# A question about fixed points and non-expansive map

Let $$K=\{x=(x(n))_n\in l_2(\mathbb{N}):\|x\|_2\le 1\ \text{ and } x(n)\ge 0 \text{ for all } n\in \mathbb{N} \}$$ and define $$T:K\to c_0$$ by $$T(x)=(1-\|x\|_2,x(1),x(2),\ldots)$$. Prove :

(1) $$T$$ is self map on $$K$$ and $$\|Tx-Ty\|_2\le \sqrt{2} \|x-y\|_2$$

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

$$\therefore \|x\|_2 =\left(\sum ^n_{n=\infty} |x(n)|^2\right)^\frac{1}{2} = \left(\sum ^n_{n=\infty} (1-\|x\|_2)^2\right)^\frac{1}{2}$$

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

how to prove (1)

• It seems that your calculation does not match the definition of $T$. – Song Mar 24 at 8:01
• @Song..now i edited correctly thank you – Inverse Problem Mar 24 at 8:13

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 $$x = (1,1,1\ldots )$$. But then clearly $$\|x\|_2 = +\infty$$ and not $$1$$ so this is a contradiction.

To show that $$T$$ is actually a map $$K \to K$$, take $$x \in K$$ and we claim that $$Tx \in K$$ as well.

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

\begin{align} \|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 \end{align}

which means $$\|Tx\|_2 \le 1$$.

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

Therefore $$Tx \in K$$.

• @mechanodroid...thank you so much but i have doubt with how to show $T$ is self map in question (1) – Inverse Problem Mar 24 at 10:27
• @InverseProblem Have a look. – mechanodroid Mar 24 at 10:42
• @mechanodroid......thank you so much ......for your help – Inverse Problem Mar 24 at 12:51
• @mechanodroid...can you give some hint this question please...math.stackexchange.com/questions/3159556/… – Inverse Problem Mar 24 at 13:09
• @InverseProblem I have added an answer, can you explain what exactly is $T$ in the definition of $T_n$? – mechanodroid Mar 24 at 15:48

You're on the right track.

As you said - $$||x||_2 =(\sum ^\infty_{n=1} |1-||x|||_2|^2)^\frac{1}{2}$$

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:

$$||Tx-Ty||_2=||(1-||x||_2,x(1),x(2),...)-(1-||y||_2,y(1),y(2),...)||_2\\=||(||y||_2-||x||_2,x(1)-y(1),x(2)-y(2),...)||_2\\=(|||y||_2-||x||_2|^2+\sum ^\infty_{n=2} |x(n)-y(n)|^2)^\frac{1}{2}\\=(|||y||_2-||x||_2|^2-|x(1)-y(1)|^2+\sum ^\infty_{n=1} |x(n)-y(n)|^2)^\frac{1}{2}\\\leq(||x-y||_2^2-|x(1)-y(1)|^2+||x-y||_2^2)^\frac{1}{2}\\=(2||x-y||_2^2-|x(1)-y(1)|^2)^\frac{1}{2}\\ \leq \sqrt2||x-y||_2$$

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

• ..can you tell me how to prove self map – Inverse Problem Mar 24 at 9:09
• I'm not familiar with that term, what does it mean? – GSofer Mar 24 at 9:11
• it means that we have to prove $T$ map from $K$ to $K$ – Inverse Problem Mar 24 at 9:13

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

• Ding....can you tell how to prove this is self map – Inverse Problem Mar 24 at 9:08
• From $\|x\|\leq 1$, the condition $x(n)\geq 0$ is preserved by $T$. Next, $\|Tx\|^2=(1-\|x\|)^2+\|x|^2\leq 1$, here one uses the elementary inequality $(1-a)^2+a^2\leq 1$ when $0\leq a\leq 1$, which can be checked by finding the maximum of $(1-a)^2+a^2$ on $[0, 1]$. – Yu Ding Mar 24 at 9:17
• what is this mean when u want to prove a map is self map it means we have to prove that map itself to a set\ – Inverse Problem Mar 24 at 9:19