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Question

Suppose that f is a continuous additive function on

R.\mathbb{R}.

If c := f (1), show that we have f(x) = cx for all

xR.x \in \mathbb{R}.

Solution

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By additivity of ff we get

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

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

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

By assumption,

f(1)=c=c1f(1)=c=c\cdot 1

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

Suppose that for some nNn\in\mathbb{N}

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

Then,

f(n+1)=f(n)+f(1)=cn+c=c(n+1)\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

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

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