Show that $\varphi \in E'$ and if $E$ is a Banach space then $\varphi \in E$ 
Problem: Let $E$ be a normed space over field $\mathbb{C}$. Fix a continuous function $f: \left[ a,b \right] \rightarrow E$ with $\left[ a,b \right] \subset \mathbb{R}$. Consider $\varphi: E' \rightarrow \mathbb{C}$ given by $\varphi(y) := \displaystyle\int_a^b (y \circ f)(t)dt, \forall y \in E'$. Show that $\varphi \in E''$ and if $E$ is a Banach space then $\varphi \in E$, in which $E'$ is the dual space of $E$.

Could you give me some hint to solve the problem.
 A: If the integral is a Lebesgue integral then 
$$
\phi(y) = \int_a^b (y\circ f)(t) dt = y\left( \int_a^b f(t)dt \right) ,
$$
where the integral on the right is a Bochner integral. Then $\int_a^b f(t)dt\in E$ is the element you are looking for.
A: $y\mapsto \varphi(y)=\int_a^b(y\circ f)(t)dt$ is linear by the linearity of the integral. It is continuous because $y_n\to y$ in $E'$ means uniform convergence on the unit ball of $E$ and hence on every multiple on the unit ball, and continuity of $f:[a,b]\to E$ implies that the range is compact and hence bounded in $E$. This shows that $\varphi\in E''$.
To show that $\varphi \in E$, if $E$ is complete, you can either use that the dual of $(E',\tau)$ is $E$ (where $\tau$ is the topology of uniform convergence on absolutely convex compact sets -- note that the argumet for the continuity of $\varphi$ actually shows that it is $\tau$-continuous), or you can show more directly that $\varphi$ is in the closure of $E$ in $E''$. This could be done by approximating $f$ uniformly on $[a,b]$ by linear combinations of functions of the form $t\mapsto I_{[\alpha,\beta]}(t) e$ for $e\in E$ and $a\le \alpha\le \beta\le b$ (these are the $E$-valued step functions for which you can calculate the corresponding $\varphi$ explicitely).
