# Analysis, silly question

1. Mar 31, 2009

### sillyus sodus

1. The problem statement, all variables and given/known data
Given a convergent sequence:
$$a_n \rightarrow a$$
and a continuous function:
$$f:\mathbb{R}\rightarrow\mathbb{R}$$
show that there exists an $$N\in\mathbb{N}$$ such that $$\forall n>N$$:
$$f(a_n)\geq\frac{f(a)}{2}$$

2. Relevant equations
Usual definitions for limit of a sequence and continuous function.

3. The attempt at a solution
I've tried playing around with it but I really don't understand what to do, I know its an easy question and I feel pretty stupid but I'm stuck. Can anyone just get me started on this one?
So far I've rearranged it so it looks like:
$$f(a)-f(a_n)\leq f(a_n)$$
Which sort of looks like a limit without the abs values...
I figure that the function of a sequence is a sequence again, btu what am I trying to say here? That it is a divergent sequence?

2. Mar 31, 2009

### gammamcc

False. Let f(x) = -1 (constant function). Let a_n =0 for all n. Here, f(a_n)=f(0)=-1 <-1/2 for all n, yet a_n:n=1,2,.... converges in the domain if f.
[Is there another condition?]

3. Mar 31, 2009

### sillyus sodus

Sorry, sorry, yes there is another condition:
$$f(a)>0$$

4. Mar 31, 2009

### gammamcc

To not give the answer away, I would just suggest that the problem is now a matter of
fitting into the \epsilon - \delta definition of continuous functions. Note that f(a)/2 < f(a).

5. Apr 1, 2009

### HallsofIvy

Take $\epsilon> a/2$ in the definition of "continuous function".

6. Apr 2, 2009

### gammamcc

whaaaa.? So much for the 'for all epsilon >0' part. Whaaa? Well, not my problem anyway....

7. Apr 2, 2009

### HallsofIvy

Yes, "For all epsilon> 0", something is true so it is true for any specific value of epsilon. Here we are given that f is continuous so we are free to choose any value of epsilon we like.

8. Apr 2, 2009

### sillyus sodus

I have an answer now, basically because $$a_n\rightarrow a$$ and f is continuous then $$f(a_n)\rightarrow f(a)$$, so there exists n bigger than N such that $$|f(a_n)-f(a)|<\epsilon$$ so then just choose $$\epsilon = f(a)/2$$ do some rearranging and the inequality pops out.

In other words, $$f(a_n)$$ is a convergent sequence.