Fn converges Uniformly to f, prove that f is continuous

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  • Thread starter Amer
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In summary, if f_n : A\rightarrow R is a sequence of continuous functions and converges uniformly to f, then f is continuous at c.
  • #1
Amer
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If [tex]f_n : A\rightarrow R [/tex] sequnce of continuous functions converges uniformly to f prove that f is continuous

My work
Given [tex]\epsilon > 0 [/tex]
fix [tex]c\in A [/tex] want f is continuous at c

[tex]|f(x) - f(c) | = |f(x) - f_n(x) + f_n(x) - f(c) | \leq |f(x) - f_n(x) | + |f_n(x) - f(c) | [/tex]
the first absolute value less that epsilon since [tex]f_n [/tex] converges uniformly to f
and since
[tex]f_n(x) [/tex] is continuous at c so there exist [tex]\delta [/tex] such that [tex]|x - c| < \delta [/tex]
then [tex]|f_n(x) - f(c) | < \epsilon [/tex]

Am i right ?
 
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  • #2
Amer said:
If [tex]f_n : A\rightarrow R [/tex] sequnce of continuous functions converges uniformly to f prove that f is continuous

My work
Given [tex]\epsilon > 0 [/tex]
fix [tex]c\in A [/tex] want f is continuous at c

[tex]|f(x) - f(c) | = |f(x) - f_n(x) + f_n(x) - f(c) | \leq |f(x) - f_n(x) | + |f_n(x) - f(c) | [/tex]
the first absolute value less that epsilon since [tex]f_n [/tex] converges uniformly to f
and since
[tex]f_n(x) [/tex] is continuous at c so there exist [tex]\delta [/tex] such that [tex]|x - c| < \delta [/tex]
then [tex]|f_n(x) - f(c) | < \epsilon [/tex]
You need three terms:
[tex]|f(x) - f(c) | \leq |f(x) - f_n(x) |+|f_n(x) - f_n(c) | + |f_n(c) - f(c) | [/tex]
 
  • #3
Plato said:
You need three terms:
[tex]|f(x) - f(c) | \leq |f(x) - f_n(x) |+|f_n(x) - f_n(c) | + |f_n(c) - f(c) | [/tex]

Thanks but how i can connect this to [/tex]| x - c | < \delta [/tex]
if i found that
[tex]|f(x) - f(c) | < \epsilon [/tex] how i can find [tex]\delta [/tex], as I said the delta come from the continuity of [tex]f_n(x) [/tex] at c ?

Thanks very much for your help
 
  • #4
Since $(f_n)$ converges uniformly to $f$, we know that exists $N_0 \in \mathbb{N}$ such that for all $n \geq N_0$ we have $|f_n(x) - f(x)| < \frac{\varepsilon}{3}$, for all $x$.

This in particular means that $|f_n(c) - f(c)| < \frac{\varepsilon}{3}$.

Since each $f_n$ is continuous, this means that for all $\varepsilon >0$ we have $\delta >0$ such that $|x-c| < \delta$ implies $| f_n(x) - f_n(c)| < \frac{\varepsilon}{3}$.

Combine these three pieces to get what you need. (Nod)
 
  • #5
Thanks very much both of you :D
 
  • #6
Fantini said:
Since each $f_n$ is continuous, this means that for all $\varepsilon >0$ we have $\delta >0$ such that $|x-c| < \delta$ implies $| f_n(x) - f_n(c)| < \frac{\varepsilon}{3}$.
Note that each $f_n$ can have its own dependence of δ on ε. It is not clear that we can select a δ such that $| f_n(x) - f_n(c)| < \frac{\varepsilon}{3}$ for all n ≥ N₀. However, this is not necessary. Given ε, we find N₀ and then select a concrete n ≥ N₀. Only then we select a $\delta >0$ such that $|x-c| < \delta$ implies $| f_n(x) - f_n(c)| < \frac{\varepsilon}{3}$. In the end we show that $| f(x) - f(c)| < \varepsilon$ if $|x-c| < \delta$, and this fact does not depend on n.
 
  • #7
Evgeny.Makarov said:
Note that each $f_n$ can have its own dependence of δ on ε. It is not clear that we can select a δ such that $| f_n(x) - f_n(c)| < \frac{\varepsilon}{3}$ for all n ≥ N₀. However, this is not necessary. Given ε, we find N₀ and then select a concrete n ≥ N₀. Only then we select a $\delta >0$ such that $|x-c| < \delta$ implies $| f_n(x) - f_n(c)| < \frac{\varepsilon}{3}$. In the end we show that $| f(x) - f(c)| < \varepsilon$ if $|x-c| < \delta$, and this fact does not depend on n.

So my work will be like this $f_n$ converges uniformly to f, $f_n$ sequence of continuous functions
choose $c \in A $ A is the domain of $f_n $
want f continuous at c, c is arbitrary so that will hold for every c in X
Given $\epsilon > 0$ since $f_n $ converges to f uniformly there exist $N$ such that $|f_n - f| < \epsilon $
for all $ N \leq n $
choose n > N , $f_n $ is continuous at c so there exist $delta > 0 $, $|x -c|<\delta$ such that $|f_n(x) - f_n(c)|< \epsilon/3 $
$|f_n(c) - f(c) | < \epsilon/3 $ because fn converges uniformly to f and that dose not depends on x
i.e it is true for all x in A, with c.
so we have, use same delta
$|f(x) - f(c) | = | f(x) - f_n(x) + f_n(x) - f_n(c) +f_n(c) - f(c) | \leq |f(x) - f_n(x)| + |f_n(x) - f_n(c) | + |f_n(c) - f(c) |< \epsilon $

how about that ?
 
  • #8
Amer said:
Given $\epsilon > 0$ since $f_n $ converges to f uniformly there exist $N$ such that $|f_n - f| < \epsilon $ for all $ N \leq n $
Should be $\epsilon/3$. The rest seems correct.
 

FAQ: Fn converges Uniformly to f, prove that f is continuous

What does it mean for a sequence of functions to converge uniformly?

When a sequence of functions converges uniformly, it means that the functions in the sequence become closer and closer to the limit function at a uniform rate, regardless of the input values. This is different from pointwise convergence, where the rate of convergence may vary for different input values.

Why is it important to prove that a function is continuous?

Proving that a function is continuous is important because it guarantees that the function is well-behaved and behaves predictably. This allows us to make accurate predictions and use the function in various mathematical and scientific applications.

How do you prove that a function is continuous using uniform convergence?

To prove that a function is continuous using uniform convergence, we first show that the sequence of functions converges uniformly to the limit function. Then, we use the definition of continuity to show that the limit function is continuous at every point in the domain, thus proving that the original function is also continuous.

Can a sequence of functions converge uniformly but not pointwise?

Yes, it is possible for a sequence of functions to converge uniformly but not pointwise. This means that the functions in the sequence get closer to the limit function at a uniform rate, but the rate of convergence may vary for different input values, leading to pointwise convergence not occurring.

What are some common techniques used to prove uniform convergence and continuity?

Some common techniques used to prove uniform convergence and continuity include the use of the Cauchy criterion, the Weierstrass M-test, and the continuity of the limit function. Other methods such as the Arzelà-Ascoli theorem and the Dini's theorem can also be used in certain cases.

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