# Normed Linear Space: Proofs

1. Dec 2, 2011

### bugatti79

If lim x_n=x n to infinity and lim y_n=y n to infinity
prove rigorously

lim n to infinity (x_n/5+10y_n)=x/5+10y.

My attempt

let ε>0. Must find $n_0 \in \mathbb{N}$ such that

$||(x_n/5+10y_n)-(x/5+10y)||<ε$ for all $n>n_0$

$||(x_n/5+10y_n)-(x/5+10y)||=||(x_n/5-x/5)||+||10y_n-10y|| \le ||(x_n/5-x/5||+||(10y_n-10y)||$

since $x_n=x$ for the limit n to infinity and similarly for y_n and given ε>0 then ε/2>0

so there exist $n_1 \in N$ such that $||x_n/5-x/5||< ε/2$ for all $n \ge n_1$

and

$||10y_n-10y||< ε/2$ for all $n \ge n_2$

Let $n_0=max{n_1,n_2}$, then for all $n \ge n_0$

implies $||(x_n/5+10y_n)||-||x/5+10y|| \le ||(x_n/5-x/5||+||(10y_n-10y)||<ε/2+ε/2=ε$........?

2. Dec 2, 2011

### Staff: Mentor

The expression after the first equals is incorrect. You are basically saying that |a + b| = |a| + |b|.

$||(x_n/5+10y_n)-(x/5+10y)||=||(1/5) (x_n -x) +10(y_n -y)|| \le (1/5)||x_n -x||+10||y_n -y||$

Now, you want the expression to the right of <= to be less than ε, and it's convenient to split it into ε/2 for the first part and ε/2 for the second part. What does that give you for ||xn - x|| and ||yn - y||?

3. Dec 3, 2011

### bugatti79

This gives
||x_n-x||<5ε/2 and ||y_n-y||<ε/20.......?

Not sure what to do next?

4. Dec 3, 2011

### Fredrik

Staff Emeritus
All you need to do is to explain how we know that n0 can be choosen such that these two inequalities hold for all positive intergers n such that n≥n0.

5. Dec 4, 2011

### bugatti79

Sorry, I dont even know how to do that...?

6. Dec 4, 2011

### Fredrik

Staff Emeritus
Sure you do. It follows almost immediately from the definition of $x_n\to x$ and $y_n\to y$. If you write down the definitions, you will see it. (If not, ask again).

7. Dec 4, 2011

### bugatti79

$$||x_n-x||<5ε/2$$ and $$||y_n-y||<ε/20$$

The LHS of each expression above would go to 0 iff ||x_n-x||=0 as n tends to infinity and similarly for ||yn-y||=0. THis implies the LHS of each exprssion will always be less than ε/2

So adding the 2 expression together would still be less than ε and hence still holds....

Is this correct...if so I dont know how to write this mathematically

8. Dec 4, 2011

### Fredrik

Staff Emeritus
I don't see you using the definition of what $x_n\to x$ means. Can you post the definition?

9. Dec 4, 2011

### bugatti79

$x_n\to x$ iff ε>0, ther is $n_0 \in \mathbb{N}$ such that $||x_n-x||<ε$ for all $n \ge n_0$....

I dont know how to employ this definition...

Thanks

10. Dec 4, 2011

### Fredrik

Staff Emeritus
There's a "for all" missing before the ε, but apart from that it's fine. Clearly, this definition implies that there's a $n_1\in\mathbb N$ such that $\|x_n-x\|<5\varepsilon/2$ for all $n\in\mathbb N$ such that $n\geq n_1$.

11. Dec 4, 2011

### bugatti79

and similarly for n2?

$n_2\in\mathbb N$ such that $\|y_n-y\|<\varepsilon/20$ for all $n\in\mathbb N$ such that $n\geq n_2$

So this is the question completed?

12. Dec 4, 2011

### Fredrik

Staff Emeritus
Right, but you still haven't explained how these two observations imply this result:
Can you find an n0 that does the job?

13. Dec 4, 2011

### bugatti79

if we let n_0={n_1,n_2}...?

14. Dec 4, 2011

### Fredrik

Staff Emeritus
n0 is a number, not a set, but maybe you actually meant n0=max{n1,n2}, because that would work.

To find the proof, you do the things we have discussed, in the order we have discussed them. But when you type it up, things get clearer if you start with the things we did last.

Let $\varepsilon>0$ be arbitrary. Let $n_1\in\mathbb N$ be such that for all $n\in\mathbb N$, $$n\geq n_1\ \Rightarrow\ \|x_n-x\|<\frac{5\varepsilon}{2}.$$ Let $n_2\in\mathbb N$ be such that for all $n\in\mathbb N$, $$n\geq n_2\ \Rightarrow\ \|x_n-x\|<\frac{\varepsilon}{20}.$$ For all $n\in\mathbb N$ such that $n\geq \max\{n_1,n_2\}$, $$\left\|\left(\frac{x_n}{5} -10y_n\right)-\left(\frac{x}{5}-10y\right)\right\| \leq \frac{1}{5} \underbrace{\|x_n-x\|}_{\displaystyle<\frac{5\varepsilon}{2}} +10\underbrace{\|y_n-y\|}_{\displaystyle<\frac{\varepsilon}{20}}<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon.$$

15. Dec 5, 2011

### bugatti79

Hi Fredrick,
I was going to reply on this thread this evening. Thanks for this excellent clarity. I will look at this and the other thread you replied to as soon as I can.

Thanks!