Is this operator continuous? Let $I=[0,1]$ and $E$ a Banach space. We note by $X:=\mathcal {C}(I,E), $ the space of all continuous functions from $I$ to $E$, with $\left \| x \right \|_X=\sup_{t\in I }\left \| x(t) \right \|_E
$.
Let $f:I\times E\rightarrow E$ a function such that:

*

*For each continuous $x\in X$, we have $f(.,x(.))$ is Pettis
integrable on $I$,


*for every $t \in I,\:\: f_t: E \rightarrow E,\:u \mapsto f_t(u):=f(t,u) \text{ is continuous}.$
Let $$T: X \rightarrow X,\:x \mapsto T(x)(t):=\int_{0}^{t}f(s,x(s)) ds$$
Claim: $T$ is continuous.

This is how I tried to solve this:

For  $t\in I,\:f_t$ is continuous, that is,
for each $u\in E$, $\forall \epsilon>0 , \exists \eta_{t,u,\epsilon}>0 \text{ such that } \forall v\in E$ $$\left \|u-v \right \| \leq \eta_{t,u,\epsilon} \Rightarrow \left \| f(t,u)-f(t,v) \right \| < \epsilon
$$

Now, let $t\in I$, $\epsilon >0$ , and $x\in X$. Let $y\in X$ such that $$\left \| x-y \right \|_X\leq \eta_{t,x(t),\epsilon}\;,$$
i.e. $$\forall  s\in I,\:\left \| x(s)-y(s) \right \|_{E\times E}\leq \eta_{t,x(t),\epsilon}\;,$$
in particular, $$\left \| x(t)-y(t) \right \|_{E\times E}\leq \eta_{t,x(t),\epsilon}\;.$$
Hence, $$\left \| f(t,x(t))-f(t,y(t)) \right \| < \epsilon \quad(*)
$$
So, $$\begin{matrix}
\left \| T(x)(t)-T(y)(t) \right \| & =   &\left \| \int_{0}^{t} f(s,x(s))-f(s,y(s)) ds\right \| \\ 
 & \leq & \int_{0}^{t} \left \| f(s,x(s))-f(s,y(s)) ds\right \| \quad(**)\\
\end{matrix}$$
unfortunately, I can't use $(*)$ in $(**)$ because it $(*)$ not uniformaly on $t$.

Is our claim true? why?
If not, what is the condition on $f_t$ that you suggest instead of continuity?
 A: I do not have a counterexample, but a strong feeling that the conjecture is false, based on the following positive proof.
If you require slightly more, namely Lebesgue integrability of $t\mapsto f(t,x(t))$ for every $x\in Z=L_\infty([0,1],E)$, then it is well-known that the superposition operator $F(x)(t)=f(t,x(t))$ is “automatically” continuous from $Z$ into $Y=L_1([0,1],E)$ (because $f$ is a Carathéodory function, and $Y$ is a regular ideal space), see e.g. Theorem 5.2.1 in M. Väth, Ideal spaces, Springer, 1997.
This implies the continuity of $T\colon Z\to X$, because if $x_n\to x$ in $X$ then $x_n\to x$ in $Z$, and so
$$\lVert T(x_n)-T(x)\rVert_X\le\lVert F(x_n)-F(x)\rVert_Y\to0\text.$$
I am not sure whether the automatic continuity of $F$ holds also in case $Z=C([0,1],E)$, but I recall that such a result was cited in case $E=\mathbb R$, and I was unable to access it. However, probably the above mentioned proof can be modified straightforwardly for that case.
Anyway, the proof in the “simpler” case $Z=L_\infty([0,1],E)$ already depends on a subtle application of Vitali's convergence theorem (much more subtle than the straightforward attempt in the question), and it is hard to imagine that there is any proof which comes without using some sort of equi-integrability of the sequence $F(x_n)$. Since the latter is exactly what you lack for Pettis integrability, I would be very surprised if the assertion would be true.
