Trjtdtk

Let $$K=x=(x(n))_nin l_2(mathbbN):$$ and define $T:Kto c_0$ by $T(x)=(1-|x|_2,x(1),x(2),ldots)$. Prove :

(1) $T$ is self map on $K$ and $|Tx-Ty|_2le sqrt2 |x-y|_2$

(2) $T $ does not have fixed points in $K$

my attempt

for (2):

suppose $T$ have fixed point i.e., $Tx=x$

then $(1-|x|_2, x(1),x(2),ldots)=(x(1),x(2),ldots)$

then $x(1)=1-|x|_2, x(2)=x(1), x(3)=x(2),ldots$

$$therefore |x|_2 =left(sum ^n_n=infty |x(n)|^2right)^frac12 = left(sum ^n_n=infty (1-|x|_2)^2right)^frac12$$

but how to prove this $x$ is not in $K$?

how to prove (1)

For $(2)$ you got that if $Tx = x$ then $$x = (1-|x|_2, 1-|x|_2, ldots)$$

so $$+infty > |x|_2^2 = sum_n=1^infty (1-|x|_2)^2$$
The only way this series converges is if $1-|x|_2 = 0$, or $|x|_2 = 1$, so $x = (1,1,1ldots )$. But then clearly $|x|_2 = +infty$ and not $1$ so this is a contradiction.

Since $|x|_2 le 1$ we have $1-|x|_2 ge 0$ so

beginalign
|Tx|_2^2 &= (1-|x|_2)^2 + sum_n=1^infty |x_n|^2 \
&= (1-|x|_2)^2 + |x|_2^2 \
&le (1-|x|_2)^2 + 2(1-|x|_2)|x|_2 + |x|_2^2 \
&= (1-|x|_2+|x|_2)^2 \
&= 1
endalign

which means $|Tx|_2 le 1$.

Also clearly all coordinates of $$Tx = (1-|x|_2, x_1, x_2, ldots)$$
are nonnegative since $x_n ge 0, forall n in mathbbN$ and $1-|x|_2 ge 0$.

Therefore $Tx in K$.

You're on the right track.

As you said - $||x||_2 =(sum ^infty_n=1 |1-||x|||_2|^2)^frac12$

There is only one possible way for this sum to converge - if and only if $||x||_2=1$. But in this case, we also get $||x||_2=0$ - a contradiction.

About (1) - let's try evaluating the required norm:

$||Tx-Ty||_2=||(1-||x||_2,x(1),x(2),...)-(1-||y||_2,y(1),y(2),...)||_2\=||(||y||_2-||x||_2,x(1)-y(1),x(2)-y(2),...)||_2\=(|||y||_2-||x||_2|^2+sum ^infty_n=2 |x(n)-y(n)|^2)^frac12\=(|||y||_2-||x||_2|^2-|x(1)-y(1)|^2+sum ^infty_n=1 |x(n)-y(n)|^2)^frac12\leq(||x-y||_2^2-|x(1)-y(1)|^2+||x-y||_2^2)^frac12\=(2||x-y||_2^2-|x(1)-y(1)|^2)^frac12\ leq sqrt2||x-y||_2$

(Every $leq$ sign is due to triangle inequality)

It seems to me one may not need the assumption $x(n)geq 0$. In fact, if $Tx=x$,
then $|Tx|^2=|x|^2$ which says $(1-|x|)^2+x(1)^2+x(2)^2...=x(1)^2+x(2)^2+...$,
so we see $|x|=1$. Then $x(1)=0$. Let $n$ be the first integer with $x(n)neq 0$,
there exists such $n$ since $|x|=1$.
So $x(n-1)=0$, but the $n$-th component in $Tx$ is $x(n-1)=0$, contradicts
$Tx=x$ and $x(n)neq 0$.

Part 1: $|Tx-Ty|^2=[(1-|x|)-(1-|y|)]^2+[x(1)-y(1)]^2+...=(|x|-|y|)^2+|x-y|^2leq |x-y|^2+|x-y|^2=2|x-y|^2$ .