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Question

# Suppose that f is a continuous additive function on$\mathbb{R}.$If c := f (1), show that we have f(x) = cx for all$x \in \mathbb{R}.$

Solution

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By additivity of $f$ we get

$f(0)=f(0+0)=f(0)+f(0) \Rightarrow \boxed{f(0)=0}$

We can prove the claim for natural numbers using Mathematical Induction:

$\text{\underline{Base of Induction:}}$      $n=1$

By assumption,

$f(1)=c=c\cdot 1$

$\text{\underline{Induction Step}}$

Suppose that for some $n\in\mathbb{N}$

$f(n)=cn \quad (*)$

Then,

\begin{align*} f(n+1)&=f(n)+f(1)\tag{Additivity}\\ &=cn+c\tag{Use (*) and f(1)=c}\\ &=c(n+1)\end{align*}

Therefore, by Principle of Mathematical Induction we conclude that

$\boxed{f(n)=cn,\hspace{0.2cm} \forall n\in\mathbb{N}}$

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