Finding a solutions for an equation

Is this operator steady? retort

Hello expensive customer to our community We will proffer you an answer to this query Is this operator steady? ,and the retort will breathe typical by way of documented info sources, We welcome you and proffer you recent questions and solutions, Many customer are questioning concerning the retort to this query.

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

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?

we are going to proffer you the answer to Is this operator steady? query by way of our community which brings all of the solutions from a number of reliable sources.

Add comment