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

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)

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

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

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 $$T^n(0)=\sum_{k=1}^n e_k \in F$$.

• how can we say that $T^n(0)=\sum_{k=1}^n e_k \in F$ – Inverse Problem Mar 24 at 5:20
• 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 $1$. – Robert Shore Mar 24 at 5:23
• thank you so much – Inverse Problem Mar 24 at 5:25
• ..i have one more question can i ask you – Inverse Problem Mar 24 at 5:32
• Just ask it. Even if I’m not around, someone else probably will be. – Robert Shore Mar 24 at 6:42