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Let $\lim_{n \rightarrow \infty} a_n = a$ and $\lim_{n \rightarrow \infty} b_n = b$ exist, then is it true that $\lim_{n \rightarrow \infty} \max \{ a_n, b_n \} = \max \{ a, b \}$?

I couldn't find this on wiki, but it seems correct. Here's my proof, can somebody check it?

Suppose $a > b$. Let $\epsilon^{\prime} > 0$. Take $0 < \epsilon < \min \{ \epsilon^{\prime}, (a - b) / 2 \}$. Then there is an $N$ s.t. $n > N \Rightarrow | a - a_n | < \epsilon \wedge | b - b_n | < \epsilon \Rightarrow a_n - b_n > (a - \epsilon) - (b + \epsilon) > 0 \Rightarrow | \max \{ a_n, b_n \} - a | = | a_n - a | < \epsilon < \epsilon^{\prime}$.

Suppose $c := a = b$. Let $\epsilon > 0$. Then there is an $N$ s.t. $n > N \Rightarrow | c - a_n | < \epsilon \wedge | c - b_n | < \epsilon \Rightarrow | c - \max \{ a_n, b_n \} | < \epsilon$.

simonzack
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    Very nicely proven. I would just change the third paragraph from "suppose $a>b$" into "suppose $a\neq b$. Then, without loss of generality, $a>b$", but that's almost nitpicking. – 5xum Mar 27 '15 at 12:38

2 Answers2

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For posterity: On the set of real numbers, $\max$ and $\min$ are given by $$ \max(a, b) = \frac{a + b + |b - a|}{2},\qquad \min(a, b) = \frac{a + b - |b - a|}{2}. $$ (Proof: "Go to the midpoint $\frac{1}{2}(a + b)$, then step to the right or left by half the distance between the numbers, $\frac{1}{2}|b - a|$.")

These functions are (almost obviously) continuous as functions of two variables, so $$ \lim_{n \to \infty} \max(a_{n}, b_{n}) = \max\bigl(\lim_{n \to \infty} a_{n}, \lim_{n \to \infty} b_{n}\bigr) = \max(a, b), $$ and similarly for $\min$.

In fact, continuity guarantees stronger "double limit" assertions: $$ \max(a, b) = \lim_{m \to \infty} \lim_{n \to \infty} \max(a_{m}, b_{n}) = \lim_{n \to \infty} \lim_{m \to \infty} \max(a_{m}, b_{n}), $$ etc.

Andrew D. Hwang
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A kind of overkill: fix $K$ compact (in this case $K=\{1,2\}$) and consider the Banach space $(\mathcal{C}(K),\|\cdot\|_\infty)$. Since the norm is continuous, then $\lim_n \|f_n\|_\infty=\|\lim_n f_n\|_\infty$ for each sequence $(f_n)$ in $\mathcal{C}(K)$, provided the limits exist.

To conclude the proof, use that $x_n \to x$ iff $x_n^+ \to x^+$ and $x_n^- \to x^-$, where $a=a^+-a^-$ is the unique decomposition into positive and negative part.

Paolo Leonetti
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  • Let $K = [0,1]$. Let $f_n(x) = nx$ for $x \leq \frac{1}{n}$, $f_n(x) = 2-nx$ for $\frac{1}{n} \leq x \leq \frac{2}{n}$, and $f_n(x) = 0$ for $x \geq \frac{2}{n}$. $f_n$ converges to $0$ pointwisely on $K$, but $||f_n||_{\infty} = 1 \neq 0$. Is this a counter-example to what you mentioned? – zxzx179 Aug 25 '19 at 01:14
  • @Stupid_Guy Nope. Your sequence $(f_n)$ is not $|\cdot|_\infty$-convergent to $f$. – Paolo Leonetti Aug 25 '19 at 12:08