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If $f:\mathbb{R}\rightarrow \mathbb{R}$ be a continuous function such that

$f(f(x))=1+x. $ Then $f'(0)$ is

what i try: $$f(f(x))=1+x$$

Replace $x\rightarrow f(x)$

$$f(f(f(x)))=1+f(x)$$

$$f(1+x)=1+f(x)$$

How do i solve it , Help me please

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jacky
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    $f(x)=x + \frac12$ satisfies the proposed relation. If the answer is the same for all functions satisfying that relation, it should be $f'(0)=1$. – PierreCarre Sep 26 '19 at 09:55
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    You're using both $f$ and $g$. Presumably they are the same? – Arthur Sep 26 '19 at 10:20
  • @jacky Can you show that $f$ is differentiable? In general, the regularity of $f\circ f$ may not transfer to $f$... If you think of the Dirichlet function $D(x)$, you have the continuity of $D(D(x)) = 1, \forall x$ but not of $D(x)$. – PierreCarre Sep 26 '19 at 10:42

2 Answers2

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This is a partial answer.

If $x$ is an integer, say $x = n$, the last relation in the OP is a linear difference equation that can be solved to obtain $f(n) = c + n$. Constant $c$ can be computed so that $f(f(n)) = 1 + n$, yielding $f(n) = n + \frac 12$. So, any such function $f$ must be linear, at least over the integers. Naturally, $f(x)=x + \frac 12$ is one possibility. Are there others? The answer is: apparently yes (see here).

It is easy to see that $f$ must be injective and, in fact, increasing. If we assume that $x_1 \ne x_2$ and $f(x_1)=f(x_2)=\alpha$ we obtain a contradiction ($f(\alpha)$ would have to be simultaneously $1+x_1$ and $1+x_2$).

Thinking about how such function would look like (periodic, increasing, coinciding with $x+\frac 12$ over the integers), I'm inclined to conclude that either $f$ is not differentiable at integer points or, being differentiable, its graphic must be tangent to $x+\frac 12$.

My conjecture is that if $f$ is differentiable at $x=0$, the derivative is $f'(0)=1$.

PierreCarre
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As long as the solution to

$$ f(x+1) = f(x)+1 $$

is

$$ f(x) = x +\Phi(x) $$

where $\Phi(x)$ is any periodic function with period $1$ we have

$$ f'(x) = 1 + \Phi'(0) $$

assuming $f(x)$ continuous at $x = 0$

Cesareo
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