Convergence of a sequence of functions

• Nov 30th 2008, 12:10 PM
akolman
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 $\displaystyle f(x)=\lim \limits_{x \to \infty} g_n(x)=\lim \limits_{x \to \infty} f_n\left(x+\frac{1}{n}\right)$

• Nov 30th 2008, 10:23 PM
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?
• Nov 30th 2008, 11:21 PM
CaptainBlack
Quote:

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
• Nov 30th 2008, 11:54 PM
akolman
Quote:

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??
• Dec 1st 2008, 02:09 AM
CaptainBlack
Quote:

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
• Dec 1st 2008, 02:18 AM
akolman
I get the difference now, so how do you do this proof?

• Dec 2nd 2008, 11:56 AM
akolman
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.