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Equivalent definition for C^1 functions
Ciobi_   2
N 4 hours ago by Alphaamss
Source: Romania NMO 2025 11.3
Prove that, for a function $f \colon \mathbb{R} \to \mathbb{R}$, the following $2$ statements are equivalent:
a) $f$ is differentiable, with continuous first derivative.
b) For any $a\in\mathbb{R}$ and for any two sequences $(x_n)_{n\geq 1},(y_n)_{n\geq 1}$, convergent to $a$, such that $x_n \neq y_n$ for any positive integer $n$, the sequence $\left(\frac{f(x_n)-f(y_n)}{x_n-y_n}\right)_{n\geq 1}$ is convergent.
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Ciobi_
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Gheorghe Țițeica 2025 Grade 11 P1
AndreiVila   3
N Mar 30, 2025 by Levieee
Source: Gheorghe Țițeica 2025
Find all continuous functions $f:\mathbb{R}\rightarrow\mathbb{R}$ such that $f(x+y)=f(x+f(y))$ for all $x,y\in\mathbb{R}$.
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AndreiVila
Mar 28, 2025
Levieee
Mar 30, 2025
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Gheorghe Țițeica 2025 Grade 11 P1
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Source: Gheorghe Țițeica 2025
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AndreiVila
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AndreiVila
#1 h  Mar 28, 2025, 9:40 PM
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Find all continuous functions $f:\mathbb{R}\rightarrow\mathbb{R}$ such that $f(x+y)=f(x+f(y))$ for all $x,y\in\mathbb{R}$.
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Mathzeus1024
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Mathzeus1024
#2 h  Mar 29, 2025, 2:00 PM
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It works for $\textcolor{red}{f(x)=x}$ and $\textcolor{red}{f(x)=C}$ (where $C$ is a real constant).
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RobertRogo
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RobertRogo
#3 h  Mar 29, 2025, 8:28 PM • 2 Y
Y by TheBlackPuzzle913, Mathzeus1024
Very beautiful!

$f(x) = x$ is good.

Assume there exists $c \in \mathbb{R}$ such that $f(c) \neq c$, say $f(c) > c$. (*)

The hypothesis can be rewritten as $f(x) = f(x + g(y))$, $\forall x, y \in \mathbb{R}$, where $g \colon \mathbb{R} \to \mathbb{R}$, $g(y) = f(y) - y$.

As $f$ is continuous, $g$ is continuous as well, and by (*) we get that either $g$ is constant (which easily gives $f(x)=x$, false) or there exist $T_2>T_1>0$ such that $[T_1, T_2] \subset \text{Im}\ g$. Notice that any number in $\text{Im}\ g$ is a period of $f$ (again, by (*)). We will now prove that $f$ is constant.

For doing so, we will show that $\forall r \in \mathbb{R}, \exists x_r, y_r \in \text{Im}\ g, \alpha_r, \beta_r \in \mathbb{Z}$ such that
\[ r = x_r \alpha_r + y_r \beta_r \]This could be rewritten as (dividing by, say, $\alpha_r$; if $\alpha_r=0$ we divide by $\beta_r$, and if $\beta_r=0$ as well then $r=0$ and this is of no interest) \[ \exists \gamma_r \in \mathbb{Q}, \exists x_r, y_r \in \text{Im}\ g \text{ such that } r = x_r+ y_r \gamma_r, \]Let $y_r \in [T_1, T_2] \subset \text{Im}\ g$ be arbitrary.

It suffices to find a $\gamma_r \in \mathbb{Q}$ such that
\[ T_1 \leq r - y_r \gamma_r \le T_2 \]
But this can be rewritten as $\exists \gamma_r \in \mathbb{Q} \text{ such that } \frac{r-T_2}{y_r} \leq \gamma_r \leq \frac{r-T_1}{y_r}$, which is obviously true as $\mathbb{Q}$ is dense in $\mathbb{R}$.


Let $x,y \in \mathbb{R}$ be arbitrary with $x<y$. By the above, there exist $x_0, y_0 \in \text{Im}\ g, \alpha_0, \beta_0 \in \mathbb{Z}$ such that
\[ y - x = x_0 \alpha_0 + y_0 \beta_0 \]
Thus $f(y) = f(x + (y - x)) = f(x + x_0 \alpha_0 + y_0 \beta_0) \overset{(*)}{=} f(x)$, and so $f$ is constant.
This post has been edited 4 times. Last edited by RobertRogo, Mar 30, 2025, 1:02 PM
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Levieee
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Levieee
#4 h  Mar 30, 2025, 7:23 AM
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Mathzeus1024 wrote:
It works for $\textcolor{red}{f(x)=x}$ and $\textcolor{red}{f(x)=C}$ (where $C$ is a real constant).

but you have to proof that these are the only two functions that work
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