# What inequality/ property of $f$ is used here?

I'm doing the following exercise:

Let $f: [1,2] \to \mathbb{F}$ be continuous and $x_n = (e_{n,k})_ {k\in \mathbb{N}}$

$$e_{n,k} = \{ 0 \text{ if } 0 \leq k \leq n-1 \text{ or } k \geq 2n + 1, \ f(k/n), \text{ if } n \leq k \leq 2n.$$

Prove that $(x_n)_n$ coverges weakly to $0 \in c_0$ and that the sequence is norm convergent iff $f=0$

In the proof of $\Rightarrow$ i.e sequence norm convergent $\Rightarrow f =0$

$x_n \to 0 \Leftrightarrow \forall \epsilon >0, \ \exists N \in \mathbb{N} : \forall n> N \ \|x_n - 0 \|_{\infty} < \epsilon.$ But $\|x_n\|_{\infty} = \max_{k, \ n \leq k \leq 2n} |f(k/n)| \Rightarrow \max_{k, \ n \leq k \leq 2n} |f(k/n)| < \epsilon$.

Now since $f$is continuous on compact interval, f is uniformly continuous. Thus $\forall \epsilon > 0 \exists \delta : \forall x,x' \in [1,2]$ we have $|x-x'| < \delta \Rightarrow |f(x) - f(x')|<\epsilon.$ Not choose $N \in \mathbb{N} : \forall n > N$ we have $\frac1n<\delta.$ Thus $\forall x \in [1,2] \ \exists k \in \mathbb{N}$ s.t $n\leq 2k\leq 2n : |x-\frac{k}{n}| < \frac{1}{n} < \delta \ \color{red}{\Rightarrow} |f(x)| \leq |f(x-\frac{k}{n})| + |f(\frac{k}{n})| \leq 2\epsilon.$

The implication marked in $\color{red}{red}$ I dont understand.

• Seems like a typo. $$\lvert f(x)\rvert = \bigl\lvert f(\tfrac{k}{n}) + \bigl(f(x) - f(\tfrac{k}{n})\bigr)\bigr\rvert \leqslant \bigl\lvert f(\tfrac{k}{n})\bigr\rvert + \bigl\lvert f(x) - f(\tfrac{k}{n})\bigr\rvert \leqslant \epsilon + \epsilon$$ was probably intended. – Daniel Fischer Apr 15 '17 at 17:05
• That makes sense, actually proves the claim. Good work! @DanielFischer – Olba12 Apr 15 '17 at 17:10
• Basically, you need to see some way or another that $\lim\limits_{n\to \infty} \lVert x_n\rVert = \sup \{ \lvert f(x)\rvert : x \in [1,2]\}$. – Daniel Fischer Apr 15 '17 at 17:12