A function with a property $f(x+y)=f(x)f(y)$

Question

A function with the property $f(x+y)=f(x)f(y)$ is well known exponential function, $f(x)=a^x$. My question is, how do you prove if there is no other function with this kind of property?

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Edit: I always find this in math contests. At first glance, it really is the exponential function. As I see in the comments section, there are many other functions.

Answer 1

The comments that suggest $f(x)=1$ and $f(x)=0$ as solutions are correct, but these are just special cases of $f(x)=a^x$ with $a=1$ or $a=0$. If $f$ is not continuous then $f$ need not be an exponential function on the irrationals, but on $\mathbb{Q}$ $f(x)=a^x$ if it satisfies this property.

To see this, let $f(1)=a$ and note that $f(n+1)=f(n)f(1)$ so that by induction $f(x)=a^n$ for all $a\in\mathbb{N}$. Assuming that $a\neq0$, from $f(x+(-x))=f(x)f(-x)$ where $f(x+(-x))=f(0)=1$ (the fact that $f(0)=1$ must be proven first from consideration of $f(1+0)$), we get that $f(x)=a^x$ for all $x\in\mathbb{Z}$. We also have that $f(\underbrace{\frac{1}{p}+\ldots+\frac{1}{p}}_\text{$p$ terms})=(f(\frac{1}{p}))^p=f(1)=a$ so that $f(\frac{1}{p})=a^\frac{1}{p}$ so it's easy to argue from this fact that $f(x)=a^x$ for all $x\in\mathbb{Q}$.
Now if we know that $f$ is continuous everywhere (I believe it is also sufficient for $f$ to be continuous merely at one point) it can be shown that $f(x)=a^x$ for all $x\in\mathbb{R}$.

Answer 2

The condition $f(x+y)=f(x)f(y)$ only implies $f(x)=a^x$ for all rational numbers $x\in\mathbb{Q}$ and for some $a\in\mathbb{R}$. You can get this equality for all real numbers if you have more conditions, for example, if $f$ is continuous in $\mathbb{R}$ or if $f$ is Lebesgue-measurable.

Answer 3

First, some properties of the function

1. $f(x)=f(\frac{x}{2}+\frac{x}{2})=f^{2}(\frac{x}{2})\geq0$ for all $x\in\mathbb{R}$
2. $f(0)=f(0+0)=f^{2}(0)$, then $f(0)=0$ or $f(0)=1$. If $f(0)=0$ then $f(x)=f(x+0)=f(x)f(0)=0$ for all $x\in\mathbb{R}$. Then we will suppose that $f(0)=1$
3. $1=f(0)=f(x-x)=f(x)f(-x)$ thus $f(-x)=\frac{1}{f(x)}$ for all $x\in\mathbb{R}$
4. $f(nx)=f^{n}(x)$ for all $n\in\mathbb{N}$ and $x\in\mathbb{R}$. This proof is by induction, because $f((n+1)x)=f(nx+x)=f(nx)f(x)=f^{n}(x)f(x)=f^{n+1}(x)$ and $f(1x)=f^{1}(x)$
5. If $q\in\mathbb{Q}^{+}$ then exist $n,m$ in $\mathbb{N}$ such that $q=\frac{m}{n}$. So if $x\in\mathbb{R}$ then $y=qx \Leftrightarrow ny=mx$ then $f(ny)=f(mx)\Leftrightarrow f^{n}(y)=f^{m}(x) \Rightarrow f(y)=f(qx)=f^{\frac{m}{n}}(x)=f^{q}(x)$. But $f((-q)x)=f(-qx)=\frac{1}{f(qx)}=\frac{1}{f^{q}(x)}=f^{-q}(x)$, thus $f(qx)=f^{q}(x)$ for all $q\in\mathbb{Q}$

Now if $H$ is a Hamel basis for $\mathbb{R}$ over $\mathbb{Q}$ then for all $x\in\mathbb{R}$ there are $x_{\alpha_{1}},...,x_{\alpha_{n}}$ in $H$ such that $x=\sum_{i=1}^{n}q_{i}x_{\alpha_{i}}$ where $q_{i}$ is a rational number then $f(x)=f(\sum_{i=1}^{n}q_{i}x_{\alpha_{i}})=\prod_{i=1}^{n}f(q_{i}x_{\alpha_{i}})=\prod_{i=1}^{n}f^{q_{i}}(x_{\alpha_{i}})$. If we arbitrarily define $b_{x}:=f(x)$ for some $b_{x}\in\mathbb{R}^{+}$ and all $x\in H$ then

1. $f$ is well defined (because all $x\in\mathbb{R}$ has a unique representation)
2. $f(x+y)=f(x)f(y)$ because if $x,y\in\mathbb{R}$ then $x=\sum_{i=1}^{n}q_{i}x_{\alpha_{i}}$ and $y=\sum_{i=1}^{m}r_{i}x_{\beta_{i}}$. Suppose the first $s$ elements of the basis are equals ($x_{\alpha_{i}}=x_{\beta_{i}}$) then $x+y=\sum_{i=1}^{s}(q_{i}+r_{i})x_{\alpha_{i}}+\sum_{i=s+1}^{n}q_{i}x_{\alpha_{i}}+\sum_{i=s+}^{m}r_{i}x_{\beta_{i}}$ thus $f(x+y)=\prod_{i=1}^{s}b^{q_{i}+r_{i}}_{x_{\alpha_{i}}}\times \prod_{i=s+1}^{n}b^{q_{i}}_{x_{\alpha_{i}}}\times \prod_{i=s+1}^{m}b^{r_{i}}_{x_{\beta_{i}}}=\prod_{i=1}^{n}b^{q_{i}}_{x_{\alpha_{i}}}\times \prod_{i=1}^{m}b^{r_{i}}_{x_{\beta_{i}}}=f(x)f(y)$

If one $b_{x}$ is distinct from the others, then $f$ is not continuous at any point over $\mathbb{R}\setminus \lbrace 0 \rbrace$ because if $x,y$ are in $H$, then existe a sequence of rational numbers $(q_{n})$ such that $\lim_{n\rightarrow\infty}q_{n}=\frac{y}{x}$ then $\lim_{n\rightarrow\infty}f(q_{n}x)=\lim_{n\rightarrow\infty}f^{q_{n}}(x)=b_{x}^{\frac{y}{x}}\neq f(y)=b_{y}$ in general, because if $z$ is other element of $H$ such that $f(z)=b_{y}$ then it will implies that $b_{x}^{\frac{y}{x}}=b_{y}=b_{z}=b_{x}^{\frac{z}{x}}$. In the other hand, if there exist a point $a\in\mathbb{R}\setminus \lbrace 0 \rbrace$ where f is continuous, then $b_{x}^{\frac{a}{x}}=f(a)\Rightarrow b_{x}^{a}=f^{x}(a)$. But $f(a)=b^{a}$ for some $b\in\mathbb{R}$ thus $b_{x}^{a}=b^{ax}\Rightarrow f(x)=b_{x}=b^{x}$ for all $x\in H$ therefore $f(x)=b^{x}$ for all $x\in\mathbb{R}$.

Answer 4

I proved at Proof of existence of $e^x$ and its properties that, if $f(x)$ is differentiable at $0$, then $f(x+y) =f(x)f(y) $ implies that $f'(x) =f'(0) f(x) $.

This leads immediately to the exponential or power function.

Answer 5

It would, assuming that $f$ is continuous and with this condition $(f(x+y)=f(x)f(y)) $ and $f(0)=1$ to prove that $\displaystyle\lim_{x\to-\infty} f(x)=0$ ?

My sketch

As $f$ is continuous, it suffices to consider a sequence $(a_n)_n$ with limit $-\infty$ and prove that $\displaystyle\lim_{n\to+\infty} f(a_n)=0$. This is true ? Because then the only option I see is to prove that this is the exponential function and then conclude that the limit is $0$.

Answer 6

Your functional equation $f(x+y)=f(x)f(y)$ can be reduced to the (perhaps more familiar) additivity equation $g(x+y)=g(x)+g(y)$.

First note that, if $f(x_0)=0$ for some $x_0$, then $f(x)=0$ for all $x$. Of course the constant function $f(x)=0$ is a solution. From now on we assume $f(x)\ne0$. Then, since $f(x)=f(x/2)^2$ (and $f$ is real-valued), $f(x)\gt0$ for all $x$. Hence we can define $g(x)=\ln f(x)$, and then $g(x+y)=g(x)+g(y)$.

Therefore, the solutions of the functional equation $f(x+y)=f(x)f(y)$ are the constant function $f(x)=0$ and the functions of the form $f(x)=e^{g(x)}$ where $g$ is an additive function.

It's known that an additive function which satisfies some weak regularity condition (e.g., bounded on a set of positive Lebesgue measure) must be a linear function, $g(x)=cx$ where $c$ is a constant. On the other hand, discontinuous additive functions can be constructed using a Hamel basis for the real numbers over the rational numbers.

Answer 7

HINT: Using the equation $f(x+y)=f(x)f(y)$,

1. If $f(x)$ is polynomial, show that the degree of L.H.S. $\not =$ degree of R.H.S. (except if $f(x)$ is the zero polynomial)
2. If $f(x)$ is trigonometric, show that it is not possible using the expansion formulae of $\sin,\cos,\tan$ of $(x+y)$.