operator norm of $f$ defined as follows $(f(u))_n = v_n = u_{n+1} - u_n\,$

Question

let $l^{\infty}$ be the vector space of bounded sequences equipped with the norm $\|u\| = \sup_{n \in \mathbb{N}} |u_n|$

and let $f$ be an endomorphism on $l^{\infty}$ defined as follows $f(u) = v$

where $v = (v_n)_{n \in \mathbb{N}}\, \text{and}\, v_n = u_{n+1} - u_n$

we have $\|f(u)\| = \sup_{n \in \mathbb{N}} |u_{n+1} - u_n| \leq 2 \sup_{n \in \mathbb{N}} |u_n| = 2\|u\|$

therefore $f$ is continuous

we also have $\|f\| \leq 2$

on the other hand, for $w_n = \frac{1}{n^k}$ we have $\frac{\|f(w)\|}{\|w\|} = \frac{2^k-1}{2^{k}}$

which for large $k$ approaches $1$

therefore $1 \leq \|f\| \leq 2$

this is the best approximation of the norm that I could get, any tips or help for computing the exact value of the norm are welcome. thanks !

Answer 1

You should try the sequence $(-1,1,0,0,0,0,0,\dots)$.