continuity & pointwise convergence

symmetry7 · #1 $Old$ June 4th, 2009, 01:57 AM

let f be any function going from R to R and for any number a in R, let f_a (such that it goes from R to R also) be the shifted function defined by the rule f_a(x) = f(x-a) for all x in R.

Show that f is continuous if and only if, whenever the sequence a_n (going from 1 to infinity) is a sequence of real numbers which converges to zero, then the (sequence of) functions f_a_n converges pointwise to f.

Who can help?

putnam120 · #2 $Old$ June 4th, 2009, 08:06 AM

I don't think the statement is true the way you have phrased it. For a counter example consider
$f(x)=1$ if $x\in\mathbb{Q}$ and $f(x)=0$ if $x\notin\mathbb{Q}$ . Then let $a_n$ be any sequence such that $a_n\to{0}$ . This has $f_{a_n}(x)\to f(x)$ but f is not continuous.

So I think you need to impose some additional conditions on either f or $f_{a_n}$ .

symmetry7 · #3 $Old$ June 4th, 2009, 05:44 PM

I appreciate your response, however this question appears in a text book written by a famous mathematician. The question appears exactly how I phrased it. I am merely trying to find an approach to showing that it is true.

siclar · #4 $Old$ June 4th, 2009, 11:32 PM

Quote:

Originally Posted by putnam120 $View Post$

I don't think the statement is true the way you have phrased it. For a counter example consider
$f(x)=1$ if $x\in\mathbb{Q}$ and $f(x)=0$ if $x\notin\mathbb{Q}$ . Then let $a_n$ be any sequence such that $a_n\to{0}$ . This has $f_{a_n}(x)\to f(x)$ but f is not continuous.

So I think you need to impose some additional conditions on either f or $f_{a_n}$ .

That counterexample doesn't work. For example, assume $x$ is rational and all $a_n$ are irrational and converge to zero. Then clearly $f(x-a_n)$ is zero, but $f(x)=1$ , so this discontinuous function indeed does not satisfy convergence for any such sequence.

To the OP, the argument from and for continuity should be straightforward. Keep in mind that one definition of continuity is all about the behavior of the function over convergent sequences. Also keep in mind that the sum of convergent sequences converges to the sum of limits.

putnam120 · #5 $Old$ June 5th, 2009, 04:21 PM

Thanks for pointing that out.

Enrique2 · #6 $Old$ June 28th, 2009, 03:21 PM

Quote:

Originally Posted by symmetry7 $View Post$

let f be any function going from R to R and for any number a in R, let f_a (such that it goes from R to R also) be the shifted function defined by the rule f_a(x) = f(x-a) for all x in R.

Show that f is continuous if and only if, whenever the sequence a_n (going from 1 to infinity) is a sequence of real numbers which converges to zero, then the (sequence of) functions f_a_n converges pointwise to f.

Who can help?

Hi, it is immediate since x-a_n goes to x iff a_n goes to 0. Then f_a_n(x)
goes to f(x) for any zero sequence a_n is equivalent to say that f(x-a_n)
goes to f(x). Use sequential definition about continuity, f is continuous at x iff f(x_n) goes to f(x) for any sequence (x_n) going to x, and take
a_n:=x_n-x

Enrique2 · #7 $Old$ June 28th, 2009, 03:32 PM

Sorry, I didn't noticed that siclair had already given the argument!