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?