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# Cauchy's Functional Equation Revisited

It is commonly known that the solution for Cauchy's Functional Equation

$f:\mathbb{Q} \rightarrow \mathbb{Q}, \quad f(x)+f(y)=f(x+y)$

has solutions $$f(x)=cx$$. (If you don't know how to prove this, I ask for you to attempt to prove this)

But now, if we instead change this function from being in the rationals to being the reals, we get a strange property.

Let $$f$$ be a function $$f:\mathbb{R} \to \mathbb{R}$$ such that

\begin{align} f(x) + f(y) &= f(x+y) \forall x, y\\ f(1) &= 1\\ \text{ The function is not }f(x) & = x \end{align}

Consider the graph of $$y = f(x)$$ on the plane. Prove that for any disc, no matter how small, there always exists a point that lies on the graph.

For this proof, you must accept the Axiom of Choice as true.

This question came from a corollary I used in APMO 2016 Q5.

Note by Sharky Kesa
3 months, 3 weeks ago

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Wikipedia :P · 3 months, 3 weeks ago

How do you know Wikipedia is correct? Anyone can edit it. · 3 months, 3 weeks ago

Anyone can edit wikipedia :v

For math homework, i usually go here if i'm confused

Because you guys are more trusted than wikipedia · 3 months, 3 weeks ago

I've seen their proof. It was right in my opinion. · 3 months, 3 weeks ago

Comment deleted 3 months ago

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Comment deleted 3 months ago

I'm saying on the graph of $$(x, f(x)$$. · 3 months, 3 weeks ago