# Analysis Question

1. Apr 29, 2017

### PsychonautQQ

1. The problem statement, all variables and given/known data
Let f_n: [0,1]-->R be a uniformly convergent sequence of continuous functions. Let f: [0,1]-->R be the uniform limit of {f_n}. Let c be an element of R. Suppose that for all n there is some x_n in [0,1] such that f_n(x_n)=c. Prove that there exists an x in [0,1] such that f(x) = c

2. Relevant equations

3. The attempt at a solution
So the idea here is that there is always some x in [0,1] such that f_n(x)=c, and since f_n converges to f, it seems quite likely that there is an x such that f(x)=c. I need to formalize this proof, I'm not great at that, but let's go!

So a key element of this problem is that f_n is a UNIFORMLY convergence sequence and f is a uniform limit, therefore when searching for our epsilon (which we will call e), we can't have it depend on x.

let e > 0, then there is an N s.t. for all n>N |f_n(x)-f(x)|<e for all x in [0,1]. Let y_n be in [0,1] s.t. f_n(y_n)=c, then for all e>0 there is an N such that if n>N then |f_n(y_n) - f(y_n)| = |c - f(y_n)| < e .....

I'm confused, am i on the right track here?

2. Apr 29, 2017

### zwierz

extract a convergent subsequence from $x_n$

3. Apr 30, 2017

### PsychonautQQ

So what you're saying is let (x_n) be the sequence of x's in [0,1] such that f_n(x_n)=c. Since this sequence is bounded, it must have a convergent subsequence by the bozo-weilstrauss theorem. Is this along the lines of what you were thinking?

On the other hand, i bet the (x_n) sequence itself would converge to the x we are looking for, why do we even need to use a subsequence?

That being said I don't know where exactly to go from here.

4. Apr 30, 2017

### PeroK

Alternatively, what can you say about $f$?

5. Apr 30, 2017

### PsychonautQQ

f is the limit of f_n as n goes to infinity?

6. Apr 30, 2017

### PeroK

I meant what properties has $f$?

7. Apr 30, 2017

### zwierz

let $\{x_{n_j}\}\subset\{x_n\}$ be such that $x_{n_j}\to x'$ as $n_j\to\infty$. Then
$$|c-f(x_{n_j})|=|f_{n_j}(x_{n_j})-f(x_{n_j})|\to 0$$ by uniform convergence of $f_{n_j}$ to $f$. Thus $f(x')=c$
We also use here that the function $f$ is continuous as a uniform limit of continuous functions.
it is not obliged to converge

Last edited: Apr 30, 2017
8. Apr 30, 2017

### PeroK

A bit of general advice. If you take a step back, what you are asked to prove is that "there exists an $x$ such $f(x) = c$".

Broadly, there are two approaches to this:

a) Find the $x$

b) Assume there is no such $x$ and obtain a contradiction.

You didn't really do either. You started with what you knew, fair enough, and began to do some analysis. But, I'm not sure you knew which strategy you had adopted.

In general, therefore, you ought to decide on your startegy first. At least, then, you know what you are aiming for.

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