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We have a sequence of functions $f_n$ that converge to $f$.

As part of a proof I am writing, if I can prove that

$$|f(x) - f(y)| = \lim_{n \to \infty}|f_n(x) - f_n(y)|$$

then I am done. Proving this requires me to show that

$$|\lim (f_n(x) - f_n(y)) | = \lim |f_n(x) - f_n(y)|$$

and I'm not sure how I can do that. I found this answer limit of absolute value of something equals absolute value of limit of something? which says it exists by 'continuity of absolute value' although I am not sure how that justifies it. Could someone elaborate?

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We note that if $x_n \to x$ and $g$ is continuous, then: $$ \lim_{n \to \infty} g(x_n) = g\left(\lim_{x \to \infty} x_n\right) $$ i.e. the limit and the function can be interchanged. The absolute value function $x \mapsto |x|$ is continuous. Therefore, in your case we have: \begin{align*} |f(x) - f(y)| &= \left|\left(\lim_{n \to \infty} f_n(x)\right) - \left(\lim_{n \to \infty} f_n(y)\right)\right| \\ &= \left|\lim_{n \to \infty} (f_n(x) - f_n(y))\right| \\ &= \lim_{n \to \infty} |f_n(x) - f_n(y)| \end{align*}

Clement Yung
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    Thanks for your answer. Just to make sure, the "note that" statement follows from definition of continuity right? –  Nov 20 '21 at 03:27
  • @a6623 There are many different definitions of continuity, but that statement is a simple consequence of most of these definitions anyway. – Clement Yung Nov 20 '21 at 03:29