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Treated with the inverse operator, one could get: $f(x)=x$

However, an other obvious solution is $f(x)=C$ (when f is not invertible). How could I reach this solution? Are there other solutions available?

This is not a homework.

To prove the uniqueness of the solution, I am trying (and currently failing) to do something similar to: Does a non-trivial solution exist for $f'(x)=f(f(x))$?

Thank you Jack and V. Your efforts help. I wonder if one could prove that the "projectors" are the only family of solution to Idempotence.

Dietrich Burde
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High GPA
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4 Answers4

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Any projection operator does the job.
For instance, let $K$ be some closed, unbounded subset of $\mathbb{R}$ and $f$ the function mapping $x$ to $\min\left( K \cap [x,+\infty)\right)$. Then $f(f(x))=f(x)$.

The general structure of an idempotent map is the following one: we have $\mathbb{R}=A\cup B$ with $A\cap B=\emptyset$ and $\forall a\in A, f(a)\in B$ together with $\forall b\in B, f(b)=b$. Then $B=\text{Im}\,f$ and $\forall r\in\mathbb{R},\,f(f(r))=f(r)$.

Jack D'Aurizio
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  • Are there any rigorous proof I could find on this? Is the operator in your solution the only family of solutions? – High GPA Jun 09 '17 at 11:12
  • @LowGPA: to check that such $f$ fulfills $ff=f$ is straightforward, but there are plenty of projection operators. For instance, you may consider the map that sets to $0$ the binary or decimal digits in certain positions, the map that does essentially the same with respect to a Hamel basis etcetera. – Jack D'Aurizio Jun 09 '17 at 11:16
  • Well, projector ($P^2=P$) seems like merely re-dictating the question itself. But your answer is great and I will accept it if no others come. – High GPA Jun 09 '17 at 11:34
  • @LowGPA: anyway, I have added a complete characterization of idempotent maps over $\mathbb{R}$. There are plenty of them since there are many ways for partitioning $\mathbb{R}$ into two components and many ways for mapping the $A$-component into the $B$-component. – Jack D'Aurizio Jun 09 '17 at 11:44
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The function $f(x) = \max(x,0).$

Also absolute value.

Mark Joshi
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If you restrict yourself to linear operators, such a mapping would be a projector. The identity is an example of a projector from $\mathbb{R}$ onto itself. A constant function, even though it is not linear, can be seen as the null application translated to your constant. If you define a proper affine subspace, then you can find a projector, so I guess there is no uniqueness

V. Traboulistopopoulos
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Any such function is a "choice of representatives" function for some equivalence relation on $\mathbb R$. In fact, this works on any set $X$, not just $\mathbb R$. Fix any equivalence relation on $X$ and let $f(x)$ map each $x$ to some fixed representative of the equivalence class of $x$, then $f$ satisfies your condition. Conversely, if $f\circ f=f$, let the equivalence relation be $x\sim y$ iff $f(x)=f(y)$.

Let's say that an $f$ corresponding to a given partition of $X$ is a projection for that partition.

If $X$ is a metric space, then a given partition admits a continuous projection if and only if every class of the partition is a closed set, in which case every projection for that partition is continuous.

So there (at least) as many continuous solutions to $f\circ f=f$ on $\mathbb R$ as there are partitions of $\mathbb R$ into closed sets.

Jack M
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