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Let $E$ be a real Banach space included in the space of functions $C^1$ of $\mathbb R\to \mathbb R $ and and $T:E\to E$ defined by $T(f)=f\circ f$. I am looking for an example of space E such that T is differentiable on E.

By developing, $T(f+h)$ we find

$$T(f(x)+h(x))=f(f(x))+h(f(x))+f'(f(x)).h(x)+h'(f(x)).h(x)+o(h(x))$$ but the term $h'(f(x)).h(x)$ is not linear on h and it seems that $DT_{f}\left(h\right)=\left(f'\circ f\right)\times h+\left(h\circ f\right)$ but we need a space where $(h'\circ f).h=o(h)$

I thought of the subspace of $C^{\infty}(\mathbb R, \mathbb R)$ formed of bounded functions of which all derivatives are bounded, provided with the norm $f\mapsto \|f\|_{\infty} + \|f'\|_{\infty}$ but this space is not complete space

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As suggested by Giorgio Metafune , I start with Gateaux differentiability . Let $E=C_b$ (bounded and continuous functions with the sup-norm). Let $g\in E$ and f be a function derivable on $\mathbb R$. By developping $$ (T(f+tg)-T(f))(x)=f(f(x)+tg(x))+tg(f(x)+tg(x))-f(f(x)) $$ then $$\lim_{t\to0}\frac{(T(f+tg)-T(f))(x)}{t}=\lim_{t\to0}\frac{f(f(x)+tg(x))-f(f(x)) }{t}+g(f(x))\\= f'(f(x)).g(x)+g(f(x)) $$

Then $$\qquad D_G(T)f(g)=g (f'\circ f)+g\circ f$$ and $g\to D_G(T)f(g)$ is linear. It's must be continuous to deduce Frechet differentiability

Update question Giorgio Metafune suggests considering $E=C_0(\mathbb R)$

Let $E=C_0(\mathbb R)$ and $T: E\to E$ defined by $T(f)=f\circ f$. Show ( if true) that T is Fréchet differentiable at all $f\in C^1(\mathbb R)$. Can some one help me or give answer

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    $\begingroup$ Is there anything preventing you from using finite dimensional examples, like the subspace of linear functions? $\endgroup$ Commented Aug 14, 2020 at 22:55
  • $\begingroup$ the linear case no interesting to me $\endgroup$
    – Paul
    Commented Aug 15, 2020 at 10:03
  • $\begingroup$ Why not use $C^1_b$, the space of all bounded diferentiable functions, with bounded derivatives, with the norm you wrote in the last line? $\endgroup$ Commented Aug 15, 2020 at 16:05
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    $\begingroup$ If by $C^1_b$ we understand the space of bounded and differentiable functions with bounded first derivatives, it Is complete under your norm. $\endgroup$ Commented Aug 15, 2020 at 22:05
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    $\begingroup$ For reasons already pointed out, you would need to control the norm of the derivative by the norm of the function (to deal with the $h'(f(x))$ term). This essentially rules out most of the usual function spaces. You can come up with examples in spaces of analytic functions. For instance, if you consider bounded analytic functions on the disk, and f maps to a smaller disk, you can indeed bound $h'(f(x))$ by the norm of h. $\endgroup$ Commented Aug 26, 2020 at 3:08

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