Finding a solutions for an equation

Is this operator steady? retort

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Is this operator steady?

Let $I=[0,1]$ and $E$ a Banach house. We level to by $X:=mathcal {C}(I,E), $ the house of all steady features from $I$ to $E$, with $left | x capable |_X=sup_{tin I }left | x(t) capable |_E

Let $f:Itimes Erightarrow E$ an obligation such that:

  • For every steady $xin X$, we maintain $f(.,x(.))$ is Pettis
    integrable on $I$,

  • for each $t in I,:: f_t: E rightarrow E,:u mapsto f_t(u):=f(t,u) textual content{ is steady}.$

Let $$T: X rightarrow X,:x mapsto T(x)(t):=int_{0}^{t}f(s,x(s)) ds$$

pretense: $T$ is steady.

This is how I attempted to decipher this:

For $tin I,:f_t$ is steady, that’s,

for every $uin E$, $forall epsilon>0 , exists eta_{t,u,epsilon}>0 textual content{ such that } forall vin E$ $$left |u-v capable | leq eta_{t,u,epsilon} Rightarrow left | f(t,u)-f(t,v) capable | < epsilon

Now, let $tin I$, $epsilon >0$ , and $xin X$. Let $yin X$ such that $$left | x-y capable |_Xleq eta_{t,x(t),epsilon};,$$

i.e. $$forall error I,:left | x(s)-y(s) capable |_{Etimes E}leq eta_{t,x(t),epsilon};,$$
specifically, $$left | x(t)-y(t) capable |_{Etimes E}leq eta_{t,x(t),epsilon};.$$

Hence, $$left | f(t,x(t))-f(t,y(t)) capable | < epsilon quad(*)

So, $$commence{matrix}
left | T(x)(t)-T(y)(t) capable | & = &left | int_{0}^{t} f(s,x(s))-f(s,y(s)) dsproper |
& leq & int_{0}^{t} left | f(s,x(s))-f(s,y(s)) dsproper | quad(**)

sadly, I can not make use of $(*)$ in $(**)$ as a result of it $(*)$ not uniformaly on $t$.

Is our pretense correct? why?

If not, what’s the situation on $f_t$ that you just indicate as an alternative of continuity?

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