Prove: f_n Convergence Implies f(x)=c

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In summary, the conversation discusses the proof of the existence of an x in [0,1] such that f(x) = c, given that f_n is a uniformly convergent sequence and f is the uniform limit of f_n. The two possible strategies to prove this are to either find the x or assume its nonexistence and obtain a contradiction. The conversation also includes some attempts at formalizing the proof, but it is advised to first decide on a strategy before proceeding.
  • #1
PsychonautQQ
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Homework Statement


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

Homework Equations

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?
 
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  • #2
PsychonautQQ said:
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
extract a convergent subsequence from ##x_n##
 
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  • #3
zwierz said:
extract a convergent subsequence from ##x_n##

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
PsychonautQQ said:
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.

Alternatively, what can you say about ##f##?
 
  • #5
PeroK said:
Alternatively, what can you say about ##f##?
f is the limit of f_n as n goes to infinity?
 
  • #6
PsychonautQQ said:
f is the limit of f_n as n goes to infinity?

I meant what properties has ##f##?
 
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  • #7
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.
PsychonautQQ said:
i bet the (x_n) sequence itself would converge
it is not obliged to converge
 
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  • #8
PsychonautQQ said:

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?

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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Related to Prove: f_n Convergence Implies f(x)=c

1. What does "f_n convergence" mean?

"F_n convergence" refers to the concept of a sequence of functions, where each function in the sequence approaches a specific value as the input variable increases. This value is known as the limit of the sequence.

2. How is f_n convergence related to the function f(x)=c?

If a sequence of functions, f_n, converges to a constant value, c, then the function f(x) must also converge to c. This means that as the input variable increases, the output of f(x) will approach c.

3. Is f_n convergence the same as pointwise convergence?

No, f_n convergence and pointwise convergence are two different concepts. F_n convergence refers to the behavior of a sequence of functions as a whole, while pointwise convergence refers to the behavior of individual points within each function in the sequence.

4. How is f_n convergence different from uniform convergence?

F_n convergence and uniform convergence are two different types of convergence. F_n convergence refers to the behavior of a sequence of functions as the input variable increases, while uniform convergence refers to the behavior of a sequence of functions as a whole.

5. Can f_n convergence be used to prove the continuity of a function?

Yes, f_n convergence can be used to prove the continuity of a function. If a sequence of continuous functions, f_n, converges to a function f, then f must also be continuous. This is known as the continuity theorem for sequences of functions.

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