# Is this operator steady? retort

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

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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