2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
3
$\begingroup$

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 $\frac{x_n+y_n+z_n}{3}\rightarrow\frac{x+y+z}{3}$.

Edit:

$$\min\{\max\{x_n,y_n\},\max\{x_n,z_n\},\max\{z_n,y_n\}\}=\min\{\frac{x_n+y_n+|x_n-y_n|}{2},\frac{x_n+z_n+|x_n-z_n|}{2},\frac{z_n+y_n+|z_n-y_n|}{2}\}$$
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.

$\endgroup$
  • $\begingroup$ Sorry. I thought wrongly! What is "mid" function $\endgroup$ – Hamid Shafie Asl Mar 21 '14 at 14:24
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.