# Additive functional equation

Find all function $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfying $$f(x+y) = f(x) + f(y)$$ and $$f(f(x)) = x$$ for all $x, y \in \mathbb{R}$

This is one problem involving additive functional equation, but I don't know how to deal with the case $x$ is an irrational number. I appreciate all help and ideas. Thank you.

P.S: Or at least from the given solution it would be nice if you can infer one of the following statements:

1. $f(x)$ is continuous on $\mathbb{R}$

2. $f(x)$ is continuous at one point

3. $f(x)$ is monotonic on $\mathbb{R}$

4. $f(x)$ is bounded (on any interval)

• The following solution is proposed assuming $f$ is continuous. Basically the condition $f(f(x))=x$ directly follows from the functional equation. – user 170039 Oct 1 '14 at 13:52
• It need only be continuous at one point. Although it has to satisfy monotonicity and boundedness on any interval. The only other option is Hamel Functions. And if you want to go there I cannot follow. – amcalde Oct 1 '14 at 13:56
• @Amcalde Can you infer at which point is $f(x)$ continuous with the given condition, please? – primitiveroot Oct 1 '14 at 14:28

## 4 Answers

It is straightforward to show that Cauchy's functional equation implies $f(qx)=q f(x)$ for all $q\in\mathbb{Q}, x\in\mathbb{R}$. Thus we can see $f$ as a $\mathbb{Q}$-linear map of the $\mathbb{Q}$-vector space $\mathbb{R}$. Like every linear map, it is determined by its values on a basis.

Let us choose a $\mathbb{Q}$-basis $B\subset\mathbb{R}$ of $\mathbb{R}$. Note that this requires the axiom of choice. That is, for every $x\in\mathbb{R}$ we can choose a coefficient function $x^*:B\rightarrow \mathbb{Q}$ such that $q(b)\not=0$ only for finitely many $b\in B$ and

$$x=\sum_{b\in B} x^*(b) b$$

Since $f$ is a linear map, it can be represented by an (infinite) $B\times B$ matrix of rational coefficients $(F_{b,b^\prime})_{b,b^\prime\in B}$ (with only finitely many non-zero terms in every column) such that

$$f(x)= F\cdot x$$

where $\cdot$ denotes multiplication of the matrix $F$ with the $\mathbb{Q}$-vector $x$, i.e.

$$f(x)^*(b) = \sum_{b^\prime\in B} F_{b,b^\prime} x^*(b^\prime)$$

$F_{b,b^\prime}$ is simply the coefficient of $b^\prime$ in the expansion of $f(b)$.

These are all solutions to Cauchy's functional equation by itself. The condition $f(f(x))=x$ now reads

$$F^2=I$$

with $I$ being the identity matrix. That is,

$$\sum_{b^{\prime\prime}\in B} F_{b,b^{\prime\prime}} F_{b^{\prime\prime},b^\prime}=\left\{\begin{array}{ll}1 & \text{if}\;b=b^\prime,\\ 0 & \text{if}\;b\not=b^\prime.\end{array}\right.$$

This characterizes all the solutions to the simultanous functional equations. The two solutions corresponding to the continuous solutions are just the cases $F=\pm I$. None of the other solutions satisfy any of your conditions $1.$ through $4.$ (since they all imply $f(x)=\pm x$).

• to be precise, you can have infinitely many nonzero entries in rows (but not in columns). I think that given a solution, you can pick a basis in which its matrix consists of blocks of $1$, $-1$, and $0 1\\ 1 0$ along the diagonal – mercio Oct 1 '14 at 15:43
• @mercio: Thank you, corrected. – J.R. Oct 1 '14 at 15:45
• Thank you very much for your answer. I am a high school student, therefore I don't understand about the vector space or matrixes. I was just guessing the answer $f(x) = \pm x$ and offer those conditions. I didn't know that there were another solutions. – primitiveroot Oct 1 '14 at 15:46
1. $f(0)=0$

2. $y=-x \implies f(-x)=-f(x)$

3. $y=x \implies f(2x)=2f(x)\implies f(3x)=3f(x) \cdots\implies f(nx)=nf(x)$ $\forall n \in \mathbb{N}$

4. $x\in \mathbb {Q} \implies \exists m,n\in \mathbb{N} \mid x=\dfrac{m}{n}\implies n$ $.$ $x=m$ $.$ $1$ $\implies f(nx)=\dfrac{m}{n}f(1)$

$\therefore$ $f(x)=f(1)x$ $\forall x \in \mathbb{Q}$

$\exists\{x_n\}\mid x_n \in \mathbb{Q}$ $\forall n\in \mathbb{N},\displaystyle \lim_{n \to \infty}x_n=x$ $\forall x\in \mathbb{R}$

$\therefore$ $f(x)=\displaystyle \lim_{x_n \to x}f(x_n)=\displaystyle \lim_{x_n \to x}f(1)x_n=f(1)x$ $\forall x\in \mathbb{R}$

• Do you think that $f(x)$ needs to be continuous to conclude $f(x) = \displaystyle \lim_{x_n \to x}f(x_n)$? – primitiveroot Oct 1 '14 at 14:18
• Not necessary but it has to be either monotonically increasing or bounded to avoid us from getting $f$ as a Hamel Function. – user 170039 Oct 2 '14 at 5:07
• Its an involution! of course not; unlike even the field automorphism equations just given strict monotonicity (or even monotonicity) you can literally infer their values off the table. Just having $F:[0,1] to [0,1]$ – William Balthes May 5 '17 at 15:45
• f(f(x))=x and F is strictly monotonic increasing specifies (i have been informed that can be reduced to monotonicity). Just think about why that is; the point reflection symmetry induced by the involution equation will mean that$F(x) \neq x$ at any point then there will exist some other value F(p)=x, where F(x)=p, and monotonicity will be violated in hit – William Balthes May 5 '17 at 15:47

It would appear that all reals values could be solved for, directly as F(x)=x and it display 1 point homogeineity over the reals (and thus 0 -point given the max value being defined).

f(x+y)=f(x)+f(y) and f(f(x))=x; Then within the context of an appropriate (QM like) entangled probabilistic system where, here satisfy the dictates of a finitely additive probability space within each system (the axioms of probability

1. F:[0,1]\to[0,1]

2.strictly monotonically increasing (and thus injective)

1. F(1)=1 for example,that is,the maximum element of the domain corresponds to the maximum element of the range, both F(max x)=F(1)=1,

2. Then so long as it holds that for all x in domain [0,1] 0<=F(x)<=1;w and 0

so long as 5. The system would presumably exhibit point reflection symmetry at all point F(x)=F-1(x) as a result of the derived F(F(x))=x (who knows how; ie for all reals).

.

ie This is handing continuity to you, essentially, on a plate, as opposed to something that would need to imposed in addition, Given a ranked entangled probabilistic structure, one could capture any irrational value simply because if F(x)=y>x, x and y being irrational then would exist F-1(x)=y>x and thus some y>x such that F(y)=x where by being strictly increasing F(y)>F(x), yet F(x)=y, F(y)=x, and thus F(y)

Having the outcomes just ranked (qualitatively-ordinally) through, >, < and =), bigger, smaller or = numerical probability value in representation, given the probability calculus would do the trick, but given the above, these would no longer be necessarily as they would be required to help one established the aforementioned conditions.

That is a strict linear order in the vertical, the inequalities would be violated in the first move as a lower irrational chance would be ascribed a higher irrational function value then some some higher irrational chance in that interval, violating the mere ordinal inequalites at the first step. So by ranked, I mean by bigger chance, iebigger x>y, bigger Frame function, credence F(x)>F(y), equal chance, x= y equal chance F(x)=F(y) and likewise for less than

At worst 6 might need to hold

6..Perhaps some form of weird, archimeadean condition or vertical countable (functional) additivity, ie would not to literally use splitting or any other such thing, to stop accelerating standard sequences in the entangled lexicographic order.

So this would be distinct from standard countable additivity (as its functional). This is as only finite dis-junctions of 'no more then three outcomes' are formed at any point (and only within individual spaces) which are used to insert the values (not outcomes into others spaces) through the logical connections or ordering betwixt spaces.

This would appear to look like one is pasting together spaces, when one is adding the values together, but of course one is adding the values together, by using the entangled relations to insert values (not inserting or splitting up the outcomes within any probability space in the system) from others spaces into, not normally solvable, pre-existent equi-probable outcomes in other spaces; due to the logical relations and the total (non atomic) order between and within all the spaces in the system.

It would just be a consistency condition for perhaps the weird (perhaps gleason like?not sure here/Luce condition (L)) like standard sequences one would form betwixt spaces, to demonstrate linearity, and the ability to capture all rational values (although given point reflection symmetry, the absolute uniqueness, linearity, and continuity of the representing probability measure one could essentially amount to a one line proof, as this could be used for the rationals as well, without the zig zagging standard sequences).

Although of course demonstrating or proving that (point reflection symmetry holds, or f(f(x)=x obtains for all real values is another matter entirely)) that would not be a one like proof!)

Ie

Perhaps Some kind of continuity proof technique might be used as a proof technique to capture all of the irrationals ( i am not sure if induction works here) but no continuity condition, as an further imposed rationality of functional condition would need to be further derived or assumed. In some sense it would falls out. The rationals by induction and the irrationals in perhaps a similar methods, simply (or rather it could apply to both), for unless F(x)=x, an inequality that is supposed to hold in the system would be violated at the first move; if these conditions under suitable constraint entails point reflection symmetry

As one could literally show for any arbitrary real probability value could capture it; that is them read (at least individually) off the table, in a finitary manner; one at a time, as it were. Even perhaps the transcendentals, if there are such entities with regards to probability values of a 3 outcome system that some connection to additivity and normalization.

Of course, one could never capture them all, applying this, 'one at a time' methods but thats just a limitation of the inability the count that which is infinite insofar that it always exists another real value (but that is somewhat like the rational numbers)) .

Just so long as the domain is [0,1] (a closed and bounded interval) one could just read it off, the table that for all rational values F(x)=x, and to likewise (i think) with a little bit more difficulty ,although with more ease then with the field equations F(xy)=F(x)F(y)).

Although I am not sure how (the multiplicative field equations) could be derived over the algebraic irrationals to begin with; and on the other hand, how they annot see how they completely resolve any arbitrary transcendental irrational number; unlike point reflection symmetry. And thus entail continuity for all extents and purposes. This being unlike point reflection symmetry, which if it could be derived over all reals, as a consequence of the the entangled ranking, probabilism, cauchy's equation, F(F(x)=x (but ultimately, F(F(x)=x and injectivity for the most part, is at the heart of it )

As (point reflection symmetry) presumably would also cover the transcendental numbers as well for the same reasons, for which some slightly stronger condition (closer to continuity would required one would think if only has the field equations and cauchys equations instead of FF(x)=x over all reals) ;

As one could literally read off, the diagram, as it were, or the equations, the values, and that for every element c in the range, [0,1] there exist one and only one c1 in the domain [0,1] such F(c1)=c, where c1=c. I don't see how it would so easily resolve the non algebraic irrationals, (i presume there are such values in [0,1] that could meet the constrains and axioms of a probabilistic kolmogorovian or kolmogorovian axiom like system in a three or more, outcome space?)

The continuity constraint used, would have already (i think) essentially derived given the above constraints and thus it would now be used (justifiably) as merely a proof technique, to capture them all real values collectively (not as an extra condition imposed, as it would already have been derived in some sense I think.) Other than the above, which could be presumably be derived, given the appropriate entangled probabilistc structure, if one could establish point reflection symmetry that is,over the entire system (not, all reals, not sure about how that would hold)

That is, the aforementioned conditions would presumably just entail,( in many cases), F(m)=m directly. For example where m is some irrational value in [0,1] that satisfies the desideratum of the probability calculus F(m)=m could be read off the table.And thus F(x)=x

T

This gives you everything in one step, in the right structure, even for two outcomes; although of course you could probably neither get the field equations, or cauchy's equation, to begin with;Let alone F(F(x)=x) (or given point reflection symmetry, which follows some additional things' at which point 'the unique-ness result would have been delivered unto your hands' to begin with).

ie, by this I mean cauchy equation even for the rational numbers would at best be compatible with the structure, and would not fall out uniquely and for other representation which are still continuous but not cauchy they could be probabilistic, and satsify all the rankings and the order, the symmetry and the three un-interesting solvable numerical values.

Not even for the factors of two, only just F(1)=1, F(0)=0 F(1/2)=1/2 which are generally trivially solvable, and sometime perhaps, quasi linear symmetry conditions, which might be able induce due to doubly cross- equimatched pairs which open the door (given some additional postualte) .

That is, if if one is committed to not something like one of the (or which I consider to be) blacklisted constraints (in a two outcome system) in 1, or 2 or 3 below. (2 or a much weakened version of one of the inequality conditions in 3 would be the best i think, but still not really easily justifiable).

,

1. The use splitting,lotteries, independence, inserting a three dimensional outcome space, singular or entire into the system,.

2. Using signed charges which invert the chances, or ordering if one puts one step out of line (to induce a form of point-reflection symmetry, somewhat of a cheating strategy). THis is better

3.or breaking the rules; t e adding up events on distinct spaces, or imposing often un-jusitifiable conditions to begin with (such as linearity or sub-linearity, or (functional) sub-additivity; to get out or even bounds onany number other than the trivial F(0)=0, F(1) and F(0.5)=0.5 resolvable=s

Otherwise the the rest of the outcomes, are some some bigger then the outcome below in the ranking, the next outcome somewhat bigger again etc(but who knows by how much, unless one puts a difference metric on it; as the sums and difference wont be ordered,even if the numerical values respect the rank. Symmetry helps here a bit, but not too much by itself); and likewise for the less than likely events.

So that all that one could say often is they are smaller than 0.5 and bigger then 0, and smaller or bigger then next outcome in the ordering, and bigger then 0.5 have F(x)>x and F(x) in [0,1] and even given uniform continuity, and certain symmetry conditions, with the function being bijective and strictly monotonically increasing from [0,1] to [0,1] ,I doubt much would come out (it might restrict the possible functions to F(x) and any number of (monotonically strictly increasing? logarithmic-like, symmetric probability functions that still satisfy the rank and have F(0.5)=0.5, F(1)=1 F(0), the rank, additivity, the partially ordered differences and sums due to inverse symmetry say, and the symmetry conditions). ( not standard normalization additivity within spaces) to begin with, which often begs the question from the start.

f(f(x))=x so long as one get F-1(x)=F(x) from F(F(x)=x, if this holds for all reals (as F(F(x)=x is claimed). Then I presume that ifthe function is injective (or bounded by and strictly monotonically increasing); and then one see the answer for the irrations in front of you, I think (at least numerically, not just how one functionally deriveds F(x)=+-x, or continuity from it, but I think one could do it)