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