# norm of Frechet derivative in point.

Let $E = \mathcal B([0,1], \Bbb R)$ with supremum norm.

Now I can define function $F:E \ni f \rightarrow ||f||^2 - f(0) \in \Bbb R$

1)Show the differentiability of $F$ in: $f_0: [0,1] \ni x \rightarrow [x] \in \Bbb R$;
2)Calculate the derivative in $f_0$ and it's norm;
3)Check, if subspace of $E$ : $H = \{ f \in E | f(0) = f(1) \}$ is a Banach space.

This is what I've done already:

$\lim _{||h|| \rightarrow 0 } \frac{| F (f_{0}+h) - F(f_0) - L(h)|}{||h||} =\lim _{||h|| \rightarrow 0 } \frac{| \ ||f_0+h||^2 - (f_0+h)(0) - ||f_0||^2 + f_0(0)- L(h)|}{||h||} = \lim _{||h|| \rightarrow 0 } \frac{| \ [1+ h(1)]^2 - h(0) - 1 - L(h)|}{||h||} = \lim _{||h|| \rightarrow 0 } \frac{| \ 1+ 2h(1) + [h(1)]^2 - h(0) - 1 - L(h)|}{||h||} = \lim _{||h|| \rightarrow 0 } \frac{| \ 2h(1) + [h(1)]^2 - h(0) - L(h)|}{||h||} = 0$
iff
(I think) $L(h) = 2h(1)-h(0)$, because $[h(0)]^2$ is too small ( something with $o(h)$ what I do not understand :( )

How to calculate the norm? I have no idea a.t.m.

About $H$: if I make a Cauchy sequence of functions ($n \in \Bbb N_+$):
$f_n(x) = \begin{cases}\frac{2^{n-1}}{2^n -1}x \text{, if } x\in[0, \frac{2^n -1}{2^n}] \\ 1-\frac{2^{n-1}}{2^n -1}x \text{, if } x \in (\frac{2^n -1}{2^n},1]\end{cases}$
This converges to a function:
$f(x)= \frac{1}{2}x$, where $f(0) = 0$ and $f(1)= \frac{1}{2}$
Is this done right?

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(1) Your derivative $L$ is correct. But you should add why you know that $\|f_0+h\| = 1+h(1)$. This is actually the better place to start:
If $\|h\|<1/2$, then $|f_0+h|<1/2$ on $[0,1)$ and $|f_0+h|>1/2$ at $1$. Hence, $\|f_0+h\|=f_0(1)+h(1)=1+h(1)$.
Then you see that $F(f_0+h) = 1+2h(1)+h(1)^2 -h(0)$ for all such $h$. The derivative is a linear approximation to $F(f_0+h)-F(f_0)$. Since $$F(f_0+h)-F(f_0)=2h(1)+h(1)^2 -h(0)$$ we take $L(h) = 2h(1) -h(0)$. The term $h(1)^2$, when divided by $\|h\|$, will have limit $0$ as $h\to 0$. This is precisely what the notation $h(1)^2 = o(h)$ means.
(2) The norm of linear function $L$ is $3$. On one hand, $$|L(h)|\le 2|h(1)|+|h(0)| \le 3\|h\| \tag{1}$$ On the other, there exist functions $h$ for which (1) turns into equality. Namely, $h(x)=2x-1$; or any function such that $h(1)=1$, $h(0)=-1$, and $\|h\|=1$.
(3) The sequence you have does not belong to $H$, because $f_n(1)\ne f_n(0)$. Moreover, no such counterexamples exist: the space $H$ is closed in $E$. To see this, note that $f\mapsto f(0)-f(1)$ is a bounded linear functional on $E$ (its norm is $2$). The kernel of a bounded linear functional is a closed subspace.