Fri, 14/11/2014 - 11:56 — asger
p(x,y) = p_{n}(y)x^{n} + … + p_{1}(y)x^{1} + p_{0}(y).

Take y_{0} in **C**. If the polynomial x → p(x,y_{0})-x is non-constant, we can find a root x_{0} so that f(x_{0})f(y_{0}) = f(p(x_{0},y_{0})) = f(x_{0}), implying f(y_{0}) = 1, i.e. y_{0} is in E. Thus for y_{0} in **C**\E, the polynomial x → p(x,y_{0})-x is constant. From this we see that p_{n}(y_{0}) = … = p_{2}(y_{0}) = 0, p_{1}(y_{0}) = 1, and since this applies to all y_{0} in **C**\E, which is infinite by lemma 5, the polynomials p_{1},…,p_{n} must be constant with p_{1} = 1, p_{2} = … = p_{n} = 0. In other words, p(x,y) = x + p_{0}(y). By symmetry we also have p(x,y) = y + q_{0}(x). Together these two facts mean that p(x,y) = x+y+c for a constant c. If we define g(x) = f(x-c), then g(x)g(y) = f(x-c)f(y-c) = f(p(x-c,y-c)) = f(x-c+y-c+c) = f(x+y-c) = g(x+y).
f(x)f(y) = f(x^{r} y^{s}) = f(a^{r'r - s's} b^{r's - rs'}) = f(a) = 1

f(y)f(x) = f(y^{r} x^{s}) = f(a^{r's - rs'} b^{r'r - s's}) = f(b) = z

This contradiction proves that r = s. Our equation now has the form f(x)f(y) = f(x^{r} y^{r}). Put x=y=1 to get f(1)f(1) = f(1), and therefore f(1) = 1.

Finally we have: f(xy) = f(xy)f(1) = f((xy)^{r} 1^{r}) = f(x^{r} y^{r}) = f(x)f(y). This finishes the proof.

I recently looked through my old math notes, and found some of my personal research projects. It's quite fun for me to read; in those days I had much more time to pursue my interest in math, and I actually managed to prove some interesting things. Here is an example:

**Theorem**. Let f be a non-constant complex function and let p be a complex polynomial in two variables. Assume f and p satifies the following equation for all x and y in **C**: f(x)f(y) = f(p(x,y)), then one of the following holds:

- f has no zeroes, and p(x,y) = x+y+c for a constant c.

Furthermore, the function g(x) = f(x-c) satisfies g(x)g(y) = g(x + y). - f has exactly one zero, and p(x,y) = k (x-c)
^{r}(y-c)^{s}+ c for complex constants k, c, and positive integers r, s.

If f is not the function x → [x≠c] (Iverson bracket), then r = s.

Furthermore, we can find constants a and b such that the function g(x) = f(ax+b) satisfies g(x)g(y) = g(xy).

The proof of the theorem uses the fact that a complex polynomial is either constant or takes all values in **C**. This follows directly from the Fundamental Theorem of Algebra, which says that every non-constant complex polynomial has a root.

Define V = {x ∈ **C**: f(x) = 0} and E = {x ∈ **C**: f(x) = 1}. Since f is not constant neither **C**\V nor **C**\E can be empty.

**Lemma 1**. Fix x_{0} in **C**. The function h: y → p(x_{0}, y) is constant if and only if x_{0} is in V.

By symmetry, for y_{0} in **C**, x → p(x,y_{0}) is constant if and only if f(y_{0}) = 0.

**Proof**: Since h is a polynomial, h is either constant or takes all values in **C**. Since f is not constant, h is constant if and only if y → f(h(y)) is constant. The function y → f(h(y)) = f(p(x_{0},y)) = f(x_{0})f(y) is constant if and only if f(x_{0}) = 0.

**Lemma 2**. p(x,y) is in V if and only if x is in V or y is in V.

**Proof**: p(x,y) is in V if and only if f(p(x,y)) = 0, which happens if and only if f(x)f(y) = 0, which again happens if and only if x is in V or y is in V.

**Lemma 3**. If y → p(x,y) is constant for infinitely many x, then p does not depend on y.

**Proof**: Write p as a polynomial in y with coefficients that are polynomials in x: p(x,y) = p_{n}(x)y^{n} + … + p_{1}(x)y^{1} + p_{0}(x). By assumption, p_{1},…,p_{n} have infinitely many roots, and so must be zero. Thus p(x,y) = p_{0}(x), which does not depend on y.

**Lemma 4**. V is finite.

**Proof**: If V is infinite, then by lemma 1 and lemma 3 we have that p(x,y) does not depend on y. Taking x' in **C**\V, we have f(y) = f(p(x',y))/f(x'), which does not depend on y, making f constant, which is a contradiction.

**Lemma 5**. If f has no zeroes, then **C**\E is infinite.

**Proof**: Assume **C**\E is finite, and let **C**\E = {x_{1},…,x_{n}}. If x is in E, then f(p(x,x_{i})) = f(x)f(x_{i}) = f(x_{i}) ≠ 1, so p(x,x_{i}) must be in **C**\E. Therefore, for any x in **C**, p(x,x_{i}) can only take the values {x_{1},…,x_{n},p(x_{1},x_{i}),…,p(x_{n},x_{i})}, which means that x → p(x,x_{i}) is constant. By lemma 1 x_{i} must be in V, which is a contradiction since V is empty.

**Lemma 6**. E is non-empty.

**Proof**: Choose x_{0} such that f(x_{0}) ≠ 0. By lemma 1, the function y -> p(x_{0},y) is not constant, so we can find a y_{0} in **C**, such that p(x_{0},y_{0}) = x_{0}. Now f(x_{0})f(y_{0}) = f(p(x_{0},y_{0})) = f(x_{0}), so f(y_{0}) = 1.

**Proof of the theorem**: For the first part of the theorem, assume that f has no zeroes. Write p as a polynomial in x with coefficients that are polynomials in y:

Take y

For the second part of the theorem, we assume that f has at least one zero. By lemma 4, V is finite, so we can write V = {x_{1},…,x_{n}}. p(x,y) is constant on each of the lines x = x_{i}, and also on the lines y = x_{i}. The lines together form a connected grid on which p is constant, so the constant along all the lines must be the same, p(x_{i},y) = p(x,x_{i}) = c for all x_{i} in V and all x,y in **C**.

Take x' in **C**\V. By lemma 1, the polynomial y → p(x',y) is not constant. Take any x_{i} in V and let y_{0} be a root of the polynomial y → p(x',y)-x_{i}. Then x_{i} = p(x',y_{0}) and 0 = f(x_{i}) = f(p(x',y_{0})), which by lemma 2 implies that y_{0} is in V. Thus x_{i} = p(x',y_{0}) = c. This proves that c is the only element of V.

Since p(x,y)-c is zero on the lines x = c and y = c, we can conclude that p(x,y)-c = (x-c)^{r} (y-c)^{s} q(x,y) for a polynomial q, where the positive multiplicities r and s are chosen such that x-c and y-c does not divide q.

We want to prove that q is constant, so let's assume that q is non-constant in y and find a contradiction. If we write q(x,y) = q_{n}(x)y^{n} + ... + q_{1}(x)y + q_{0}(x), where q_{n} is not identically zero for an n>0, we see that the polynomial y → q(x,y) can only be constant if q_{n}(x) = 0, which only happens for finitely many x. Using polynomial division, we can also write q(x,y) = u(x,y)(y-c) + r(x). If q(x,y) = 0, we must have x=c or y=c, so if we take an x ≠ c such that y → q(x,y) is not constant, then c must be a root, and 0 = q(x,c) = r(x). Every such x must be a root in r(x), so r must be identically zero, and so y-c divides q. This contradicts the assumption, so q must be constant in y. The same argument proves that q is constant in x, so we have p(x,y) = k(x-c)^{r} (y-c)^{s} + c for a constant k.

We now prove that we can choose g(x) = f(ax+b) such that g(x)g(y) = g(xy). First of all, we can assume without loss of generality that c = 0, since we can define f_{1}(x) = f(x+c), such that f_{1}(x)f_{1}(y) = f(x+c)f(y+c) = f(p(x+c,y+c)) = f(k x^{r} y^{s} + c) = f_{1}(k x^{r} y^{s}), and substitute this function for f. We can also assume k = 1, by substituting the function f_{2}(x) = f(x/k^{t}), t = 1/(r+s-1), for f: f_{2}(x)f_{2}(y) = f(x/k^{t})f(y/k^{t}) = f(k^{1-rt-st} x^{r} y^{s}) = f(k^{-t} x^{r} y^{s}) = f_{2}(x^{r} y^{s}).

By lemma 6 we know that 1 is contained in f(**C**), so f(**C**) will at least contain {0,1}. If f(**C**) = {0,1}, then there is only one possibility for f, since we must have f(x)=0 if and only if x=0. This function can be written x → [x≠0]. It is obvious that f(x)f(y) = f(xy).

Now we can assume that f(**C**) contains an element z not in {0,1}. First we prove that r = s: Assume r > s, and choose a with f(a)=1 and b with f(b)=z:

Set r' = r/(r^{2}-s^{2}) and s' = s/(r^{2}-s^{2}). Notice that r'r - s's = 1 and r's - rs' = 0. Set x = a^{r'} b^{-s'}, and y = a^{-s'} b^{r'}. We have:

This contradiction proves that r = s. Our equation now has the form f(x)f(y) = f(x

Finally we have: f(xy) = f(xy)f(1) = f((xy)

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