Proving Limit of a Sequence of Periodic Functions with Continuous Functions

[Jncik]

Homework Statement

suppose that {h_n(x), n=1,2...} is a sequence of 2π periodic and positive functions

and

and suppose there is a sequence [URL]http://latex.codecogs.com/gif.latex?{a_{n},%20n%20=%201,2..}[/URL] so that h_n(x) = 0 for |x|>a_n and [URL]http://latex.codecogs.com/gif.latex?\lim_{n-%3Eoo}a_{n}%20=%200[/URL]

show that for every 2π periodic and continuous function f we have

http://imageshack.us/m/151/4133/asdjc.gif

Homework Equations

it is known that

for every continuous function f(x) in [-π,π] we have: for every ε>0 there is a n such that for every x,t in (-a_n,a_n)=> |f(x-t) - f(x)|< ε

i have no idea how to do it...

our professor never showed anything similar...

 

[Ray Vickson]

You say the function h_n(x) is zero for |x| > a_n and is positive for -a_n < x < a_n. This contradicts your statement that h_n is periodic. Also, when you say that {h_n(x), n=1, 2, ...} is a sequence of 2n periodic functions, what does this mean? Do you mean that you have a sequence {h_k, k=1, 2, ..., 2n} of periodic functions (all with the same period 2*pi?), or do you mean that the function f_n has period 2n, or what?

RGV