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In this proof of the completeness of $(C(K), \| \cdot \|_\infty)$ they use the following inequality:

$ \sup_{x \in K} \lim_{m \to \infty} | f_n(x) - f_m(x) | \leq \liminf_{m \to \infty}\, \sup_{x \in K} | f_n(x) - f_m(x) |.$

Can someone explain to me why this is true? Presumably, they first write $ \sup_{x \in K} \lim_{m \to \infty} | f_n(x) - f_m(x) | = \sup_{x \in K} \,\liminf_{m \to \infty} | f_n(x) - f_m(x) |$ and then swap but the exact argument why this gives the inequality is unclear to me.

Many thanks for your help.

1 Answers 1

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We have for all $x\in K$, $m,n$ integers: $|f_n(x)-f_m(x)|\leq\sup_{x\in K}|f_n(x)-f_m(x)|$ and taking the $\liminf$: $\forall n\in\mathbb N,\forall x\in K\quad\liminf_m|f_n(x)-f_m(x)|\leq\liminf_m\,\sup_{z\in K}|f_n(z)-f_m(z)|,$ and now taking the supremum over $x\in K$ we get $\forall n\in\mathbb N: \quad\sup_{x\in K}\,\liminf_m|f_n(x)-f_m(x)|\leq\liminf_m\,\sup_{z\in K}|f_n(z)-f_m(z)|,$ hence $\forall n\in\mathbb N: \quad\sup_{x\in K}\,\lim_m|f_n(x)-f_m(x)|\leq\liminf_m\,\sup_{z\in K}|f_n(z)-f_m(z)|,$ since we got the convergence of $\{f_n(x)-f_n(x)\}_m$.