Ronan Howard

Let $\mathbb{R}$ define the set of real numbers. Find all continuous functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that $f(f(x+y))=f(x)+f(y)$ .

Reveal Answer

Claim:

Solutions are $f(x)=0$ and $f(x)=x+c$ where $c\in\mathbb{R}$

Proof:

Let $y=0$ ,

\begin{aligned} f(f(x)) = f(x) + f(0) \end{aligned}

Replacing $x$ with $x+y$ and using the original equation,

\begin{aligned} f(f(x+y)) &= f(x+y) + f(0)\\ f(x) + f(y) &= f(x+y) + f(0)\\ f(x+y) &= f(x) + f(y) - f(0) \end{aligned}

Therefore $f(x+y)$ is related to $f(x)+f(y)$ in addition to some constant. This should remind you of Cauchy's Functional Equation.

Let $g(x) = f(x) - f(0)$ , then,

\begin{aligned} g(x+y) &= f(x+y) - f(0)\\ &= f(x) + f(y) - f(0) - f(0)\\ &= [f(x)-f(0)] + [f(y)-f(0)]\\ &= g(x) + g(y) \end{aligned}

Since $f$ is continuous, $g$ must also be continuous. Therefore $g(x+y)=g(x)+g(y)$ is Cauchy's equation where all solutions are given by $g(x) = kx$ where $k\in\mathbb{R}$ .

Proof for g(x)=kx

Claim:

If $g(x+y)=g(x)+g(y)$ , then $g(x)=kx$ where $g:\mathbb{R}\rightarrow\mathbb{R}$ is continuous and $x,y,k\in\mathbb{R}$ .

Proof:

Let $y=0$ , then,

\begin{aligned} g(x) &= g(x) + g(0)\\ g(0) &= 0 \end{aligned}

Let $y=x$ , then,

\begin{aligned} g(x+x) &= g(x) + g(x)\\ g(2x) &= 2g(x)\\ \end{aligned}

By letting $y=x,2x,3x,...$ and repeating the step above, we can prove by induction that $g(kx)=kg(x)$ for any positive integer $k$ .

Expanding this idea to negative integers, let $y=-x$ ,

\begin{aligned} g(x-x) &= g(x) + g(-x)\\ g(0) &= g(x) + g(-x)\\ 0 &= g(x) + g(-x)\\ g(-x) &= -g(x)\\ \end{aligned}

Expanding this idea to rational numbers, let $x=\frac{m}{n}$ ,

\begin{aligned} g(\frac{m}{n}) &= mg(\frac{1}{n})\\ g(\frac{1}{n}) &= \frac{1}{n} g(1)\\ \end{aligned}

Let $u=\frac{1}{n}$ , then $g(u)=ug(1)$ and so $g(x)$ must be of the form $kx$ for all rationals.

Since $f$ is continuous, we can define a rational sequence $x_n$ such that $\lim_{n \to \infty}x_n=x$ where $x$ is any real number.

\begin{aligned} \lim_{n \to \infty}[g(x_n) &= x_n g(1)]\\ g(x) &= x g(1) \qquad \Box\\ \end{aligned}

Since $g(x)=kx$ , we can substitute that into the original equation $g(x)=f(x)-f(0)$ to work out that $f(x)=kx+f(0)$ .

Since $f(0)$ is a constant, let $f(0)=c$ so that we can write $f(x)=kx+c$ . Then substitute into the left and right hand side of the original function so that we can compare.

\begin{aligned} \tag{LHS} f(f(x+y)) &= f(k(x+y)+c) \\ &= k(k(x+y)+c)+c\\ &= k^2(x+y)+ck+c \end{aligned}

\begin{aligned} \tag{RHS} f(x) + f(y) &= kx+c+ky+c \\ &= k(x+y)+2c \end{aligned}

By equating terms, we get the equations $k^2=k$ and $2c=ck+c$ . $k^2=k$ tells us that $k$ equals either $0$ or $1$ . If we analyze both of these cases, we can find all solutions.

Assume $k=0$ , then $f(x)=c$ ,

\begin{aligned} \tag{LHS} f(f(x+y)) = f(c) = c \end{aligned}

\begin{aligned} \tag{RHS} f(x) + f(y) &= c+c = 2c \end{aligned}

Equating the two, we find that $c=2c$ which is only possible if $c=0$ . Therefore, if $k=0$ , then $c=0$ . This implies that $f(x)=0$ is a solution.

Assume $k=1$ , then $f(x)=x+c$ ,

\begin{aligned} \tag{LHS} f(f(x+y)) = f(x+y+c) = x+y+2c \end{aligned}

\begin{aligned} \tag{RHS} f(x) + f(y) &= x+c+y+c = x+y+2c \end{aligned}

Since the right and left hand side are equal, we have shown that $f(x)=x+c$ is also a solution. $\Box$

Nested Functions