Exercise 4.3.13

Question

Let $f$ be a function defined on all of $\mathbf{R}$ that satisfies the additive condition $f(x+y)=f(x)+f(y)$ for all $x, y \in \mathbf{R}$.
  \begin{enumerate}[label=(\alph*)] 
  \item Show that $f(0)=0$ and that $f(-x)=-f(x)$ for all $x \in \mathbf{R}$.
  \item Let $k=f(1)$. Show that $f(n)=k n$ for all $n \in \mathbf{N}$, and then prove that $f(z)=k z$ for all $z \in \mathbf{Z} .$ Now, prove that $f(r)=k r$ for any rational number $r$.
  \item Show that if $f$ is continuous at $x=0$, then $f$ is continuous at every point in $\mathbf{R}$ and conclude that $f(x)=k x$ for all $x \in \mathbf{R}$. Thus, any additive function that is continuous at $x=0$ must necessarily be a linear function through the origin.
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
  \item $f(0+0) = f(0) + f(0)$ implies $f(0) = 0$ and thus $f(x + (-x)) = 0 = f(x) + f(-x)$ meaning $f(-x) = -f(x)$.
  \item $f(n) = f(1) + \dots + f(1) = k + \dots + k = kn$. Now since $f(-n) = -f(n) = -kn$ we have $f(z) = kz$ for all $z \in \mathbf{Z}$. Finally let $r = p/q$ for $p \in \mathbf{Z}$ and $q \in \mathbf{N}$. notice that $f(qr) = k(qr)$ and since $f(qr) = f(r + \dots + r) = f(r) + \dots + f(r) = qf(r)$ we have $f(r) = k(qr)/q = kr$.
  \item Assume $f$ is continuous at $0$ and let $x \in \mathbf{R}$ be arbitrary. Let $x_n$ be a sequence approaching $x$. since $(x-x_n) 	o 0$ we have $f(x-x_n) 	o 0$ because $f$ is continuous at zero. Now since $f$ is additive $f(x-x_n) = f(x) - f(x_n) 	o 0$ implies $f(x_n) 	o f(x)$ meaning $f$ is continuous at $x \in \mathbf{R}$ by the sequential characterization of continuity.

Now to see that $f(x) = kx$ for $x \in \mathbf{I}$ simply take a limit of rationals approaching $x$.
   \end{enumerate}

Exercise 4.3.13

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Exercise 4.3.13

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