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In doing an exercise on functional analysis I am asked to come up with a linear bounded operator $T: \ell_0\rightarrow \ell_0$ (where $\ell_0=\{x=(x_1,x_2,...)\in\ell^\infty: \lim\limits_{n\rightarrow\infty}x_n=0\}$ is equipped with the sup norm $\| \cdot \|_\infty$) s.t. $$ \|Tx\|_\infty < \|T\| $$ holds for all $x\in\partial B_1(0)=\{x\in\ell_0: \|x\|_\infty=1\}$.

Obviously, an operator which in some way gets rid of all $x_i=\pm 1$ in $x=(x_1,x_2,...)$ would do the job because every $x\in\partial B_1(0)$ can only contain finitely many 1s. But every operator I have constructed so far fails at some property required for $T$. Either it's not linear, not bounded or $\mathrm{ran}\ T\not\subset\ell_0$...

Does someone have a hint what operators to consider?

Or do operators satisfying the necessary properties in the given space even exist (I do think that one exists because after all, the norm is a continuous function to $\mathbb R$ and the unit ball in $\ell_0$ is not compact so there is no reason for $\|Tx\|_\infty$ to attain its maximum in the unit ball)?

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  • $\begingroup$ It is more common to use $\mathcal{c}_0$ for the space you are describing. $L_0$ ($\ell_0(\mathbb{N})$ in your case) is use in analysis for the space of all measurable functions ($\ell_0$ would be the space of all sequences). $\endgroup$
    – Mittens
    May 14, 2023 at 17:03

2 Answers 2

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Hint : Consider a pointwise multiplication operator with a sequence $a_n$ with $|a_n| < 1$, $a_n \to 1$.

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Let $$(Tx)_n=\begin{cases}\displaystyle \sum_{k=1}^\infty {x_k\over 2^k} & n=1\\ \displaystyle \sum_{k=1}^\infty (-1)^k{x_k\over 2^k} & n=2\\ 0& n\ge 3 \end{cases} $$ Then $T$ maps $\ell^\infty$ into $\ell_0$ (actually into the sequences which vanish for $n\ge 3$) and $\|T\|=1,$ as $\sum_{k=1}^\infty 2^{-k}=1.$ However the norm of $\|T\|$ is not attained even in the unit ball of $\ell^\infty,$ hence neither in the unit ball of $\ell_0.$

Remark The operator can be simplified to $(Tx)_n=0$ for $n\ge 2.$ Then the norm is not attained in the unit ball of $\ell_0,$ but is attained in $\ell^\infty$ on the sequence with all terms equal $1.$

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