let $$T:(C([0,1]),\lvert \lvert . \rvert\rvert_\infty) \rightarrow \mathbb R, \ f \rightarrow \int_0^1f(s)^2 \,ds $$
$(i)$ Calculate the directional derivative of $T$ in direction $g \in C([0,1])$ and point $f \in C([0,1])$
$(ii)$ What difference it makes, if we change the norm $\mid \lvert.\rvert\rvert_\infty$ with another norm?
For (i) i tried to use $$\lim_{t \to 0} \frac {1}{t} . \left(T(f+gt)-T(f)\right)$$
so i did :
$$\lim_{t \to 0} \frac {1}{t} \left(\int_0^1 f+gt(s)^2\,ds- \int_0^1 f(s)^2\,ds)\right)$$ $$=\lim_{t \to 0} \frac {1}{t} \left(\int_0^1 f(s)^2 \,ds+ \int _0^1 gt(s)^2\,ds- \int_0^1 f(s)^2\,ds\right)$$ $$=\lim_{t \to 0} \frac {1}{t}\left(\int _0^1 t\cdot g(s)^2\,ds\right)$$ $$= \int _0^1 g(s)^2\,ds$$
Is it correct, what i did here?
Does it make any difference if we change the norm as in (ii) ? I think it does not make any difference because $\int_0^1g(s)^2\,ds \in \mathbb R$ so it is not important what the norm of $C([0,1])$ is. But i am not sure !!
In order to compute the Gateaux derivative, note that for any $f,g \in C[0,1]$ we have
\begin{align} \lim_{t \to 0} \frac{1}{t} (T(f+ tg)-T(f)) &= \lim_{t \to 0} \frac{1}{t} \int_0^1 (f + tg)^2(s) ds - \int_0^1 f^2(s) ds \\ &= \lim_{t \to 0} \frac{1}{t} \int_0^1 f^2(s) + 2 t f(s) g(s) + t^2 g^2(s) - f^2(s) ds \\ &= \lim_{t \to 0} \frac{1}{t} \int_0^1 2 t f(s) g(s) + t^2 g^2(s) ds \\ &= \lim_{t \to 0} \frac{1}{t} \int_0^1 2 t f(s) g(s) ds + \int_0^1 t^2 g^2(s) ds \\ &= 2 \int_0^1 f(s) g(s) ds + \lim_{t \rightarrow 0} t \underbrace{\int_0^1 g^2(s) ds}_{< \infty} \\ &= 2 \int_0^1 f(s) g(s) ds + 0 \\ &= 2 \int_0^1 f(s) g(s) ds = 2 \langle f,g \rangle_{L^2} \end{align}
Regarding the Gateaux derivative, I am not sure what difference should occur when the norm is changed, since Gateaux derivative only depends on the norm on the target - here $(\mathbb{R}, \vert \cdot \vert)$.
However, it may have an influence on the existence of the Frechet derivative.