Exercise: comparing two norms of a sequence

Question

Let $V:=\{(x_n)\in \ell^{\infty}(\mathbb R):x_0=0\}$ equipped with the norm $$\|(x_n)\|_{\infty}:=\sup_m|x_m|\qquad\forall(x_n)\in V.$$ Let the mapping $N:V\to\mathbb R$ be defined by $$N((x_n)):=\sup_{m\in\mathbb N}|x_{m+1}-x_m|\qquad\forall(x_n)\in V.$$

a) Show that $\|(x_n)\|_{\infty}\le2N((x_n))\quad\forall(x_n)\in V$.

b) Show equality holds in (a) for the sequence $(x_n)$ defined by: $x_0=0$ and $x_n=(-1)^n$ for every $n\ge1$.

Why should (a) be true when the sequence $(x_n):=\left\{\begin{array} &n&\text{if }n\le 10\\ 10&\text{otherwise} \end{array}\right.$ has $\|(x_n)\|_{\infty}=10$ but $2N((x_n))=2$?

Why should (b) be true when $\|(x_n)\|_{\infty}=1$ but $2N((x_n))=4$?

Answer 1

The inequality in a) is not correct !

We have $|x_{m+1}-x_m| \le |x_{m+1}|+|x_m| \le 2||(x_n)||_{\infty}$, thus

$N((x_n)) \le 2||(x_n)||_{\infty}$

Answer 2

This is probably a typo, since (a) indeed does not hold but $$N((x_n)) \leqslant 2 \left \| (x_n) \right \|_\infty\qquad\forall(x_n)\in V$$ does.

In order to see it, since $\left \| (x_n) \right \|_\infty = \sup_m|x_m|$, we have that $\forall(x_n)\in V$ $$-\left \| (x_n) \right \|_\infty\ \leqslant - x_n \leqslant \left \|(x_n) \right \|_\infty\ \qquad\forall n \in \mathbb{N}$$

Also $$-\left \|(x_n) \right \|_\infty\ \leqslant x_{n+1} \leqslant \left \|(x_n) \right \|_\infty\ \qquad\forall n \in \mathbb{N}$$

Adding the two up we get $$-2\left \|(x_n) \right \|_\infty\ \leqslant x_{n+1} - x_n \leqslant 2 \left \|(x_n) \right \|_\infty\ \qquad\forall n \in \mathbb{N}$$

meaning

$$|x_{n+1} - x_n| \leqslant 2 \left \|(x_n) \right \|_\infty\ \qquad\forall n \in \mathbb{N}$$

which gives

$$\sup_n|x_{n+1} - x_n| \leqslant 2 \left \|(x_n) \right \|_\infty$$ or equivalently $$N((x_n)) \leqslant 2 \left \| (x_n) \right \|_\infty$$

Now (b) is easy to verify, since $\left\| (x_n) \right\|_\infty = 1$ and $N((x_n)) = 2$.