# Sampled Bits

## In theory there is no difference between theory and practice; in practice there is

1 October 2018

by Alexandra Albu

Not long ago, a colleague asked me whether I prefer maths problems with long and detailed requirements or problems with short ones. His question reminded me of a problem I saw several years ago which has a very short requirement and a really nice and somehow surprising solution.

Determine all differentiable functions $f: \mathbb{R} -> \mathbb{R}$ satisfying $f \circ f = f$.

Romanian Mathematical Olympiad 2014, Final Round

It is not hard to see that the constant and the identity functions satisfy the requirements. Further, we’ll prove that these are the only solutions.

As the function $f$ is differentiable, it is very tempting - or at least, I fell into this trap - to differentiate the given relation and see then what you can find out about $f$. However, this seems to be a dead end.

Ok, since this doesn’t work, what other options do we have? Let’s look again at the relation in the requirement. One thing you may think of is trying to find something about the image of $f$, since $f$ composed with itself gives still $f$. The official solution goes in this direction, noticing that because $f$ is continuous - being differentiable - its image is an interval $I$, included in $\mathbb{R}$.

If the interval contains a single value, that the function is constant and we’re done. If it isn’t, then denote by $m=\inf{I}$ and $M=\sup{I}$. We have that $m \lt M$ and $m, M \in \overline{\mathbb{R}}$ The relation in the requirement translates to:

$f(x)=x, \forall x \in I \tag{1}$

Now, we’re trying to prove that in this case, $f$ must be the identity function, meaning that $I=\mathbb{R}$.

We’ll suppose the contrary, i.e. that $a$ is real - so not $-\infty$. Since $f$ is continuous, from relation (1) we get that $f(a)=a$ and, further:

$f'(a)=\lim_{x \to a, x>a} \frac{f(x)-f(a)}{x-a}= \lim_{x \to a, x>a} \frac{x-a}{x-a}=1$

However, $f(a)=a=\inf{I}=\inf{f(x)}|x\in \mathbb{R}$, so $f$ reaches its minimum at $a$. From Fermat’s theorem, we have that $f’(a)=0$, so we obtained a contradiction. Therefore, $a=-\infty$. In a similar way, it can be proved that $b=\infty$.

Source: official solution

tags: mathematical analysis - functional equations