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Consider the functional equation $$f(x+y) = f(x)g(y)+f(y)g(x)$$ valid for all complex $x,y$. The only solutions I know for this equation are $f(x)=0$, $f(x)=Cx$, $f(x)=C\sin(x)$ and $f(x)=C\sinh(x)$.

Question $1)$ Are there any other solutions ?


If we set $x=y$ we can conclude that if there exists a $g$ for a given $f$ then $g$ must be $g(x)=\frac{f(2x)}{2f(x)}$. By using this result, I tried setting $y=2x$ yielding $$f(x+2x)=f(x)g(2x)+g(x)f(2x)=\dfrac{f(x)f(4x)}{2f(2x)}+\dfrac{f^2(2x)}{2f(x)}$$

Question $2)$ Are $f(x)=0$,$f(x)=Cx$,$f(x)=C\sin(x)$ and $f(x)=C\sinh(x)$ the only solutions to $f(3x)=\dfrac{f(x)f(4x)}{2f(2x)}+\dfrac{f^2(2x)}{2f(x)}$? If not, what are the other solutions?

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  • $\begingroup$ @vadim123 No that is not true. $f(x)=\sin(x)$ and the sine has zero's. $\endgroup$ – mick Jul 15 '13 at 21:30
  • $\begingroup$ In that case you take the limit. @vadim123 $\endgroup$ – mick Jul 15 '13 at 21:34
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    $\begingroup$ @mick What about $f(x) = A\cdot \sin(Bx)$? $\endgroup$ – Erick Wong Jul 15 '13 at 21:44
  • $\begingroup$ I think that question (2) should be split into a separate question. $\endgroup$ – Malper Dec 31 '13 at 21:15
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As leshik pointed out, this equation has plenty of discontinuous solutions (e.g. for $g=1$ it becomes Cauchy's functional equation), so let's just consider continuous solutions. $f=0$ is the trivial solution; from now on we will assume that $f$ is not identically zero. We consider two cases:

  1. If $f(x)$ and $g(x)$ are linearly dependent, then there is some $\lambda \neq 0$ such that $g(x)=\lambda f(x)$. Since we can replace $f(x)$ with $c f(x)$ and the original equation is still satisfied, we may assume without loss of generality that $g(x)=\frac{1}{2}f(x)$, so the equation becomes $f(x+y)=f(x)f(y)$. Now since $f(x+0)=f(x)f(0)$ and $f$ is not identically zero, we must have $f(0) \neq 0$. Then since $f(0)=f(0+0)=f(0)^2$, $f(0)=1$. By continuity $F(t) = \ln (2f(t))$ is defined in a neighborhood of $0$ and satisfies $F(x+y)=F(x)+F(y)$. This is Cauchy's functional equation, so since $F$ is continuous $F(x)=rx$, and therefore $f(x)=e^{rx}$ and $g(x)=\frac{1}{2}e^{rx}$. Therefore the general solution in this case is:

    • $f(x)=ke^{rx}$, $g(x)=\frac{1}{2}e^{rx}$.
  2. Now suppose that $f(x)$ and $g(x)$ are linearly independent. Since $f(0)=f(0+0)=2f(0)g(0)$, we either have $f(0)=0$, or else $f(0) \neq 0$ and $g(0)=\frac{1}{2}$. In the latter case we would have $f(x)=f(x+0)=\frac{1}{2}f(x)+f(0)g(x)$, so $f(x)$ and $g(x)$ would be linearly dependent. Therefore $f(0)=0$.

    Since $f$ is not identically zero, there is no $x$ for which both $f(x)$ and $g(x)$ are zero, since otherwise $f(x+y)=f(x)g(y)+f(y)g(x)$ would be identically zero. Then by continuity there must be some $p$ such that $f(p) \neq 0$ and $g(p) \neq 0$. It follows that $g(x)=\frac{1}{f(p)}f(x+p)-\frac{g(p)}{f(p)}f(x)$, so $g$ is a linear combination of $f$ and its translate $f_p$ (where $f_p(x)=f(x+p))$. Since $f$ and $g$ are linearly independent, it follows that $f$ and $f_p$ are linearly independent. Then since $f(x+y)=f(x)g(y)+f(y)g(x)$ for all $x,y$, every translate $f_y$ of $f$ is a unique linear combination of $f$ and $g$, and therefore every translate $f_y$ of $f$ is a unique linear combination of $f$ and $f_p$.

    Since all translates of $f$ are linearly combinations of the linearly independent functions $f$ and $f_p$, this implies that $f$ is differentiable as described in the answer to this question. Then by the answers to this question, $f$ is a solution to the ODE $\frac{d^{2}f}{dx^{2}}+B\frac{df}{dx}+Cf=0$ for some $B,C \in \mathbb{R}$, other than the solution $ce^{rx}$. With the boundary conditions $f(0)=0$ it follows that there are only three families of solutions:

    • $f(x) = kxe^{rx}$, $g(x)=e^{rx}$

    • $f(x) = ke^{rx}\sin(sx)$, $g(x)=e^{rx}\cos(sx)$

    • $f(x)= ke^{rx}\sinh(sx)$, $g(x)=e^{rx}\cosh(sx)$

    Of course, these are valid solutions because of the following addition formulae $$(x+y)=x\cdot 1 + y \cdot 1$$ $$\sin(x+y)=\sin(x)\cos(y)+\cos(x)\sin(y)$$ $$\sinh(x+y)=\sinh(x)\cosh(y)+\sinh(y)\cosh(x)$$

We have found all the continuous solutions of the given functional equation. QED

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You can take $g=1,$ then your equation reduces to $f(x+y)=f(x)+f(y).$ Without any additional assumption on $f,$ this functional equation has plenty of discontinuous solutions. See here

Of course, once you add some regularity condition (such as continuity, differentiability, boundedness), you can get much more using what Vadim wrote in his post. As to your question 2 - the same arguments apply, since this is a consequence of the general equation.

Since I do not know which condition you prefer to add, let me just mention that if $g$ is continuous, then equation $g(x+y)=2g(x)g(y)$ has solution $g(x)=0$ or $g(x)=2^{ax-1}.$ To see that, note that if $g(x_0)=0$ at some point then $g(x+x_0)=0$ everywhere. Alternatively, $g(2x)=2g^2(x)>0$ for all $x$ implies that the function $h(x)=\ln g(x)-1$ satisfies $h(x+y)=h(x)+h(y)$ and $h$ is continuos. So it is linear. Note, that I assumed that $f$ maps $\mathbb{C}$ into $\mathbb{R}$ but the situation does not change much if you switch to functions with complex range.

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  • $\begingroup$ Thanks but "my master" forbids AC. $\endgroup$ – mick Jul 15 '13 at 22:26
  • $\begingroup$ @vadim123 Taking $f(x)=2x$ for rational $x$ and $f(x)=0$ for irrational $x$ won't solve the equation, because $x+y$ could be rational and non-zero while both $x$ and $y$ are irrational. $\endgroup$ – Andreas Blass Jul 16 '13 at 0:39
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    $\begingroup$ It is consistent with the standard (ZF) axioms of set theory without the axiom of choice that the only solutions of $f(x+y)=f(x)+f(y)$ on the reals are the linear functions. $\endgroup$ – Andreas Blass Jul 16 '13 at 0:41
  • $\begingroup$ @Andreas Blass: I assume Vadim's answer applies to the Question 2 instead. $\endgroup$ – leshik Jul 16 '13 at 0:59
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Some partial results:

Assume henceforth that neither $f$ nor $g$ is constant and $0$. Take $y=0$, we get $$f(x)=f(x)g(0)+f(0)g(x),\textrm{ or }f(x)(1-g(0))=f(0)g(x)$$ then either $g(0)=1$ and $f(0)=0$, or

$$f(x)=\frac{f(0)}{1-g(0)}g(x)$$ Substituting $x=0$ gives $g(0)=1-g(0)$ so $g(0)=0.5$, and hence $f(x)=2f(0)g(x)$. Substituting into the original gives $$2f(0)g(x+y)=2f(0)g(x)g(y)+2f(0)g(y)g(x)$$ whence $$g(x+y)=2g(x)g(y)$$ As Kunnysan points out, $g(x)=e^x/2$ satisfies this. Taking $x=n-1, y=1$ lets us prove inductively that $g(n)=2^{n-1}g(1)^n$.

Suppose now that $f(0)=0$ and $g(0)=1$.

Now, taking $y=-x$, we have $$0=f(0)=f(x)g(-x)+f(-x)g(x)$$ Rearranging, we have (provided both expressions exist)$$\frac{f(x)}{f(-x)}=-\frac{g(x)}{g(-x)}$$

Hence if $f$ is even, then $g$ is odd and vice versa.

I think you'll have a hard time proving that the families you have are everything, unless you have some sort of differentiability assumption.

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    $\begingroup$ $g(x+y)=2g(x)g(y)$ and that would give you a solution $g(x)=\frac{e^x}{2}$ $\endgroup$ – Kunnysan Jul 15 '13 at 22:07
  • $\begingroup$ Nice, but im a bit confused now. Why doesnt $\sin$ pop out of these equations ? $\endgroup$ – mick Jul 15 '13 at 22:23
  • $\begingroup$ Sine pops out of the "general" case where $f(0)=0$, $g(0)=1$. The gray box is the "special" case where $f$ is a multiple of $g$. $\endgroup$ – vadim123 Jul 15 '13 at 22:30
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Be $B$ a basis of $\mathbb R$ as vector space over $\mathbb Q$, then it is immediately clear that for any $b,b'\in B$, we can independently choose $f(b)$ and $f(b')$ because the addition theorem doesn't relate values in different $\mathbb Q$-subspaces of $\mathbb R$.

We can remove this ambiguity by additionally demanding that $f$ is continuous. Then the values of $f$ on the rational numbers determine the values on the irrational numbers by continuity. Therefore in the following I'll assume $f$ is continuous.

First, I'll put forward a few general properties of the solutions:

Observation 1: For any fixed $g(x)$, the set $V_g:=\{f:f(x+y)=f(x)g(y)+f(y)g(x)\}$ is a vector space.

Proof:

  • The function $f(x)=0$ obviously is in $V_g$.
  • if $f_1\in V_g$ and $f_2\in V_g$ then also $f=f_1+f_2\in V_g$: $$\begin{align} f(x+y) &= f_1(x+y)+f_2(x+y)\\ &= f_1(x)g(y)+f_1(y)g(x) + f_2(x)g(y)+f_2(y)g(x)\\ &= (f_1(x)+f_2(x))g(y) + (f_1(y)+f_2(y))g(x)\\ &= f(x)g(y) + f(y)g(x) \end{align}$$
  • If $f\in V_g$, then also $F=\alpha f\in V_g$: $$F(x+y)=\alpha f(x+y) = \alpha (f(x)g(y)+f(y)g(x)) = \alpha f(x)g(y) + \alpha f(y)g(x) = F(x)g(y)+F(y)g(x)$$

This observation means that we only have to find a basis for each $g$, and can then get the other functions as linear combinations.

Note that in proving the vector space property, we've actually already proved the

  • First solution: $\color{red}{f(x)=0}$

Observation 2: If $f(x)$ is a solution with $g(x)$ as "companion function", then for any $\alpha$, $f_\alpha(x):=f(\alpha x)$ is a solution with $g_\alpha(x):=g(\alpha x)$ as companion function.

Proof: $$f_\alpha(x+y) = f(\alpha(x+y)) = f(\alpha x+\alpha y) = f(\alpha x)g(\alpha y)+f(\alpha y)g(\alpha x) = f_\alpha(x)g_\alpha(y)+f_\alpha(y)g_\alpha(x)$$

This scaling property will prove quite handy.

Observation 3: If $f(x)$ is a solution with the function $g(x)$, then $f_\lambda(x):=f(x)\mathrm e^{\lambda x}$ is a solution with the function $g_\lambda(x):=g(x)\mathrm e^{\lambda x}$.

Proof: $f_\lambda(x+y) = f(x+y)\mathrm e^{x+y} = (f(x)g(y)+f(y)g(x))\mathrm e^{x+y} = f(x)\mathrm e^xg(y)\mathrm e^y + f(y)\mathrm e^yg(x)\mathrm e^x = f_\lambda(x)g_\lambda(y)+f_\lambda(y)g_\lambda(x)$.

Note that this observation already allows to derive some new solutions in addition to those found in the question, for example $\mathrm e^x\sin x$ is another such function.

Observation 4: If $g(0)=0$, then $f(x)=0$ is the only element of $V_g$

Proof: We have $f(x) = f(0+x) = f(0)g(x) + g(0)f(x)$. Since by assumption $g(x)=0$, we get $f(x) = f(0)g(x)$. But for $x=0$ this means $f(0) = f(0)g(0) = 0$, therefore $f(x)=0$ for all $x$.

Observation 4 means that for any non-trivial solution, we have $g(0)\ne 0$.

Observation 5: If for any $x_0$ we have $f(x_0)=0$, then $f(0)=0$.

Proof: We have $0 = f(x_0) = f(0 + x_0) = f(0)g(x_0) + f(x_0)g(0) = f(0)g(x_0)$. So either $f(0)=0$ or $g(x_0)=0$. In the latter case, $f(x+x_0) = f(x)g(x_0)+f(x_0)g(x)=0$ for all $x$, that is, we would have the trivial solution, which of course also has $f(0)=0$ (as can be seen also directly by inserting $x=-x_0$).

Observation 5 means we have at most three types of solutions:

  1. Solutions which are zero at some $x_0\ne 0$ (and additionally at $x=0$).
  2. Solutions which are zero only at $x=0$.
  3. Solutions that are nowhere zero.

Let's first consider the case that there's an $x_0\ne0$ so that $f(x_0)=0$. Since we have proved that $g(x_0)=0$ only admits the trivial solution, we can in the following assume $g(x_0)\ne 0$.

Observation 6: If $f(x_0)=0$ and $f(y_0)=0$, then $f(x_0+y_0)=0$.

Proof: This follows directly from the sum formula: $f(x_0+y_0) = f(x_0)g(y_0) + g(x_0)f(y_0) = 0$

From this observation one gets immediately:

Observation 7: If $f(x_0)=0$, then $f(n x_0)=0$ for all $n\in\mathbb N$.

Proof by induction:

  • For $n=0$, we have proved it in observation 4.
  • For $n=1$, it is just the assumption.
  • If we have proven it for $n$, $f((n+1) x_0) = f(n x_0 + x_0) = 0$ by observation 5.

Observation 8: If $f(x_0)=0$, then $f(-x_0)=0$.

Proof: $0 = f(-x_0 + x_0) = f(-x_0)g(x_0)+f(x_0)g(-x_0) = f(-x_0)g(x_0)$. Since $g(x_0)\ne 0$, we must have $f(-x_0)=0$

Observations 7 and 8 together mean that if $f(x_0)=0$, then $f(nx_0)=0$ for all $n\in\mathbb Z$.

Observation 9: There's a minimal $x_0$ so that $f(x_0)=0$.

Proof: If there were no such minimal $x_0$, then for any $\epsilon>0$ we would find an $x_1<\epsilon$ so that $f(x_1)=0$. But with that and the fact that with $x_1$, also all $nx_1$ are zeros of $f$, it is easy to create a series of zeros converging to an arbitrary real number, so we would have $f(x)=0$ for all $x$.

To simplify our notation, we can use the scaling property (observation 2) to assume without loss of generality that the first positive zero of $f(x)$ is $1$. Then the previous result mean that $f(n)=0$ for all $n\in\mathbb Z$.

Observation 10: There are no non-integer values $x_0$ so that $f(x_0)=0$.

Proof: If there were such a value, then $f(x_0-\lfloor x_0\rfloor)=0$. But $0<x_0-\lfloor x_0\rfloor<1$, contrary to the assumption that $1$ is the smallest positive zero of $f$.

Observation 11: $f(x+n) = f(x)g(1)^n$

Proof: We have $f(x+1)=f(x)g(1)$, which implies $f(x+n)=f(x)g(1)^n$. On the other hand, a direct application of the sum formula gives $f(x+n)=f(x)g(n)$. Thus $g(n) = g(1)^n$.

Observation 12: $g$ vanishes for all $n+\frac12$.

Proof: For $n$ odd, we get $0 = f(n) = f(\frac n2+\frac n2) = 2f(\frac n2)g(\frac n2)$. Since $n$ is odd, $f(\frac n2)\ne 0$, and therefore $g(n/2)=0$.

To simplify the further considerations, use the vector space property (observation 1) together with the knowledge that $f(\frac12)\ne0$ to fix without loss of generality $f(\frac12)=1$.

Observation 13: $g(x)=f(x+\frac12)$.

Proof: By the addition property, $f(x+\frac12)=f(x)g(\frac12)+g(x)f(\frac12)$. Bur $g(\frac12)=0$ (observation 12) and $f(\frac12)=1$ (we just fixed that value).

Observation 14: $g(x+y)=g(x)g(y)+f(x+1)f(y)$

Proof: $g(x+y) = f(\frac12+x+y) = f(\frac12+x)g(y) + f(y)g(\frac12+x) = g(x)g(y)+f)y=f(\frac12+\frac12+x)$.

Observation 15: $g(n) = g(1)^n$

Proof: By induction: $g(n+1) = g(n)g(1) + f(n+1)f(1) = g(n)g(1)$.

Note that the same recursion formula can be used also backwards, so it is true for all $n$.

Now we can use observation 3 (the exponential factor property) to reduce our considerations to the two cases $g(1)=1$ and $g(1)=-1$. Then of course $g(n)=(\pm 1)^n$. Especially $g(-1)=g(1)$ and $g(2)=1$.

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  • $\begingroup$ All those observations should rather be propositions! =D $\endgroup$ – Pedro Tamaroff Dec 31 '13 at 20:27

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