Uniform convergence of a sequence of functions

In summary: Then f'' is undefined at x = 0 because the limit does not exist. In this case, we say that the derivative does not exist at x = 0. Similarly, $$f''=\frac{1}{2}$$Then f'' is undefined at x = -1 because the limit does not exist. In this case, we say that the derivative does not exist at x = -1.
  • #1

Homework Statement



This is a translation so sorry in advance if there are funky words in here[/B]
f: ℝ→ℝ a function 2 time differentiable on ℝ. The second derivative f'' is bounded on ℝ.
Show that the sequence on functions $$ n[f(x + 1/n) - f(x)] $$ converges uniformly on f'(x) on ℝ.

Homework Equations


In my attempt I used the definition of differentiation:
$$\lim_{x\to\infty} \frac{f(x+h) - f(x)}{h} $$

and I used a criteria for uniform convergence of sequences of functions:

fn fonverges to f uniformely on A
if and on if
for all ε > 0, ∃ N ∈ ℕ, for which
$$\lim_{x\to\infty} sup|fn - f(x)| \leqslant \varepsilon$$
for all n≥ N, for all x ∈ A

The Attempt at a Solution


[/B]
I arrived to an answer but I fear I got sidetracked somewhere because I did not use the bounded second derivative.

I rewrote
$$ n[f(x + 1/n) - f(x)] = \frac{[f(x + 1/n) - f(x)]}{1/n} $$

Now this looks awefully like the derivative of fn for all x which is:
$$\lim_{n\to\infty} \frac{[f(x + 1/n) - f(x)]}{1/n}$$

And now I applied the definition of the uniform convergence which is:
$$\lim_{n\to\infty} sup| \frac{[f(x + 1/n) - f(x)]}{1/n} - f'(x)| \leqslant \varepsilon $$

And therefore, I proved the uniform convergence to f'(x) on ℝ.

(I am missing a few for all ε belonging to...and stuff, I just wanted to write it quickly)
Thank you
 
Last edited:
Physics news on Phys.org
  • #2
1) Your definition of differentiation is wrong. It should have h -> 0.
2) You have not used the bound on f''. I think you need that to prove uniform convergence.
 
  • #3
FactChecker said:
1) Your definition of differentiation is wrong. It should have h -> 0.
2) You have not used the bound on f''. I think you need that to prove uniform convergence.

Thanks for the reply.
You're right my definition is not good when I wrote down the definitions at the start. But in the problem I think it is.
I did not write it but here I implied that h = (1/n)
So n -> ∞ ⇒ h -> 0 because
 
  • #4
We need to be somewhat careful about whether or not f' is well defined. We have to be certain that the limit exists before we claim that it is equal to anything. The existence of some limits is trivial, while others are a little more questionable. Suppose, for example,
$$f''=\frac{1}{\sqrt{x}}$$
 

1. What is uniform convergence of a sequence of functions?

Uniform convergence of a sequence of functions is a type of convergence in which the functions in the sequence converge to a single limit function at the same rate, regardless of the input value. This means that the difference between the values of the functions and their limit function becomes smaller and smaller as the input value increases.

2. How is uniform convergence different from pointwise convergence?

Uniform convergence and pointwise convergence are two different types of convergence for sequences of functions. Pointwise convergence only requires that each individual function in the sequence converges to the limit function, while uniform convergence requires that the convergence happens at the same rate for all functions in the sequence.

3. What are the conditions for uniform convergence of a sequence of functions?

In order for a sequence of functions to uniformly converge, the functions must converge pointwise and satisfy the condition that for any given positive number, there exists an index in the sequence after which the difference between the function and the limit function is always less than that number.

4. How is uniform convergence related to continuity?

Uniform convergence is closely related to continuity because if a sequence of functions uniformly converges to a limit function, and all the functions in the sequence are continuous, then the limit function will also be continuous. This means that uniform convergence can guarantee the continuity of the limit function.

5. Why is uniform convergence important in analysis and applications?

Uniform convergence is important in analysis and applications because it ensures that the limit function of a sequence of functions is well-behaved and can be used to approximate other functions. This can be particularly useful in numerical analysis and approximation methods, where finding an accurate approximation of a function is crucial.

Suggested for: Uniform convergence of a sequence of functions

Replies
5
Views
382
Replies
10
Views
841
Replies
3
Views
261
Replies
2
Views
464
Replies
1
Views
887
Replies
23
Views
2K
Replies
7
Views
1K
Replies
11
Views
1K
Back
Top