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Exercise :

Let $T:\ell^\infty \to \ell^\infty$ be an operator such that : $$T(x_1,x_2,\dots) = \bigg(\frac{1}{2}x_2,\frac{1}{3}x_3,\dots\bigg)$$ Show that for all $y \in \ell^\infty$, the equation $$x = y + Tx$$ has a unique solution.

Attempt :

I have proved that $T$ is a linear operator. Now, $\ell^\infty$ is the space defined as : $$\ell^\infty = \{x =(x_n) : \|x\|< \infty\} \quad \|x\| :=\sup|x_n|$$ From the definition of the norm over $\ell^\infty$, we can observe that

$$\|T(x_1,x_2,\dots)\|<\|(x_1,x_2,\dots)\|$$

This means that there exists an $M<1$, such that :

$$\|T(x_1,x_2,\dots)\|\leq M\|(x_1,x_2,\dots)\|$$

Thus, $T$ is a bounded linear operator $T \in B(\ell^\infty)$ with $\|T\| \leq M <1$.

Now, it is

$$x = y + Tx \Leftrightarrow x-Tx = y \Leftrightarrow(1-T)x=y$$

where $1$ is the identity operator.

But $\ell^\infty$ is a Banach space and since $\|T\| <1$, then it is :

$$(1-T)^{-1}=\sum_{n=0}^\infty T^n \Leftrightarrow (1-T)^{-1}y=\sum_{n=0}^\infty T^ny$$

Thus $x = \sum_{n=0}^\infty T^ny$ is a unique solution to the equation $x=y+Tx$ for all $y \in \ell^\infty$.

Question : Is my approach correct and rigorous enough ?

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  • $\begingroup$ You are correct and the proof looks sufficiently rigorous. However, $\|T\|$ can be computed exactly. That is, $\|T\|=\dfrac12$. $\endgroup$
    – Batominovski
    Nov 17, 2018 at 18:01
  • $\begingroup$ @Batominovski I guess that stems from the fact that $T(x_n) = \frac{x_{n+1}}{n+1}$ ? $\endgroup$
    – Rebellos
    Nov 17, 2018 at 18:04
  • $\begingroup$ Yes, see the answer below. $\endgroup$
    – Batominovski
    Nov 17, 2018 at 18:14
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    $\begingroup$ Oh, just one more thing. I think the proof of existence of $M$ is a bit muddly. I didn't see notice it at first. $\endgroup$
    – Batominovski
    Nov 17, 2018 at 18:18
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    $\begingroup$ The thing is there are some cases where you have an operator $T$ such that $\big\|T(x)\big\|< M\|x\|$ for all $x\neq 0$, but it turns out that $\|T\|=M$. See for example this thread: math.stackexchange.com/questions/2997663/…. The operator $\mathcal{T}$ in that example satisfies $\big\|\mathcal{T}(f)\big|<\|f\|$ for all nonzero $f\in\mathcal{C}$, but $\|\mathcal{T}\|=1$ (it is not proven there, but it is not too difficult to show that the equality does not happen unless $f\equiv 0$). $\endgroup$
    – Batominovski
    Nov 17, 2018 at 19:47

1 Answer 1

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You are correct and the proof looks sufficiently rigorous. However, $\|T\|_\text{op}$ can be computed exactly. That is, $\|T\|_\text{op}=\dfrac12$. To show this, let $z=(z_1,z_2,z_3,\ldots)\in\ell^\infty$. Then, $$T(z)=\left(\frac{z_2}{2},\frac{z_3}{3},\frac{z_4}{4},\ldots\right)$$ so that $$\big\|T(z)\big\|_{\infty}=\sup\left\{\frac{|z_k|}{k}\,\Big|\,k=2,3,4,\ldots\right\}\leq \sup\left\{\frac{\|z\|_\infty}{k}\,\Big|\,k=2,3,4,\ldots\right\}=\frac{\|z\|_\infty}{2}\,.$$ Note that the equality holds for $z=(0,1,0,0,0,\ldots)$. This implies $\|T\|_{\text{op}}= \dfrac{1}{2}$.

You can write an explicit solution $x\in\ell^\infty$ to $x=y+T(x)$. That is, $$x=(1-T)^{-1}y=\left(\sum_{k=1}^\infty\,\frac{y_k}{k!},\sum_{k=2}^\infty\,\frac{2!y_k}{k!},\sum_{k=3}^\infty\,\frac{3!y_k}{k!},\ldots\right)$$ Nonetheless, you did sufficient and good work. I was just making additional comments.

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