# Finding quotient norm

I have to show that $\|(\alpha)+c_0\|_0=\limsup\limits_{n\to \infty}|\alpha_n|$ for all $(\alpha_n)\in \ell^{\infty}$.

We know that $\|(\alpha)+c_0\|_0=\inf\{\|(\alpha_n)+(x_n)\|:(x_n)\in c_0\}\leq \sup\limits_{n\in \mathbb N}|\alpha_n|$. I can't do anything else. Please suggest anything.

This can be proven by showing the two inequalities.

First, let $(x_n)\in c_0$. We have $$\limsup |a_n| \leq \limsup (|a_n|-|x_n|) \leq \limsup |a_n + x_n| \leq \sup |a_n + x_n| = \| (a_n)+(x_n) \|.$$ Taking the infimum over all $x_n \in c_0$ yields "$\geq$".

For the second inequality we can use $$\limsup_{k\to\infty} = \lim_{k\to\infty} \sup_{n\geq k} .$$

We choose $(x_n^{(k)}) \in c_0$ for each $k\in\mathbb N$ as follows: $$x_n^{(k)} = -a_n \mathbf 1_{n\leq k}$$

Then $$\|(a)+c_0\|_0 \leq \| (a_n) + (x_n^{(k)}) \| = \sup_{n\geq k} |a_n|$$

By taking the limit $k\to infty$ we get the desired inequality.

• How did you get $\limsup |a_n| \leq \limsup (|a_n|-|x_n|)$? Isn't it true that $|a_n|-|x_n| \leq |a_n|$? Consider $a_n = 1$ and $x_n = 1$. Nov 10, 2022 at 12:07