Let $f:[a,b]\to V$ be a Hilbert valued function which is $L^2$. By expanding the norm in terms of the inner product, I got the following identity. Is it true?
$\Vert \int_a^b f(s) ds\Vert^2=(b-a) \int_a^b \Vert f(s)\Vert^2 ds.$
Here is the calculation: \begin{align*} \Vert \int_a^b f(s) ds\Vert^2 &= \langle \int_a^b f(s) ds, \int_a^b f(s) ds\rangle \\ & = \int_a^b (\langle f(s), \int_a^b f(s) ds\rangle)ds\\ & = \int_a^b (\int_a^b \langle f(s), f(s)\rangle ds)ds\\ & =(b-a) \int_a^b \Vert f(s)\Vert^2 ds. \end{align*} Where is the mistake?
$\Vert \int_a^b f(s) ds\Vert^2 = \langle \int_a^b f(s) ds, \int_a^b f(t) dt\rangle \\ = \int_a^b (\langle f(s), \int_a^b f(t) dt\rangle)ds\\ = \int_a^b (\int_a^b \langle f(s), f(t)\rangle dt)ds\\$.
Further simplification is not possible.