# Example of an operator which does not attain its operator norm in the unit ball in $\ell_0\subset\ell^\infty$ (set of zero sequences)

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)?

• 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). May 14, 2023 at 17:03

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