# Error with the proof that all solutions to the Cauchy Functional Equation are linear

If $f(x)$ is continuous, it is known that $f(x+y)=f(x)+f(y)$ implies that $f(x)$ is linear, and non-continuous solutions are discussed in these links. (1, 2,3, 4)

However, what is wrong with this proof that all solutions to the Cauchy Functional Equation are of the form $f(x)=cx$?

If $x$ is rational, it is known that $f(x)=cx$ for some fixed constant $c$, as seen here.

If $x$ is irrational let us assume that $x=n+\alpha$, where $0 \le \alpha <1$.

$f(x)=f(n+\alpha)=f(n)+f(\alpha)$.

Because of the upper result, $f(n)=cn$.

Let the decimal expansion of $\alpha$ be $\sum _{ i=1 }^{ \infty }{ \frac { { a }_{ i } }{ { 10 }^{ i } } }$

Note that $\frac { { a }_{ i } }{ { 10 }^{ i } }$ is rational.

Then, $$f(\alpha)=f(\sum _{ i=1 }^{ \infty }{ \frac { { a }_{ i } }{ { 10 }^{ i } } })=\sum _{ i=1 }^{ \infty }{ f(\frac { { a }_{ i } }{ { 10 }^{ i } } ) } =c\sum _{ i=1 }^{ \infty }{ \frac { { a }_{ i } }{ { 10 }^{ i } } }=c\alpha$$

Therefore $f(x)=cn+c\alpha=cx$. What did I do wrong?

• How do you prove that $f(\sum _{ i=1 }^{ \infty }{ \frac { { a }_{ i } }{ { 10 }^{ i } } })$ equals $\sum _{ i=1 }^{ \infty }{ f(\frac { { a }_{ i } }{ { 10 }^{ i } } ) }$ without assuming $f$ continuous? – ForgotALot Feb 6 '16 at 5:48
• Exactly. Let $b_n=\sum_{j=1}^n a_j10^{-j}.$ Then $f(\sum_{j=1}^{\infty}a_j10^{-j})=\sum _{j=1}^{\infty}f(a_j10^{-j})$ is equivalent to $f(\lim_{n\to \infty}b_n)=\lim_{n\to \infty}f(b_n)$. This assumes $f$ is continuous at $\alpha$. – DanielWainfleet Feb 6 '16 at 6:02

How do you prove that $f(\sum _{ i=1 }^{ \infty }{ \frac { { a }_{ i } }{ { 10 }^{ i } } })$ equals $\sum _{ i=1 }^{ \infty }{ f(\frac { { a }_{ i } }{ { 10 }^{ i } } ) }$ without assuming $f$ continuous?
Exactly. Let $b_n=\sum_{j=1}^n a_j10^{-j}$. Then $f(\sum_{j=1}^{\infty}a_j10^{-j})=\sum _{j=1}^{\infty}f(a_j10^{-j})$ is equivalent to $f(\lim_{n\to \infty}b_n)=\lim_{n\to \infty}f(b_n)$. This assumes $f$ is continuous at $\alpha$.