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fn converges Uniformly to f, prove that f is continuous

Amer

Active member
Mar 1, 2012
275
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 ?
 

Plato

Well-known member
MHB Math Helper
Jan 27, 2012
196
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]
 

Amer

Active member
Mar 1, 2012
275
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
 

Fantini

"Read Euler, read Euler." - Laplace
MHB Math Helper
Feb 29, 2012
342
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)
 

Amer

Active member
Mar 1, 2012
275
Thanks very much both of you :D
 

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,493
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.
 

Amer

Active member
Mar 1, 2012
275
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 ?
 

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,493
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.