could anyone help me calculating the derivative of this function:
$G:C[0,1]\rightarrow\mathbb{R},x\mapsto \left(\int_0^1 x(s)ds\right)^2$
I tried using the Gateaux Differential - working out
$\frac{1}{s}(G(u+sh)-G(u)$ to then go $s\mapsto 0$
but this leads nowhere. I have the feeling this needs some "trick" regarding the squared integral. Could I write this differently?
Thank you in advance.
Setting $G(x) = \left( \int_0^1 x(s)\,ds \right)^2$ gives $$ G(x+h\xi) = \left( \int_0^1 (x(s) + h\xi(s))\,ds \right)^2 = \left( \int_0^1 x(s)\,ds + h \int_0^1 \xi(s)\,ds \right)^2 \\= \left( \int_0^1 x(s)\,ds \right)^2 + 2h \left( \int_0^1 x(s)\,ds \right)\left( \int_0^1 \xi(s)\,ds \right) + h^2 \left( \int_0^1 \xi(s)\,ds \right)^2 $$ so $$ \frac{G(x+h\xi) - G(x)}{h} = 2 \left( \int_0^1 x(s)\,ds \right)\left( \int_0^1 \xi(s)\,ds \right) + h \left( \int_0^1 \xi(s)\,ds \right)^2 \\ \to 2 \left( \int_0^1 x(s)\,ds \right)\left( \int_0^1 \xi(s)\,ds \right) $$ as $h\to0.$
Thus, $$ \lim_{h\to0} \frac{G(x+h\xi) - G(x)}{h} = 2 \left( \int_0^1 x(s)\,ds \right)\left( \int_0^1 \xi(s)\,ds \right) . $$