those whose singular values converge to $0$.

Posted by: Mark Meckes on December 10, 2010 2:23 PM | Permalink | Reply to this

### Re: Pri la Funkcia Ekvacio f(x + y) = f(x) + f(y)

Yes, it’s a great analogy. The algebra of bounded operators is the noncommutative analogue of $\ell^\infty$, and the algebra of compact operators is the noncommutative analogue of the algebra of sequences $c_0$ that you mentioned. The algebra of compact operators is the closure in the operator norm of the algebra of finite-rank operators, just as the algebra $c_0$ is the closure in the $\ell^\infty$ norm of the algebra of sequences with only finitely many nonzero terms.

In fact it’s more than analogy. We can let $\ell^\infty$ act as multiplication operators on $\ell^2$, getting an inclusion of $\ell^\infty$ in the bounded operators on $\ell^2$. Then the intersection of $\ell^\infty$ with the compact operators is $c_0$, and the intersection of $\ell^\infty$ with the finite rank operators is the sequences with only finitely many nonzero terms. And, as you hint, the intersection of $\ell^\infty$ with the $p$th-Schatten-class operators is $\ell^p$.

Posted by: John Baez on December 11, 2010 1:33 PM | Permalink | Reply to this

### Re: Pri la Funkcia Ekvacio f(x + y) = f(x) + f(y)

I hadn’t heard of Banach limits.

For anyone too lazy to click: a Banach limit is a recipe for taking the limit of any bounded sequence of reals. (I’ll say this for the reals rather than the complex numbers, because I only know the definition in the real case.) Formally, a Banach limit is a continuous linear map $\phi: \ell^\infty \to \mathbf{R}$ such that

• if $x$ converges then $\phi(x) = \lim_{n\to\infty} x_n$
• $\phi(x_2, x_3, \ldots) = \phi(x_1, x_2, \ldots)$
• $\phi(x) \geq 0$ if $x_n \geq 0$ for all $n$.

Elsewhere on this thread it’s been mentioned that without some form of the axiom of choice, it’s impossible to construct any continuous linear map $\ell^\infty \to \mathbf{R}$ apart from those of the form $x \mapsto \sum_n c_n x_n$ where $c \in \ell^1$. It’s easy to see that no element of $\ell^1$ gives a Banach limit. Thus, the existence of Banach limits depends on the axiom of choice.

It struck me that this was very similar to the situation in a post of mine a few months ago: means. For example, there’s no nontrivial recipe for taking the “mean” of binary digits without using the axiom of choice. (In standard terminology: the existence of a nonprincipal ultrafilter on an infinite set depends on the axiom of choice.) Similarly, to prove the amenability of any infinite abelian group, you need the axiom of choice.

A Banach limit is a kind of mean. For example, if $x$ is a genuinely convergent sequence, you might think of its limit as a kind of “average of $x_1, x_2, \ldots$, ignoring any finite segment at the beginning”. In the means post, I also talked about Arrow’s Theorem; and you might also think of a Banach limit as a voting system, whereby $x_1, x_2, \ldots$ get to vote for what their limit should be.

Posted by: Tom Leinster on December 13, 2010 3:18 PM | Permalink | Reply to this

### Re: Pri la Funkcia Ekvacio f(x + y) = f(x) + f(y)

Applying a Banach limit (for $N\to \infty$) of averages like $\frac{1}{2N+1} \sum_{j=-N}^N f(j)$ is one of the quickest-and-dirtiest ways to get a mean on the Banach space of bounded functions on $\mathbb{Z}$, and hence it is not too far from Banach limits to amenability. (I sadly suspect that it is now possible to write papers on variants of amenability without ever having seen this particular motivating case, but that’s getting off-topic.) By the way, if you’re interested in showing that additive functions have to be linear (under certain constraints), then what about showing that “almost additive” functions have to be “nearly linear”? For one interpretation of almost additive this leads to the industry that is Hyers-Ulam stability; but a slightly different interpretation, which I personally prefer, will lead to the question of when certain second bounded cohomology groups are constant.
Posted by: Yemon Choi on December 14, 2010 2:20 AM | Permalink | Reply to this

### Re: Pri la Funkcia Ekvacio f(x + y) = f(x) + f(y)

Interesting, thanks.

What are almost additive and nearly linear functions? Are you speaking asymptotically?

Posted by: Tom Leinster on December 14, 2010 9:53 AM | Permalink | Reply to this

### Re: Pri la Funkcia Ekvacio f(x + y) = f(x) + f(y)

Tom wrote:

(I’ll say this for the reals rather than the complex numbers, because I only know the definition in the real case.)

There are lots of equivalent ways to define a Banach mean in the complex case.

One is to take the definition you gave in the real case, change $\mathbf{R}$ to $\mathbf{C}$, reinterpret $\ell^\infty$ to mean the complex version of $\ell^\infty$, and don’t do anything else.

(This assumes your reader has the wits to know a complex number $x$ has $x \ge 0$ iff it’s real and nonnegative in the usual sense for real numbers.)

Another is to reinterpret $\ell^\infty$ to mean the complex version of $\ell^\infty$ and say $\phi : \ell^\infty \to \mathbb{C}$ is a Banach mean iff it is a continuous linear function whose restriction to the real version of $\ell^\infty$ is a Banach mean in the sense you defined.

Posted by: John Baez on December 14, 2010 10:25 AM | Permalink | Reply to this

### Re: Pri la Funkcia Ekvacio f(x + y) = f(x) + f(y)

Thanks. I was hoping that the definition would as simple as that. But since it isn’t clear to me how the list of axioms was arrived at, I didn’t have a basis on which to decide what definition would be “right”.

Posted by: Tom Leinster on December 14, 2010 10:33 AM | Permalink | Reply to this

### Re: Pri la Funkcia Ekvacio f(x + y) = f(x) + f(y)

OK, here’s what I think would be a more satisfactory way of stating the definition over $\mathbf{C}$.

We replace the third axiom in the real case ($x_n \geq 0$ for all $n$ $\implies$ $\phi(x) \geq 0$) by:

if $V \subseteq \mathbf{C}$ is closed and $x_n \in V$ for all $n$ then $\phi(x) \in V$.

Taking $V$ to be the right-hand half $[0, \infty)$ of the real axis gives the version of the definition that John mentioned first. We have to show that in the presence of the other two axioms, this special case implies the general case.

I’m halfway to a proof in my head, but it’s getting a bit messy. Maybe before I put more effort into it, some expert can tell me: is it true?

Posted by: Tom Leinster on December 14, 2010 10:58 AM | Permalink | Reply to this

### Re: Pri la Funkcia Ekvacio f(x + y) = f(x) + f(y)

Actually, if there exists a non-zero (complex-linear, bounded linear) functional $\psi$ on complex $\ell^\infty$ which is left-invariant, then we can find another one which takes the value 1 on the constant sequence/function $1$, and is positive in the sense that it takes non-negative values when evaluated against elements of $\ell^\infty$ whose coefficients are non-negative. Key words: “Hahn-Jordan decomposition of measures”.

So the insistence on having an invariant mean is not quite as restrictive as one might first think; you can always try to produce a non-zero invariant functional, and then sit back and invoke standard machinery to get an invariant mean.

Posted by: Yemon Choi on December 14, 2010 10:03 PM | Permalink | Reply to this

### Re: Pri la Funkcia Ekvacio f(x + y) = f(x) + f(y)

Coming a little late to the conversation, I couldn’t figure out where to put this reply, so I’ll just start a new top-level comment.

It is consistent with ZF that the infinite-dimensional quotient Banach space $\ell^\infty / c_0$ has a trivial (topological) dual. This and some related issues were well hashed out in this MathOverflow question. See in particular Greg Kuperberg’s answer for a good summary, and Gerald Edgar’s answer for some more interesting points about the relationship between Hahn-Banach and the Axiom of Choice.

The reason I referred to the “prospect” of such a pathological Banach space is simply that its existence is not guaranteed by rejecting AC. I have no idea, for example, what the Axiom of Determinacy implies about such a thing.

Posted by: Mark Meckes on December 10, 2010 2:03 PM | Permalink | Reply to this

### Re: Pri la Funkcia Ekvacio f(x + y) = f(x) + f(y)

Just to be absolutely clear, $c_0$ is the space of sequences which converge to $0$, with the sup norm. In other words, the quotient space I mentioned is precisely John’s second suggestion here.
Posted by: Mark Meckes on December 10, 2010 2:08 PM | Permalink | Reply to this

### Re: Pri la Funkcia Ekvacio f(x + y) = f(x) + f(y)

Cool! I’m pleased to hear that the properties of $\ell^\infty/c_0$ have been pondered in different axiomatic systems. And it’s especially intriguing to hear that a relatively consistent extension of ZF implies that $(\ell^\infty)^* = \ell^1$.

As students of analysis all know, the way $(L^p)^* = L^q$ for $1/p + 1/q = 1$ breaks down in the limiting case $p = \infty$ is an irksome annoyance, good for losing points on exams. Back in the days when I did analysis, I used to know a nice description of $(L^\infty)^*$. Yes, it’s here. But I never noticed that $(\ell^\infty)^* - \ell^1$ consisted of stuff that might not exist in ZF set theory! Is that also true for $(L^\infty(\mathbb{R}))^* - L^1(\mathbb{R}))$?

The Wikipedia article I just linked to claims that “except in rather trivial cases, the dual of $L^\infty$ is much bigger than $L^1$”. This seems rather misleading to me now, given how axiom-dependent it is! Maybe I’ll fix it.

Posted by: John Baez on December 10, 2010 2:54 PM | Permalink | Reply to this

### Re: Pri la Funkcia Ekvacio f(x + y) = f(x) + f(y)

Is that also true for $(L^\infty(\mathbb{R}))^* - L^1(\mathbb{R}))$?

I think so. I believe that the story is the same for $L^p(X,\mu)$ whenever $L^1(X,\mu)$ is infinite dimensional and separable (which can of course easily be characterized by simple properties of the measure space $(X,\mu)$, but I’m too busy/lazy to remember/reproduce those right now). The details are probably in Schechter’s book, which is cited in the MathOverflow post I referred to, and which I don’t have a copy of to check.

Posted by: Mark Meckes on December 10, 2010 3:05 PM | Permalink | Reply to this

### Re: Pri la Funkcia Ekvacio f(x + y) = f(x) + f(y)

Is there a nice reason that we always talk about $c_0$ and never about the space of convergent sequences?

Posted by: Tom Ellis on December 10, 2010 5:17 PM | Permalink | Reply to this

### Re: Pri la Funkcia Ekvacio f(x + y) = f(x) + f(y)

“Always” and “never” are overstatements, as evidenced by the fact that the space of convergent sequences also has a standard name: $c$. But $c_0$ certainly is discussed more often, partly because it shows up more in the wild (as for example here), partly because it fits more simply into the family of other classical spaces ($c_0^* \cong \ell^1$, with the duality given in the familiar way), and partly because from a Banach space geometric point of view $c$ and $c_0$ are essentially the same thing anyway ($c$ and $c_0$ are isomorphic, and $c_0$ is a 1-codimensional 1-complemented subspace of $c$).

Posted by: Mark Meckes on December 10, 2010 5:51 PM | Permalink | Reply to this

### Re: Pri la Funkcia Ekvacio f(x + y) = f(x) + f(y)

For what it’s worth, $c_0$ is a commutative C*-algebra without unit, and $c$ — I’d never heard that name, but it’s nice! — is the corresponding C*-algebra with unit.

If someone hands you a C*-algebra without unit, say $A_0$, you can give it a unit by forming

$A = A_0 \oplus \mathbb{C}$

and giving $A$ a multiplication in the more or less obvious way, where $A_0$ has the same multiplication as before, but $1 \in \mathbb{C}$ is decreed to be the multiplicative unit.

In the case at hand,

$c = c_0 \oplus \mathbb{C}$

means that every convergent sequence can be uniquely written as a sequence that converges to zero, plus a constant sequence.

Now, a commutative $C^*$-algebra with unit is always isomorphic to the algebra of all continuous functions on some compact Hausdorff space.

Puzzle: what is this space for $c$?

Puzzle: what is this space for $\ell^\infty$?

These puzzles should be familiar to people who have studied this stuff a bunch, but they might be fun for people who haven’t.

Posted by: John Baez on December 11, 2010 1:45 PM | Permalink | Reply to this

### Re: Pri la Funkcia Ekvacio f(x + y) = f(x) + f(y)

I like these puzzles, and I hadn’t seen them before.

Gur p bar vf na byq sevraq. Gur fcnpr jr arrq vf gur bar-cbvag pbzcnpgvsvpngvba K bs gur angheny ahzoref A. (Vs lbh yvxr, gung’f gur fhofrg bs E pbafvfgvat bs bar, bar unys, bar dhnegre, rgp, naq mreb.) V unccra gb unir gubhtug nobhg K n ybg: vg’f gur “havirefny pbairetrag frdhrapr”, zrnavat gung n pbagvahbhf znc sebz K gb nabgure gbcbybtvpny fcnpr L vf n frdhrapr va L gbtrgure jvgu n cbvag gb juvpu vg pbairetrf.

Gur ryy-vasgl bar gbbx zr ybatre. Ohg urer vg vf: vg’f gur Fgbar-Prpu pbzcnpgvsvpngvba M bs A. Fgbar-Prpu pbzcnpgvsvpngvba vf yrsg nqwbvag gb gur sbetrgshy shapgbe sebz (pbzcnpg Unhfqbess fcnprf) gb (gbcbybtvpny fcnprf), juvpu zber be yrff qbrf vg. V unq gb qb n yvggyr ovg bs znabrhievat gb gnxr pner bs gur snpg gung P vfa’g pbzcnpg, ohg bayl n yvggyr ovg.

Abj V frr jul lbh nfxrq gubfr gjb dhrfgvbaf gbtrgure.

Urer’f n dhrfgvba: qbrf gur P-* nytroen bs pbagvahbhf shapgvbaf ba gur Pnagbe frg unir n anzr?

Posted by: Tom Leinster on December 11, 2010 2:18 PM | Permalink | Reply to this

### Re: Pri la Funkcia Ekvacio f(x + y) = f(x) + f(y)

Tom Leinster wrote:

Great! I’m glad you got why I asked those two puzzles together.

By the way, Tom, I like the idea of encoding answers to puzzles using rot13. But encoding your new question in rot13 seems a bit much! So I’ll take the liberty of decoding it for our readers:

Here’s a question: does the C-* algebra of continuous functions on the Cantor set have a name?

I don’t know a name for it…

Posted by: John Baez on December 12, 2010 7:35 AM | Permalink | Reply to this

### Re: Pri la Funkcia Ekvacio f(x + y) = f(x) + f(y)

Weeell, I rot13-ed it because it makes it clear that neither of the puzzle answers is “the Cantor set”. (I’ll admit that I spent a while wondering whether that was the answer to the second one.) But that’s OK; there’s no harm in giving out a clue.

Posted by: Tom Leinster on December 12, 2010 7:41 AM | Permalink | Reply to this

### Re: Pri la Funkcia Ekvacio f(x + y) = f(x) + f(y)

Google finds a number of hits for the phrase “algebra of continuous functions on the Cantor set” (so it is something that people have thought about), but none that I looked at gave a more succinct name. So I think it’s probably just called the algebra of continuous functions on the Cantor set.

Posted by: Mark Meckes on December 15, 2010 1:15 AM | Permalink | Reply to this

### Re: Pri la Funkcia Ekvacio f(x + y) = f(x) + f(y)

So going back to the equation f(x+y)=f(y)+f(y), how is it obvious that f(x)=Ax is the only solution? Is there an elementary explanation for a beginner as I am?

Posted by: professorjamal on December 10, 2010 7:09 PM | Permalink | Reply to this

### Re: Pri la Funkcia Ekvacio f(x + y) = f(x) + f(y)

Prof. J, I don’t think it is obvious… but it’s true.

Here’s the argument under the relatively strong hypothesis that $f$ is continuous. So: we have a continuous function $f: \mathbf{R} \to \mathbf{R}$, satisfying $f(x + y) = f(x) + f(y)$ for all $x, y \in \mathbf{R}$. We have to prove that there exists $A \in \mathbf{R}$ such that $f(x) = A x$ for all $x \in \mathbf{R}$.

Put $A = f(1)$. By induction, whenever $n$ is a positive integer, $f(n) = n\cdot f(1) = A n$. Also $f(1) = f(0 + 1) = f(0) + f(1),$ so $f(0) = 0$. Hence for positive integers $n$, $0 = f(0) = f(n + (-n)) = f(n) + f(-n) = A n + f(-n),$ so $f(-n) = A\cdot(-n)$. So we have now shown that $f(x) = A x$ for all integers $x$ (positive, zero or negative).

Also by induction, $f(n x) = n f(x)$ for all positive integers $n$ and real numbers $x$. Hence $f(y/n) = f(y)/n$ for all positive integers $n$ and real numbers $y$ (by putting $x = y/n$ and rearranging). In particular, this is true when $y$ is an integer $m$—but we already know that for integers $m$, we have $f(m) = A m$. Hence $f(m/n) = f(m)/n = A m/n$ for all integers $m$ and positive integers $n$. In other words, $f(x) = A x$ for all rational numbers $x$.

Now let $x$ be a real number. We may choose a sequence $(x_k)$ of rational numbers converging to $x$. Since $f$ is continuous, $f(x) = \lim_{k \to \infty} f(x_k) = \lim_{k \to \infty} A x_k = A \lim_{k \to \infty} x_k = A x.$ So $f(x) = A x$ for all real $x$, as required.

That proves it under the hypothesis of continuity. If you want to see a proof under the weaker hypothesis of measurability, I can do no better than to point you to Example 3 at this Tricki page.

Posted by: Tom Leinster on December 11, 2010 3:01 AM | Permalink | Reply to this

### Re: Pri la Funkcia Ekvacio f(x + y) = f(x) + f(y)

Professor Jamal wrote:

So going back to the equation $f(x+y)=f(x)+f(y)$, how is it obvious that $f(x)= A x$ is the only solution? Is there an elementary explanation for a beginner as I am?

Tom wrote:

Prof. J, I don’t think it is obvious… but it’s true.

I know you know this perfectly well, Tom, but just so Prof. J. doesn’t get confused:

It’s actually not true that the only solution is $f(x) = A x$, unless we either assume that the function $f$ is slightly ‘nice’ (e.g. continuous, or measurable), or add some axioms to our mathematics that eliminate the possibility of ‘nasty’ solutions.

Lots of mathematicians believe in a certain principle called the axiom of choice, and this makes it easy to see that there are other solutions — at least if you know the fancy kind of math that makes this sort of thing easy! Think of $\mathbb{R}$ as a vector space over $\mathbb{Q}$, and choose a basis $c_i$ for this vector space. Then we get a solution with

$f(c_i) = a_i$

where $a_i \in \mathbb{R}$ are whatever numbers we like.

But all this is only if you use the axiom of choice. With other axioms you can show there are no other solutions besides the obvious ones, $f(x) = A x$.

So, it’s quite intriguing: an incredibly simple equation, but different people who believe in different axioms of math will have different opinions on how many solutions it has!

Posted by: John Baez on December 11, 2010 2:00 PM | Permalink | Reply to this

### Re: Pri la Funkcia Ekvacio f(x + y) = f(x) + f(y)

A random thought: can one say anything constructively about localic Banach spaces? I mean, if instead of the set of continuous linear functionals you take the locale of such, then can that locale be useful and nontrivial even in cases where it doesn’t have enough points? I.e. is the situation anything like Tychonoff’s theorem, where you need AC to make some infinite-dimensional space have “points,” but the infinite-dimensional space behaves perfectly well constructively as a space as long as you don’t demand that it have enough (or even any) points?

Posted by: Mike Shulman on December 11, 2010 6:47 AM | Permalink | Reply to this

### Re: Pri la Funkcia Ekvacio f(x + y) = f(x) + f(y)

Nice question! I wish I knew enough constructive analysis to make some progress on it just by thinking about it for a few minutes. Maybe someone out there can?

Here’s my best attempt: there’s a constructive version of the Gelfand-Naimark theorem. The ordinary version relates compact Hausdorff spaces to commutative unital C*-algebras. In the ordinary version, points in the space correspond to homomorphisms from the algebra to $\mathbb{C}$. All continuous linear functionals are limits in the weak-$*$ topology of finite linear combinations of such homomorphisms.

So what happens in the constructive version when you start with a space with no points? Do you get a commutative C*-algebra with no continuous linear functionals on it?

People like Thierry Coquand, Klaas Landsmann and Bas Spitters could help us out here…

Posted by: John Baez on December 12, 2010 7:55 AM | Permalink | Reply to this

### Re: Pri la Funkcia Ekvacio f(x + y) = f(x) + f(y)

Gajan Kristnaskon! Or if you think it’s too late to be saying that, bonan novjaron!

By chance, I just found a multidimensional generalization of Fréchet’s theorem.

Theorem  If $f: \mathbf{R}^n \to \mathbf{R}$ is a measurable function satisfying the identity $f(x + y) = f(x) + f(y)$, then $f = \langle \xi, - \rangle$ for some $\xi \in \mathbf{R}^n$.

Here $\langle - , - \rangle$ is the standard inner product.

This is (essentially) Theorem 8.19 of Gerald Folland’s book Real Analysis.

Actually, the generalization to higher dimensions is very easy. (Think about the restriction of $f$ to each coordinate axis.) But I think it’s a nice result. Also, he takes care of the $1$-dimensional case in ten lines, in English.

Posted by: Tom Leinster on December 28, 2010 3:33 AM | Permalink | Reply to this

### Re: Pri la Funkcia Ekvacio f(x + y) = f(x) + f(y)

For that matter, it’s easy to deduce from Fréchet’s Theorem that any biadditive function $f: \mathbf{R}^n \times \mathbf{R}^n \to \mathbf{R}$ is $\mathbf{R}$-bilinear.

Posted by: Tom Leinster on December 29, 2010 4:38 AM | Permalink | Reply to this

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