On the state of the second part of Hilbert’s fifth problem

Abstract

In the second part of his fifth problem Hilbert asks for functional equations “In how far are the assertions which we can make in the case of differentiable functions true under proper modifications without this assumption.” In the case of the general functional equation

$$\begin{aligned} f(x)=h\Bigl (x,y,\bigl (g_1(x,y)\bigr ),\ldots ,\bigl (g_n(x,y)\bigr )\Bigr ) \end{aligned}$$

for the unknown function f under natural condition for the given functions it is proved on compact manifolds that $f\in C^{-1}$ implies $f\in C^{\infty }$ and practically the general case can also be treated. The natural conditions imply that the dimension of x cannot be larger than the dimension of y. If we remove this condition, then we have to add another condition. In this survey paper a new problem for this second case is formulated and results are summarised for both cases.

1 Introduction

In his celebrated address to the 1900 International Congress of Mathematicians, in his fifth problem Hilbert [9] asked (in the language of present day mathematics) whether it is true, that every locally Euclidean group is a Lie group? If $\varphi$ is a homeomorphism mapping ${{{\mathbb {R}}}^n}$ onto a neighbourhood of the unit element e of the group G, such that $\varphi (0)=e$, then we can copy the group operation locally to ${{{\mathbb {R}}}^n}$:

$$\begin{aligned} (x,y)\mapsto \psi (x,y)=\varphi ^{-1}\bigl (\varphi (x)\cdot \varphi (y)\bigr ). \end{aligned}$$

The group axioms give a system of functional equations to $\psi$, for example associativity becomes

$$\begin{aligned} \psi \bigl (x,\psi (y,z)\bigr )=\psi \bigl (\psi (x,y),z\bigr ). \end{aligned}$$

In this way, roughly speaking, the problem asks whether we can replace $\varphi$ with an equivalent homeomorphism so that the corresponding $\psi$ is analytic and satisfies the same system of functional equations. In the second part of his fifth problem Hilbert goes on as follows:

“Moreover, we are thus led to the wide and interesting field of functional equations which have been heretofore investigated usually only under the assumption of the differentiability of the functions involved. In particular the functional equations treated by Abel (Oeuvres, vol. 1, pp. 1, 61, 389) with so much ingenuity $\ldots$ and other equations occurring in the literature of mathematics, do not directly involve anything which necessitates the requirements of the differentiability of the accompanying functions. $\ldots$ In all these cases, then, the problem arises: In how far are the assertions which we can make in the case of differentiable functions true under proper modifications without this assumption?” (Hilbert’s emphasis).

After this, Hilbert quotes a result of Minkowsky, that under certain conditions, the solutions of the functional inequality

$$\begin{aligned} f(x+y)\le f(x)+f(y)\qquad x,y\in {{{\mathbb {R}}}^n}\end{aligned}$$

are partially differentiable, and remarks, that certain functional equations, for example the system of functional equations

$$\begin{aligned} \begin{aligned} f(x+\alpha )-f(x)&=g(x),\\ f(x+\beta )-f(x)&=0, \end{aligned} \end{aligned}$$

where $\alpha ,\beta$ are given real numbers, may have solutions f which are continuous but not differentiable, even if the given function g is analytic. In general, the situation is similar, if there is no “free variable” in the equations.

Regularity for functional equations in a wider sense of course, is also very important. Problems XIX., XX. of Hilbert consider similar questions for calculus of variation and elliptic partial differential equations, respectively. There is a connection with the regularity of functional equations in both directions. On the one side the analyticity of solutions of functional equations usually comes from the regularity of solutions of differential equations. On the other side, tools in the regularity theory of partial differential equations such as Campanato spaces and Morrey’s lemma, were used to prove regularity results for functional equations.

2 General strategy

A general strategy to solve some kind of functional equations is the following:

• (A) “Weak” regularity implies “strong” regularity;

• (B) Get a differential equation (integral equation, etc.);

• (C) Solve this equation.

There are large classes of functional equations for which this strategy does not work, namely, equations without “free variables”: the number of the real variables is equal to the number of places in the unknown function(s). These equations are also called equations with one (maybe vector) variable: See Hilbert’s counterexample above. These are not considered here.

For functional equations with “free variables” step (A) may be applicable. We say for such equations that they have regularity property. The “weak” regularity assumption may be continuity (to be in ${\mathcal {C}}$), Lebesque measurability or Baire measurability (having Baire property). For simplicity, the union of these two classes will be denoted by ${\mathcal {C}}^{-1}$. The most important “strong” regularity property is to be in ${{\mathcal {C}}^\infty }$, which implies that steps (B) and (C) can be applied.

The class of functional equations with “free” variables can be divided into two classes: non-iterated (or non-composite) and iterated (or composite) equations. Equations where the unknown functions are substituted into each other belong to this second class. A simple example is the Aczél–Benz equation:

$$\begin{aligned} f\bigl (x+f(y)\bigr )=f(x)+f\bigl (x+y-f(x)\bigr ). \end{aligned}$$

The continuous real solutions were found by Z. Daróczy in 1980. The function $x\mapsto \frac{1}{2}\bigl (x\pm |x|\bigr )$ is a solution, which is not differentiable. There is an excellent survey paper by Páles [42] (see also [7]) about the regularity of composite equations, containing also open problems. Therefore we shall not treat this type here.

Roughly speaking the general strategy works for non-composite equations with “free variables”. Nevertheless, we need some care. In the case of the “most general” equation without iteration

$$\begin{aligned} H\Bigl (x,y,f\bigl (G(x,y)\bigr ),f_0\bigl (G_0(x,y)\bigr ), \ldots ,f_n\bigl (G_n(x,y)\bigr )\Bigl )=0 \end{aligned}$$

where the f’s are the unknown functions, there is no regularity phenomenon. We have to formulate precisely under what conditions we expect regularity results.

In the survey paper [3] J. Aczél collected the functional equations investigated by Abel (mentioned by Hilbert, see the citation above). Among these, one has no “free variable”. Another one is a composite equation of associativity–commutativity. A third, composite equation was solved under continuity by Sablik [43,44,45]. All other equations are non-composite, and that ${\mathcal {C}}^{-1}\implies {{\mathcal {C}}^\infty }$ can be proved using the results below.

Usually the following steps are used:

• I. Lebesgue measurability implies continuity;

• II. Baire measurability implies continuity;

• III. Continuity implies local Lipschitz property hence differentiability almost everywhere;

• IV. Differentiability almost everywhere implies ${{\mathcal {C}}^1}$;

• V. ${\mathcal {C}}^p$ implies ${\mathcal {C}}^{p+1}$ for $p=1,2\ldots$.

3 The main problem with “enough free variables”

The first general method for non-composite linear or quasilinear euqations with free variables is given in the book [1] of Aczél. The method comes from papers by Andrade, 1900 and Kac, 1937. This makes it possible to prove that continuous real function solutions are in ${{\mathcal {C}}^\infty }$. This is generalised in five senses:

• 1. Starting from ${\mathcal {C}}^{-1}$, i.e., from Lebesgue measurability or from Baire measurability;

• 2. The unknown function(s) has several (real) variables;

• 3. The unknown function(s) is vector-valued;

• 4. Strongly nonlinear equations;

• 5. The equation is satisfied only almost everywhere.

The following problem was first formulated in my paper [11] and also included in [2] between the most important open problems on functional equations in a somewhat different form.

Problem 1

Let X, Y, and Z be open subsets of ${{\mathbb {R}}}^r$, ${{\mathbb {R}}}^s$, and ${{\mathbb {R}}}^t$, respectively, and let D be an open subset of $X\times Y$ and let W be an open subset of $D\times Z^n$. Let $f:X\rightarrow Z$, $g_i:D\rightarrow X$ $(i=1,2,\ldots ,n)$, and $h:W\rightarrow Z$ be functions. Suppose that

• (FE) if $(x,y)\in D$ then

  $$\begin{aligned} \Bigl (x,y,f\bigl (g_1(x,y)\bigr ),\ldots ,f\bigl (g_n(x,y)\bigl )\Bigr )\in W \end{aligned}$$

  and

  $$\begin{aligned} f(x)=h\Bigl (x,y,f\bigl (g_1(x,y)\bigl ),\ldots ,f\bigl (g_n(x,y)\bigr )\Bigr ); \end{aligned}$$

• (S) h and $g_i$ $(i=1,2,\ldots ,n)$ are in ${{\mathcal {C}}^\infty }$;

• (RC) for each $x\in X$ there exists a y for which $(x,y)\in D$ and $\frac{\displaystyle \partial g_i}{\displaystyle \partial y}(x,y)$ has rank $r={{\,\textrm{dim}\,}}(X)$ $(i=1,2,\ldots ,n)$.

Is it true that every f, which is in ${\mathcal {C}}^{-1}$ (i.e., measurable or has the Baire property) is in ${{\mathcal {C}}^\infty }$?

Terminology is the same as in [5, 40, 41]. Notice the “rank condition” (RC) that ${{\,\textrm{rank}\,}}\frac{\displaystyle \partial g_i}{\displaystyle \partial y}=r$, which implies ${{\,\textrm{dim}\,}}(Y)=s\ge r$, i.e. that we have “enough free variables”.

The transfer principle. Seemingly, the equation in the above problem is more special than the “most general equation”. Not only is it explicit (which is necessary), but there is only one unknown function. Why? If we have a general non-composite functional equation with several variables and several unknown functions, and we can express each unknown function from it, then (after writing different variables in each equation) we may consider a vector-valued function having the different unknown functions as coordinates and use results concerning equations with only one, but vector-valued unknown function. This shows how important it is to discuss vector-valued unknown functions with vector variables. This “transfer principle” — which is not so simple as it seems to be — is treated in detail in the book [28].

The above problem is well formed. None of the conditions of the problem above can be omitted without introducing new conditions; see [28] for examples. This supports our calling this problem “the main regularity problem of non-composite functional equations with several variables”.

4 The main results with “enough free variables”

The above Problem 1, as it stands, is not completely solved. The problem corresponding to steps I., II., IV. and V. are completely solved, even in much larger generality, see [28] and the original papers [8, 10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27, 29,30,31,32,33,34,35, 37]. The problem can be reformulated to manifolds. This is solved affirmately in the case the domain of f is a compact ${{\mathcal {C}}^\infty }$ manifold [24]. For simplicity, we shall not consider here equations on manifolds. Moreover, in general, we shall not consider results treated in the survey paper [35], and restrict ourselves only to results about the above Problem 1. There are affirmative partial answers, as follows:

Partial solution 1

If in Problem 1 equation (FE) is quasilinear:

$$\begin{aligned} f(x)=\sum _{i=1}^n h_i\Bigl (x,y,f\bigl (g_i(x,y)\bigr )\Bigr ), \end{aligned}$$

where the functions $g_i$ and $h_i:D\times Z\rightarrow {{\mathbb {R}}}^t$ are in ${{\mathcal {C}}^\infty }$, then every solution $f\in {\mathcal {C}}^{-1}$ is in ${{\mathcal {C}}^\infty }$.

Proof

This follows from Theorem 1.30 in [28]. $\square$

Partial solution 2

If in Problem 1 in equation (FE) we have $n=2$, $g_1(x,y)\equiv y$ and f is real valued, then every ${\mathcal {C}}^{-1}$ solution f is in ${{\mathcal {C}}^\infty }$.

Proof

This follows from Theorem 1.29 in [28]. $\square$

Partial solution 3

If in Problem 1f has locally bounded variation, then f is in ${{\mathcal {C}}^\infty }$.

Proof

This follows from Theorem 1.27 in [28]. $\square$

Partial solution 4

If in Problem 1 besides other conditions, there exists a compact subset C of X such that for each $x\in X$ there exists a $y\in Y$ satisfying $g_i(x,y)\in C$, then every ${\mathcal {C}}^{-1}$ solution f is in ${{\mathcal {C}}^\infty }$.

Proof

This follows from Theorem 1.28 in [28]. $\square$

Note that the additional compactness condition in this Partial solution 4 is retained by the transfer principle. Remark that the compactness condition is not so restrictive, and can usually be overcome by a simple trick considering a sequence of increasing open sets with union ${{\,\textrm{dmn}\,}}f$. This was used in [32]. Sometimes we need more refined tricks; see the paper [37]. I do not know any regularity problem which would be answered by Problem 1, but not solved using the partial solutions. So may be say that Problem 1 is practically solved. The Partial solutions 1–4 and some further results (see [28]) make it hard to imagine a counterexample, so Problem 1 may also be called a conjecture. Nevertheless, refinements of the problem, especially to do step III under as weak conditions as possible are very important because of composite equations, see [41].

Today in some cases even with a computer we can prove regularity: such programs were developed by Sándor Czirbusz [4].

It is possible to add a last step to proving regularity:

• VI. ${{\mathcal {C}}^\infty }$ implies analiticity.

This step does not seem so important, and only some initial steps are done in this direction; see [36]. Functional equations satisfied only almost everywhere are very important because of applications in probability theory and other areas: see [31] and the references there.

5 Problem and results with “few free variables”

Let us consider the Sincov equation

$$\begin{aligned} f(x_1,x_2)=f(x_1,y)+f(y,x_2),\qquad x_1,x_2,y\in {{\mathbb {R}}}\end{aligned}$$

with the unknown function $f:{{\mathbb {R}}}^2\rightarrow {{\mathbb {R}}}$. This equation plays a role in the introduction of the thermodynamical temperature scale. As it is easy to see, the general solution is $f(x_1,x_2)=g(x_1)-g(x_2)$, where $g:{{\mathbb {R}}}\rightarrow {{\mathbb {R}}}$ is an arbitrary function, hence there is no regularity phenomenon in this case. This equation shows that if the number of “free variables” is less than the number of places in the unknown function then there is no regularity phenomenon in general.

On the other side, general regularity results which overcome the difficulty that the “rank condition” is not satisfied were given by Światak [46] (see also [39]). She applied the distribution method — probably first suggested by Fenyő [6] — to the equation

$$\begin{aligned} \sum _{i=1}^n h_i(x,y)f\bigl (g_i(x,y)\bigr )=h_0(x,y), \quad x\in {{{\mathbb {R}}}^r},\,y\in Y\subset {{{\mathbb {R}}}^s}\end{aligned}$$

with unknown function f. The essence of her method is to differentiate the distribution solution. If we obtain a hypoelliptic partial differential equation for f then the solutions are in ${{\mathcal {C}}^\infty }$ and hence it is proved that ${\mathcal {C}}$ implies ${{\mathcal {C}}^\infty }$.

This motivated the search for general methods along the line of steps I.–V. The results were published in three papers [19, 22, 25], altogether almost 80 pages. They are using new function spaces, which — roughly speaking — interpolate between ${\mathcal {C}}^{-1}$ and ${\mathcal {C}}^0$ and between ${\mathcal {C}}^p$ and ${\mathcal {C}}^{p+1}$, $p=0,1,\ldots$. Although these general results solve several problems, they are definitely not easy to use. This was the motivation for [30] in which three Corollaries are given which are less general but easy to use. The Corollaries led me to formulate the following open problem first published here:

Problem 2

Let $X\subset {{\mathbb {R}}}^r$ be an open set and $f:X\rightarrow {{{\mathbb {R}}}^m}$ a function. Suppose that

• (ES) we have

  $$\begin{aligned} \Bigl (x,y,f\bigl (g_{i,1}(x,y)\bigr ), \ldots ,f\bigl (g_{i,n_i}(x,y)\bigr )\Bigr )\in W_i \end{aligned}$$

  and the functional equation

  $$\begin{aligned} f(x)=h_i\Bigl (x,y,f\bigl (g_{i,1}(x,y)\bigr ), \ldots ,f\bigl (g_{i,n_i}(x,y)\bigr )\Bigr ) \end{aligned}$$

  is satisfied, whenever $i\in I$, $(x,y)\in D_i$ (here I is an index set), moreover

• (S) $D_i\subset X\times Y_i$ is an open set, $Y_i$ is a Euclidean space, $W_i$ is an open subset of $D_i\times ({{\mathbb {R}}}^m)^{n_i}$, all the functions $h_i:W_i\rightarrow {{{\mathbb {R}}}^m}$ and $g_{i,j}:D_i\rightarrow X$ are in ${{\mathcal {C}}^\infty }$, moreover

• (D) for each $x\in X$ and for each proper linear subspace V of ${{\mathbb {R}}}^r$ there exists an $i\in I$ and a y such that $(x,y)\in D_i$ and

  $$\begin{aligned} {{\,\textrm{dim}\,}}\left( \frac{\displaystyle \partial g_{i,j}}{\displaystyle \partial x}(x,y)(V)+{{\,\textrm{rng}\,}}\frac{\displaystyle \partial g_{i,j}}{\displaystyle \partial y}(x,y)\right) >{{\,\textrm{dim}\,}}(V) \end{aligned}$$

  whenever $1\le j\le n_i$, where ${{\,\textrm{rng}\,}}$ is the range of the partial differential as a linear operator.

Is it true that $f\in {\mathcal {C}}^{-1}$ implies $f\in {{\mathcal {C}}^\infty }$?

Remark that equations in the equation system (FES) need not be independent. To the contrary, they may be obtained from the same equation expressing different terms from it and introducing new variables: see the example below. The smoothness condition (S) is natural enough. Let us consider the dimension condition (D). Observe, that if ${{\,\textrm{dim}\,}}(Y_i)>0$, then the dimension condition (D) is satisfied “in general”, because “in general”

$$\begin{aligned} \begin{aligned} \det \left( \frac{\displaystyle \partial g_{i,j}}{\displaystyle \partial x}(x,y)\right)&\not =0,\\ {{\,\textrm{dim}\,}}\left( \frac{\displaystyle \partial g_{i,j}}{\displaystyle \partial x}(x,y)(V)\right)&={{\,\textrm{dim}\,}}(V),\\ {{\,\textrm{rank}\,}}\left( \frac{\displaystyle \partial g_{i,j}}{\displaystyle \partial y}(x,y)\right)&=\min \bigl \{r,{{\,\textrm{dim}\,}}(Y_i)\bigr \}>0. \end{aligned} \end{aligned}$$

Hence it is natural to expect that (D) is satisfied “in general”.

Critical subspaces. Let $X\subset {{\mathbb {R}}}^r$ be an open set and for each $i\in I$ let $D_i\subset X\times Y_i$ be an open set, where $Y_i$ is a Euclidean space and let the functions $g_{i,j}:D_i\rightarrow X$, $1\le j\le n_i$ be in ${\mathcal {C}}^1$. Suppose, that for each $x\in X$ there is $i\in I$ and y such that $(x,y)\in D$. For a proper linear subspace V of ${{\mathbb {R}}}^r$ we will say that it is a critical subspace at x if for each $i\in I$ and for each y for which $(x,y)\in D_i$ we have

$$\begin{aligned} {{\,\textrm{dim}\,}}\left( \frac{\displaystyle \partial g_{i,j}}{\displaystyle \partial x}(x,y)(V)+{{\,\textrm{rng}\,}}\frac{\displaystyle \partial g_{i,j}}{\displaystyle \partial y}(x,y)\right) \le {{\,\textrm{dim}\,}}(V) \end{aligned}$$

for some $1\le j\le n_i$.

It is clear, that the dimension condition (D) can be formulated in a way that there is no critical subspace for any $x\in X$. It is also clear that if a linear subspace V is critical then any proper linear subspace of ${{{\mathbb {R}}}^r}$ containing V is critical too. Hence it is enough to consider minimal critical linear subspaces. We have to investigate whether they exist or not.

Let us see the known partial solutions to Problem 2.

Partial solution 5

In Problem 2, if $f\in {\mathcal {C}}^1$, then $f\in {{\mathcal {C}}^\infty }$.

Proof

This follows from Corollary 2 in [30]. Remark, that there is a typo in Corollary 2, condition (SI): instead of $\partial _t^{\alpha _0}$ should have been $\partial _x^{\alpha _0}$. $\square$

Partial solution 6

If in Problem 2 instead of (ES) and (S) we have that

• (LS) the linear functional equations

  $$\begin{aligned} f(x)=h_{i,0}(x,y)+\sum _{j=1}^{n_i} h_{i,j}(x,y)f\bigl (g_{i,j}(x,y)\bigr ) \end{aligned}$$

  are satisfied, whenever $i\in I$, $(x,y)\in D_i$ (here I is an index set), moreover

• (S) $D_i\subset X\times Y_i$ is an open set, $Y_i$ is a Euclidean space, the functions $h_{i,0}:D_i\rightarrow {{{\mathbb {R}}}^m}$, $h_{i,j}:D_i\rightarrow {{\mathbb {R}}}$ and $g_{i,j}:D_i\rightarrow X$ are in ${{\mathcal {C}}^\infty }$,

then $f\in {\mathcal {C}}^0$ implies $f\in {{\mathcal {C}}^\infty }$.

Proof

This follows from Corollary 1 in [30]. $\square$

Partial solution 7

If $K\subset \{0,1,\ldots ,r\}$ contains 0 and r and in Problem 2 instead of the dimension condition (D) the somewhat stronger constant dimension condition

• (CD) for each $x_0\in X$ and for each proper linear subspace $V_0$ of ${{\mathbb {R}}}^r$ with $k_0={{\,\textrm{dim}\,}}(V_0)\in K$ there exist an $i\in I$ and a $y_0$ such that $(x_0,y_0)\in D_i$ and

  $$\begin{aligned} {{\,\textrm{dim}\,}}\left( \frac{\displaystyle \partial g_{i,j}}{\displaystyle \partial x}(x,y)(V)+{{\,\textrm{rng}\,}}\frac{\displaystyle \partial g_{i,j}}{\displaystyle \partial y}(x,y)\right) \end{aligned}$$

  is the same constant $k\in K$, $k>k_0$ for $1\le j\le n_i$ whenever x is close enough to $x_0$, y is close enough to $y_0$ and V is close enough to $V_0$ in the Grassmann manifold ${\mathcal {G}}(r,k)$

is satisfied, then $f\in {\mathcal {C}}^{-1}$ implies $f\in {\mathcal {C}}^0$.

Here again if ${{\,\textrm{dim}\,}}(Y_i)>0$, then the “constant dimension” condition (CD) is satisfied “in general” but not if there is a critical subspace for some $x\in X$.

Proof

This follows from Corollary 3 in [30]. $\square$

Example

Let us consider the functional equation

$$\begin{aligned} f(x,y)=\frac{f(x+t,y)+f(x,y+t^2)}{2},\quad x,y,t\in {{\mathbb {R}}}. \end{aligned}$$

For this equation a proper linear subspace V is critical if and only if

$$\begin{aligned} {{\,\textrm{dim}\,}}\Bigl (V+\bigl \{(u,0):u\in {{\mathbb {R}}}\bigr \}\Bigr )\le 1, \end{aligned}$$

i. e. $V=V_1=\bigl \{(u,0):u\in {{\mathbb {R}}}\bigr \}$, or

$$\begin{aligned} {{\,\textrm{dim}\,}}\Bigl (V+\bigl \{(0,2tu):u\in {{\mathbb {R}}}\bigr \}\Bigr )\le 1 \end{aligned}$$

for all $t\in {{\mathbb {R}}}$, i. e. $V=V_2=\bigl \{(0,u):u\in {{\mathbb {R}}}\bigr \}$.

Substituting $x+t$ with x, for the new equation

$$\begin{aligned} f(x,y)=2f(x-t,y)-f(x-t,y+t^2),\quad x,y,t\in {{\mathbb {R}}}\end{aligned}$$

a proper linear subspace V is critical if and only if

$$\begin{aligned} {{\,\textrm{dim}\,}}\Bigl (V+\bigl \{(-u,2tu):u\in {{\mathbb {R}}}\bigr \}\Bigr )\le 1 \end{aligned}$$

for all $t\in {{\mathbb {R}}}$ or

$$\begin{aligned} {{\,\textrm{dim}\,}}\Bigl (V+\bigl \{(-u,0):u\in {{\mathbb {R}}}\bigr \}\Bigr )\le 1. \end{aligned}$$

Hence for these two equations only $V_1$ remains critical.

Substituting $x+t^2$ with x, for the new equation

$$\begin{aligned} f(x,y)=2f(x,y-t^2)-f(x+t,y-t^2),\quad x,y,t\in {{\mathbb {R}}}\end{aligned}$$

a proper linear subspace V is critical if and only if

$$\begin{aligned} {{\,\textrm{dim}\,}}\Bigl (V+\bigl \{(u,-2tu):u\in {{\mathbb {R}}}\bigr \}\Bigr )\le 1 \end{aligned}$$

or

$$\begin{aligned} {{\,\textrm{dim}\,}}\Bigl (V+\bigl \{(0,-2tu):u\in {{\mathbb {R}}}\bigr \}\Bigr )\le 1 \end{aligned}$$

for all $t\in {{\mathbb {R}}}$. Hence for all three equations no critical subspace remains.

Sincov equation. For the Sincov equation

$$\begin{aligned} f(x_1,x_2)=f(x_1,y)+f(y,x_2),\qquad x_1,x_2,y\in {{\mathbb {R}}}\end{aligned}$$

we find the critical subspaces $V_1$ and $V_2$ as above, but in this case we cannot remove these with substitutions. Let us observe the connection of the critical subspaces with the general solution $f(x_1,x_2)=g(x_1)-g(x_2)$.

Final remarks. There is a simple method to reduce the regularity problem of an equation with “few free variables” to the regularity problem of a functional equations with “enough free variables”. This consists in expressing some terms from the original equation and after some substitutions writing back these new equations into the original one. The author suggested this in his talk [27]. The connection between the two methods was studied by István Kovácsvölgyi [38] in a simple case and in this case he found them to be equivalent. Nevertheless, the method of interpolating spaces seems to be essential in more complicated cases, for example in the study [29] of the complex valued measurable solutions of equation

$$\begin{aligned} f(x)f(y)=G(|x|+|y|,x+y),\quad x,y\in {{\mathbb {R}}}^3,\ x\times y\not =0. \end{aligned}$$