# Thread: Convergence of a sequence of functions

1. ## Convergence of a sequence of functions

Suppose $\displaystyle \{f_n\}$ is a sequence of functions converging uniformly to$\displaystyle f$. Define $\displaystyle g_n(x)=f_n\left(x+\frac{1}{n}\right)$ . Show that the sequence $\displaystyle \{g_n\}$ converges pointwise to f.

So I need to show that $\displaystyle f(x)=\lim \limits_{x \to \infty} g_n(x)=\lim \limits_{x \to \infty} f_n\left(x+\frac{1}{n}\right)$

2. I am pretty confused, does pointwise convergence means
$\displaystyle f(x)=\lim \limits_{x \to \infty} g_n(x)=\lim \limits_{x \to \infty} f_n\left(x+\frac{1}{n}\right)$ ???

Also, I noticed that if the all $\displaystyle {f_n}$ are continuous, then the proof wouldn't be hard. However, they are not mentioning anything, is continuity necessary? How can I prove that?

3. Originally Posted by akolman
I am pretty confused, does pointwise convergence means
$\displaystyle f(x)=\lim \limits_{x \to \infty} g_n(x)=\lim \limits_{x \to \infty} f_n\left(x+\frac{1}{n}\right)$ ???

Also, I noticed that if the all $\displaystyle {f_n}$ are continuous, then the proof wouldn't be hard. However, they are not mentioning anything, is continuity necessary? How can I prove that?
No it means:

$\displaystyle f(x)=\lim \limits_{n \to \infty} g_n(x)=\lim \limits_{n \to \infty} f_n\left(x+\frac{1}{n}\right)$

CB

4. Originally Posted by CaptainBlack
No it means:

$\displaystyle f(x)=\lim \limits_{n \to \infty} g_n(x)=\lim \limits_{n \to \infty} f_n\left(x+\frac{1}{n}\right)$

CB
How different is that from uniform continuity??

5. Originally Posted by akolman
How different is that from uniform continuity??
Where did contiuity enter the question?

The difference between uniform and pointwise convergence is that:

Pointwise: For all $\displaystyle \varepsilon>0$ there exists an $\displaystyle N_x$ such that:

$\displaystyle |f_n(x)-f(x)|<\varepsilon ,\ \ \forall n>N_x$

but $\displaystyle N_x$ may depend on $\displaystyle x$

Uniform:For all $\displaystyle \varepsilon>0$ there exists an $\displaystyle N$ independednt of $\displaystyle x$ such that:

$\displaystyle |f_n(x)-f(x)|<\varepsilon ,\ \ \forall n>N$.

CB

6. I get the difference now, so how do you do this proof?

7. If I let $\displaystyle x_n=x+\frac{1}{n}$, then $\displaystyle g_n(x)=f_n(x_n)$. I know that $\displaystyle x_n$ converges uniformly to $\displaystyle x$

$\displaystyle \exists N_1$ such that for $\displaystyle n>N_1 , |x_n-x|<\varepsilon$

Since $\displaystyle f_n$ converges uniformly to $\displaystyle f(x)$,
$\displaystyle \exists N_2$ such that for $\displaystyle n>N_2, |f(x)-f_n(x)|<\varepsilon$,

Is this right?
If I let $\displaystyle N=max(N_1,N_2)$ ,
$\displaystyle n>N \Rightarrow |x_n-x| < \varepsilon$
$\displaystyle \Rightarrow |f_n(x_n)-f(x_n)| <|f_n(x)-f(x)|<\varepsilon$

I am sorry if I pushing this question too much, but I am still clueless.