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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)

mechanodroid
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Inverse Problem
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3 Answers3

2

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)

GSofer
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2

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
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1

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

Yuval
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  • Ding....can you tell how to prove this is self map – Inverse Problem Mar 24 '19 at 09:08
  • 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 $[0, 1]$. – Yuval Mar 24 '19 at 09:17
  • 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 '19 at 09:19