Tuesday, February 7, 2017

sequences and series - 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?



Answer



The answer is in the comments:


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 $\alpha$.


No comments:

Post a Comment

analysis - Injection, making bijection

I have injection $f \colon A \rightarrow B$ and I want to get bijection. Can I just resting codomain to $f(A)$? I know that every function i...