# Convergence in $\sup$ norm $\Rightarrow$ Cauchy in $\sup$

Let $(f_n)$ be a sequence of bounded functions on a set $E \subseteq \mathbb R$ and suppose that $f$ is a bounded function such that $\|f_n - f\|_{\infty} \to 0$ as $n \to \infty$. Prove that $(f_n)$ is a Cauchy sequence in the $\sup$ norm.

My Thoughts

Method 1:

Since $(f_n) \to f$ in the $\sup$ norm, we have $$\tag{*} \lim_{n \to \infty} \|f_n - f \|_{\infty} = 0$$ By a previous theorem, we have that $(*)$ is true iff for all $\epsilon >0$ there exists an $N$ so that for all $n \ge N$, $\|f_n - f\|_{\infty} \le \epsilon$.

Now, $$\tag{\dagger} \|f_n - f\|_{\infty} = \lim_{m \to \infty} \|f_n - f_m\|_{\infty}$$ My book brings a proof of the converse of this statement to this point and then essentially claims (in more formal language) "since the left hand side of $(\dagger)$ is equal to the right hand side, this implies $$\forall \epsilon > 0 \ \exists M\ \forall n, m > M \Big[\|f_n - f_m\|_{\infty} \le \epsilon\Big]$$

Is this faulty reasoning or not? If so, how else would I prove this?

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There's nothing special about the supremum norm here. In any metric space, a convergent sequence is a Cauchy sequence. It's just the triangle inequality: $d(f_n, f_m) \le d(f_n, f) + d(f, f_m)$.
I see. So since $(f_n)$ is convergent, I have that for $\epsilon >0$ there is an $N$, respectively an $M$, so that for $n \ge N$, $m \ge M$, we have $\|f_n -f \|_{\infty} < {\epsilon \over 2}$ and $\|f_m -f \|_{\infty} < {\epsilon \over 2}$. Therefore we have $\|f_n - f_m\|_{\infty} < \epsilon$. Thank you for this succinct explanation –  Zvpunry Aug 26 '12 at 22:27