If $(x_n)$, $(y_n)$ and $(z_n)$ are 3 convergent sequences then show that, the sequence $(w_n)$ defined by $(w_n)$=mid{$(x_n),(y_n),(z_n)$} is also convergent. Does anyone have a proof for this? Please help.

  • $\begingroup$ I suppose mid is the term that is neither the max or the min ? If that's the case suppose the limits are X, Y and Z and X<Y<Z. What do you think would be the limit of $(w_n)$ ? $\endgroup$ – T_O Mar 19 '14 at 16:44
  • $\begingroup$ i don't think that $mid$ is a standard terminology...you should probably explain that. $\endgroup$ – wanderer Mar 19 '14 at 16:48
  • $\begingroup$ Since $\text{mid}(a,b,c) = a+b+c - \max(a,b,c) - \min(a,b,c)$ and $\max()$ and $\min()$ are continuous functions from $\mathbb{R}^3$ to $\mathbb{R}$, so does $\text{mid}()$. $\endgroup$ – achille hui Mar 19 '14 at 16:51
  • $\begingroup$ Yea....mid is neither the min nor the max....it is something inbetween or in the middle of min and max...@wanderer @T_O $\endgroup$ – Naive Mar 19 '14 at 16:52

Let $x_n\rightarrow x$,$y_n\rightarrow y$,$z_n\rightarrow z$ and suppose that $\varepsilon>0$ is arbitrary positive number. Then there exist $N_x,N_y,N_z\in\mathbb{N}$ such that $$\forall n\geq N_x, \forall n\geq N_y,\forall n\geq N_z,\quad |x_n-x|<\varepsilon,|y_n-y|<\varepsilon,|z_n-z|<\varepsilon$$ Set $N=\max\{N_x,N_y,N_z\}$. Now for any $n\geq N$ we have $$|\frac{x_n+y_n+z_n}{3}-\frac{x+y+z}{3}|=|(\frac{x_n}{3}-\frac{x}{3})+(\frac{y_n}{3}-\frac{y}{3})+(\frac{z_n}{3}-\frac{z}{3})|\leq\\ |(\frac{x_n}{3}-\frac{x}{3})|+|(\frac{y_n}{3}-\frac{y}{3})|+|(\frac{z_n}{3}-\frac{z}{3})|\leq \frac{1}{3}|x_n-x|+\frac{1}{3}|y_n-y|+\frac{1}{3}|z_n-z|<\\ \frac{1}{3}\varepsilon+\frac{1}{3}\varepsilon+\frac{1}{3}\varepsilon=\varepsilon$$ Hence $\frac{x_n+y_n+z_n}{3}\rightarrow\frac{x+y+z}{3}$.


and also if we set $\alpha=\frac{x_n+y_n+|x_n-y_n|}{2},\beta=\frac{x_n+z_n+|x_n-z_n|}{2},\gamma=\frac{z_n+y_n+|z_n-y_n|}{2}$ then $$\min\{\alpha,\beta,\gamma\}=\min\{\min\{\alpha,\beta\},\gamma\}=\min\{\frac{\alpha +\beta-|\alpha-\beta|}{2},\gamma\}= \frac{\frac{\alpha+\beta-|\alpha-\beta|}{2}+\gamma-|\frac{\alpha+\beta-|\alpha-\beta|}{2}-\gamma|}{2}$$

So by continuity we have the answer.

  • $\begingroup$ Sorry. I thought wrongly! What is "mid" function $\endgroup$ – Hamid Shafie Asl Mar 21 '14 at 14:24

Hint: If all 3 limits are distinct, show that the mid converges to the middle limit. If 2 limits are equal, show that the mid converges to value of the 2 equal limits. If all 3 limits are equal, show that the mid converges to the same limit.


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