# fn converges Uniformly to f, prove that f is continuous

#### Amer

##### Active member
If $$f_n : A\rightarrow R$$ sequnce of continuous functions converges uniformly to f prove that f is continuous

My work
Given $$\epsilon > 0$$
fix $$c\in A$$ want f is continuous at c

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

Am i right ?

#### Plato

##### Well-known member
MHB Math Helper
If $$f_n : A\rightarrow R$$ sequnce of continuous functions converges uniformly to f prove that f is continuous

My work
Given $$\epsilon > 0$$
fix $$c\in A$$ want f is continuous at c

$$|f(x) - f(c) | = |f(x) - f_n(x) + f_n(x) - f(c) | \leq |f(x) - f_n(x) | + |f_n(x) - f(c) |$$
the first absolute value less that epsilon since $$f_n$$ converges uniformly to f
and since
$$f_n(x)$$ is continuous at c so there exist $$\delta$$ such that $$|x - c| < \delta$$
then $$|f_n(x) - f(c) | < \epsilon$$
You need three terms:
$$|f(x) - f(c) | \leq |f(x) - f_n(x) |+|f_n(x) - f_n(c) | + |f_n(c) - f(c) |$$

#### Amer

##### Active member
You need three terms:
$$|f(x) - f(c) | \leq |f(x) - f_n(x) |+|f_n(x) - f_n(c) | + |f_n(c) - f(c) |$$
Thanks but how i can connect this to [/tex]| x - c | < \delta [/tex]
if i found that
$$|f(x) - f(c) | < \epsilon$$ how i can find $$\delta$$, as I said the delta come from the continuity of $$f_n(x)$$ at c ?

Thanks very much for your help

#### Fantini

MHB Math Helper
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.

#### Amer

##### Active member
Thanks very much both of you

#### Evgeny.Makarov

##### Well-known member
MHB Math Scholar
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
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
$|f(x) - f(c) | = | f(x) - f_n(x) + f_n(x) - f_n(c) +f_n(c) - f(c) | \leq |f(x) - f_n(x)| + |f_n(x) - f_n(c) | + |f_n(c) - f(c) |< \epsilon$

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.