2
$\begingroup$

Let $K=\{x=(x(n))\in c_0:0\le x(n)\le 1$ for all $n\in \mathbb{N}\}$. Define $T:K\to c_0$ by $T(x)=(1,x(1),x(2),x(3),...).$ Prove :

(a) $T$ is a self map on $K$ and $||Tx-Ty||_\infty=||x-y||_\infty $

(b) If $0_{c_0}\in F$ and $T(F)\subseteq F.$ where $F$ is a closed convex subset of $K$ then $x_n=e_1+e_2+...+e_n\in F$, for all $n\in \mathbb{N}$

(c) $T$ does not have any fixed point in $K$

My try:

$||.||_\infty = \sup_{i \geq 1} |a_n|$

im trying to calculate $||Tx-Ty||_{\infty}=||((0,x_1-y_1,.....)||_{\infty}=\sup_{n\ge 1}|x_n-y_n||=||x-y||_\infty$ is i am correct?

for (c) if $Tx=x$ then $(1,x_1,x_2,...)=(x_1,x_2,...)$ then $x_1=1=x_2=x_3..$

but $x_n $ not in $c_0$

and i dont know how to prove that $T$ is self MAP and i dont know how to prove (b)

$\endgroup$
  • $\begingroup$ What are the $e_n$? Cheers! $\endgroup$ – Robert Lewis Mar 24 at 4:31
  • $\begingroup$ @RobertLewis.. i think $e_n$ are the like $e_n=(0,0,....1....)$ $\endgroup$ – Inverse Problem Mar 24 at 4:34
  • $\begingroup$ OK, thanks for the input! $\endgroup$ – Robert Lewis Mar 24 at 4:35
  • $\begingroup$ What is $c_0$? Infinite sequences with only finitely many non-zero elements? $\endgroup$ – Robert Shore Mar 24 at 5:00
  • $\begingroup$ yes and they converges to 0 $\endgroup$ – Inverse Problem Mar 24 at 5:01
1
$\begingroup$

Herein I take $c_0$ to be the sequences $(x(n))$, $n \in \Bbb N$, such that

$\displaystyle \lim_{n \to \infty} x(n) = 0; \tag 1$

that is, $c_0$ is the set of sequences which converge to $0$. I also take it that "$T$ is a self map on $K$" means that the range of $T$ lies in $K$; that is,

$T:K \to K. \tag 2$

These things being said, for

(a): if $x = (x(n)) \in K$, we are given that

$Tx = (1, x(1), x(2), \ldots ); \tag 3$

that is,

$(Tx)(1) = 1, \tag 4$

$(Tx)(n) = x(n - 1), \; \; 2 \le n \in \Bbb N; \tag 5$

since

$0 \le x(n) \le 1, \; \forall n \in \Bbb N, \tag 6$

we have

$0 \le T(x(n)) \le 1, \forall n \in \Bbb N \tag 7$

as well; thus,

$T(x(n)) \in K, \tag 8$

i.e, $T:K \to K$; $T$ is a self-map on $K$; also, with

$y = y(n) \in K, \tag 9$

$\Vert Tx - Ty \Vert_\infty = \Vert (0, x(1) - y(1), x(2) - y(2), \ldots ) \Vert_\infty$ $= \sup \{0, \vert x(n) - y(n) \vert, \; n \in \Bbb N \} = \sup \{ \vert x(n) - y(n) \vert, \; n \in \Bbb N \}, \tag{10}$

where this last equality binds by virtue of the fact that

$\vert x(n) - y(n) \vert \ge 0, \; \forall n \in \Bbb N; \tag{11}$

now

$\sup \{ \vert x(n) - y(n) \vert, \; n \in \Bbb N \} = \Vert x - y \Vert_\infty; \tag{12}$

combining (10) and (12) yields

$\Vert Tx - Ty \Vert_\infty = \Vert x - y \Vert_\infty; \tag{13}$

as for

(b): with

$0_{c_0} = (0, 0, 0 \ldots) \in F \tag{13}$

and

$T(F) \subseteq F, \tag{14}$

we have

$e_1 = (1, 0, 0, \ldots) = T0_{c_0} \in F, \tag{15}$

$e_1 + e_2 = (1, 1, 0, 0, \ldots) = Te_1 = T^20_{c_0} \in F, \tag{16}$

$e_1 + e_2 + e_3 = (1, 1, 1, 0, 0, \ldots) = T(e_1 + e_2) = T^30_{c_0} \in F; \tag{17}$

if we denote by $E_k \in K$ the sequence consisting of $k$ leading $1$s and every other element $0$, that is

$E_k(j) = 1, \; 1 \le j \le k, \tag{18}$

$E_k(j) = 0, \; j > k, \tag{19}$

we may formulate an inductive hypothesis

$e_1 + e_2 + \ldots + e_k = E_k = T(e_1 + e_2 + \ldots + e_{k - 1}) = T^k 0_{c_0} \in F, \tag{20}$

of which (15)-(17) are the first three cases $k = 1, 2, 3$; applying $T$ to this equation yields

$T(e_1 + e_2 + \ldots + e_k) = TE_k = T^2(e_1 + e_2 + \ldots + e_{k - 1}) = T^{k + 1}0_{c_0} \in F, \tag{21}$

where (14) implies the rightmost assertion

$T^{k + 1} 0_{c_0} \in F; \tag{22}$

it is evident from (20) and the definitions that

$TE_k = E_{k + 1} = E_k + e_{k + 1} = e_1 + e_2 + \ldots + e_k + e_{k + 1}, \tag{23}$

and we may use this to transform (21) into

$e_1 + e_2 + \ldots + e_k + e_{k + 1} = E_{k + 1} = T(e_1 + e_2 + \ldots + e_{k - 1} + e_k) = T^{k + 1} 0_{c_0} \in F, \tag{24}$

completing the induction; thus (20) holds for every $1 \le k \in \Bbb N$; we have thus completed the demonstration of part (b); finally, we turn to

(c): a fixed point $x \in K$ of $T$ satisfies

$T(x(n)) = x(n), \tag{25}$

whence

$T^2(x(n)) = TT(x(n)) = T(x(n)) = x(n), \tag{26}$

$T^3(x(n)) = TT^2(x(n)) = T(x(n)) = x(n); \tag{27}$

and indeed if

$T^k(x(n)) = x(n), \tag{28}$

we have

$T^{k + 1}(x(n)) = TT^k(x(n)) = T(x(n)) = x(n); \tag{29}$

thus we see inductivly that (28) binds for all $1 \le k \in \Bbb N$.

Now for

$(x(n)) \in K \subset c_0, \tag{30}$

for any $\epsilon > 0$ there exists

$0 < N \in \Bbb N \tag{31}$

such that

$j > N \Longrightarrow x(j) < \epsilon; \tag{32}$

now if we choose $k$ in (20) with

$k > N, \tag{33}$

we see by means of (18)-(19) that

$\exists n > N, \; E_k(n) = 1; \tag{34}$

but this contradicts (32); thus $T$ has no fixed points in $K$.

Note: Apparently we do not need the hypotheses that $F$ is closed and convex to attain (b). End of Note.

To Prove $T $ is a self map and $T$ have no fixed points

$\endgroup$
2
$\begingroup$

Let $x=(x_n)$ be a sequence that converges to $0$ with $\max_n x_n \leq 1.$ Let $y=(y_n)=T(x)$ Then $x_n=y_{n+1}$ and $y_0=1$, so $\max_n y_n =1$ and $\lim_{n\to \infty} y_n=\lim_{n \to \infty} x_n=0$, so $T$ maps $K$ to $K$.

If $0 \in F$ and $T(F) \subseteq F$, then $T^n(0)=\sum_{k=1}^n e_k \in F$.

$\endgroup$
  • $\begingroup$ how can we say that $T^n(0)=\sum_{k=1}^n e_k \in F$ $\endgroup$ – Inverse Problem Mar 24 at 5:20
  • $\begingroup$ Because you know the formula for $T$. Work out the first few examples and you’ll see what happens. All T does is shift the entire sequence one position to the right, and then fills in the first coordinate with a $1$. $\endgroup$ – Robert Shore Mar 24 at 5:23
  • $\begingroup$ thank you so much $\endgroup$ – Inverse Problem Mar 24 at 5:25
  • $\begingroup$ ..i have one more question can i ask you $\endgroup$ – Inverse Problem Mar 24 at 5:32
  • $\begingroup$ Just ask it. Even if I’m not around, someone else probably will be. $\endgroup$ – Robert Shore Mar 24 at 6:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.