Suppose $x_n \rightarrow x$ and $y_n \rightarrow y$. Show that $\min\{x_n, y_n\} \rightarrow \min\{x, y\}$

I'm not really sure how to show this. If $x_n$ converges to $x$ and $y_n$ converges to $y$, then wouldn't it already trivially follow that $\min\{x_n, y_n\}$ converges to $\min\{x, y\}$?

I know that I'm missing something, and any hints would be appreciated.

  • $\begingroup$ This is not trivial at all. Consider separately the three cases $x<y$, $x=y$ , $x>y$. $\endgroup$ – Crostul Apr 5 '17 at 10:45
  • $\begingroup$ @Crostul well in each of the cases the min would be x, x or y, and y respectively, but where does the convergence come in? The reason I currently see it as trivial is because with the way I'm conceptualizing it right now $x_n$ and $x$ and $y_n$ and $y$ are practically interchangeable so the min will always be the same in all those cases. $\endgroup$ – abc Apr 5 '17 at 10:50

You just need a nice statement about the min, and you are done.

Lemma : $\min\{ a,b\} = \dfrac{a+b - |a-b|}{2}$.

Proof : If $a \geq b$, then $|a-b| = a-b$, so that $a+b-a+b = 2b$. Otherwise, $a < b$, so $|a-b| = b-a$, and we have $a+b + a - b = 2a$. After division by $2$, we see this corresponds to the minimum function.

Note that $|\min\{x_n,y_n\} -\min\{x,y\}|= \left|\frac{(x_n + y_n - |x_n - y_n|) - (x + y - |x-y|)}2 \right|$.

We use triangle inequality: $$ \left|\frac{(x_n + y_n - |x_n - y_n|) - (x + y - |x-y|)}2 \right| \leq \left|\frac{x_n-x}2\right| + \left|\frac{y_n-y}2\right| + \left|\frac{|x_n-y_n| - |x-y|}2\right| $$

Observe the terms on the right hand side. The first can be made small, since $x_n \to x$. The second can be made small, since $y_n \to y$. The third canbe made small, since $x_n - y_n \to x-y$.

Hence, we have convergence.

Inductively, I encourage you to show this corollary:

For any finite of number of convergent sequences $s_1...s_n$, $\min_{1 \leq i \leq n}\{s_i\} $ will converge to the minimum of the limits.

Alternately, you can also show this for maximum, because for the maximum of two quantities, there is a similar formula: $$ \max\{a,b\} = \frac{a + b + |a-b|}{2} $$

Hence, the same argument will follow.

  • $\begingroup$ There's no process, really. The triangle inequality, iterated, states that $|a+b+c+...| \leq |a| + |b| + |c| + ...$, so it was all about breaking the term inside the first modulus, into lots of pieces with which I could work with. It's all about deciding which $a,b,c$ etc. you are going to use, and the inequality writes itself. $\endgroup$ – астон вілла олоф мэллбэрг Apr 5 '17 at 11:10

We have that $$ \min\{x_n,y_n\}=\frac{x_n+y_n-|x_y-y_n|}2. $$ Using the properties of limits and the continuity of the absolute value, we get $$ \frac{x_n+y_n-|x_y-y_n|}2\to\frac{x+y-|x-y|}2=\min\{x,y\} $$ as $n\to\infty$.

  • $\begingroup$ Which properties of limits? $\endgroup$ – abc Apr 5 '17 at 12:06
  • $\begingroup$ @abc We need to use the facts that $\lim_{n\to\infty} (a_n \pm b_n) = \lim_{n\to\infty} a_n \pm \lim_{n\to\infty} b_n$ and $\lim_{n\to\infty} c a_n = c \cdot \lim_{n\to\infty} a_n$. $\endgroup$ – Cm7F7Bb Apr 5 '17 at 12:29

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