Example illustrating the importance of Hamel bases for description of additive functions

Question

Can anyone please explain and show (step by step) that how $ f \big( f ( x ) \big) = x $ in the following example?

One can show that for any Hamel basis $ H $ there exists a bijection $ \varphi : [ 0 , 1 ] \to H $. If we set $ \varphi ( t ) = h _ t $, then $ H = \{ h _ t | t \in [ 0 , 1 ] \} $. This observation leads to constructions of additive functions having interesting properties.

Example. Let $ f : \mathbb R \to \mathbb R $ be the additive function determined by the function $ s : H \to \mathbb R $ defined by: $$ s ( h _ t ) = \begin {cases} h _ t & \text {if } t \in \left[ 0 , \frac 1 2 \right] \\ - h _ t & \text {if } t \in \left( \frac 1 2 , 1 \right] \text . \end{cases} $$ Then it is easy to check that $ f \big( f ( x ) \big) = x $ for any $ x \in \mathbb R $.

Also, please explain and show (step by step) that how $ f \big( f ( x ) \big) = f ( x ) $ in the following example?

Another interesting example of a discontinuous additive function is obtained by considering the function $ s : H \to \mathbb R $ defined by $$ s ( h _ t ) = \begin {cases} h _ t & \text {if } t \in \left[ 0 , \frac 1 2 \right] \\ 0 & \text {if } t \in \left( \frac 1 2 , 1 \right] \text . \end{cases} $$ Then $ f \big( f ( x ) \big) = f ( x ) $ for any $ x \in \mathbb R $.

I was unable to do it and I am completely clueless.

Answer 1

For the first one, for all $h_t\in H$ we have

\begin{align} f(f(h_t)) = f(s(h_t)) &= \begin{cases} f(h_t) & t\le \frac 1 2 \\ f(-h_t) & t>\frac 1 2\end{cases} \\ &= \begin{cases} s(h_t) & t\le \frac 1 2 \\ -s(h_t) & t>\frac 1 2\end{cases} \\ &= \begin{cases} h_t & t\le \frac 1 2 \\ -(-h_t) & t>\frac 1 2\end{cases} \\ &= h_t. \end{align} Since $f(f(x))=x$ on a Hamel basis and both $f\circ f$ and $\operatorname{id}_{\mathbb R}$ are additive, it also holds on all of $\mathbb R$.

For the second one, for all $h_t\in H$ we have \begin{align} f(f(h_t)) = f(s(h_t)) &= \begin{cases} f(h_t) & t\le \frac 1 2 \\ f(0) & t>\frac 1 2\end{cases} \\ &= \begin{cases} s(h_t) & t\le \frac 1 2 \\ 0 & t>\frac 1 2\end{cases} \\ &= \begin{cases} h_t & t\le \frac 1 2 \\ 0 & t>\frac 1 2\end{cases} \\ &= s(h_t) = f(h_t). \end{align} Since $f(f(x))=f(x)$ on a Hamel basis and both $f\circ f$ and $f$ are additive, it also holds on all of $\mathbb R$.

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In reply to questions raised in the comments:

• Since $t\in[0,1]$ anyway, we do indeed have $t\in[0,\tfrac 1 2]$ equivalent to $t\le \tfrac 1 2$ and $t\in(\tfrac 1 2,1]$ equivalent to $t>\tfrac 1 2$, I just prefer writing it that way.
• The map $\operatorname{id}_{\mathbb R}\colon \mathbb R\to\mathbb R$ is the identity map on $\mathbb R$, it satisfies $\operatorname{id}_{\mathbb R}(x)=x$ for all $x\in\mathbb R$.
• I can't recommend a book on Hamel bases, I just picked up what I needed to answer your question.
• Whenever you have two additive functions $\varphi,\psi\colon\mathbb R\to\mathbb R$ that agree on some Hamel basis $H\subseteq\mathbb R$, that is $\varphi(h)=\psi(h)$ for all $h\in H$. It follows that $\varphi(x)=\psi(x)$ for all $x\in\mathbb R$, since any map $H\to\mathbb R$ has a unique additive extension to $\mathbb R$.
• $f$ being additive means that $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb R$. In particular for $x=y=0$ you get $f(0)=f(0)+f(0)$ so $f(0)=0$ and for $y=-x$ you get $f(0)=f(x)+f(-x)$, so $f(-x)=-f(x)$ ($f$ is odd). That is what I used to get things like $f(-h_t)=-f(h_t)=-s(h_t)$.