1. ## Sequence of functions

Consider the sequence of functions $\displaystyle f_n(t)=\cos^{2n}(n!\pi t),$ on the space $\displaystyle (\mathcal C[0,1],d_\infty).$ Is $\displaystyle f_n(t)$ convergent?

Well I think this is a limit problem, but I don't know how to compute it, any ideas?

2. Originally Posted by Connected
Consider the sequence of functions $\displaystyle f_n(t)=\cos^{2n}(n!\pi t),$ on the space $\displaystyle (\mathcal C[0,1],d_\infty).$ Is $\displaystyle f_n(t)$ convergent?

Well I think this is a limit problem, but I don't know how to compute it, any ideas?
No. Note that if it was convergent then $\displaystyle \displaystyle \lim f_n(t)$ would be continuous...yet I don't think it's hard to show that $\displaystyle \lim f_n(t)=\mathbf{1}_{\mathbb{Q}}$

3. Originally Posted by Drexel28
yet I don't think it's hard to show that $\displaystyle \lim f_n(t)=\mathbf{1}_{\mathbb{Q}}$
What do you mean with the right notation?

How do you solve the problem?

4. Originally Posted by Connected
What do you mean with the right notation?

How do you solve the problem?
It is the indicator function--and no I can't show you how to solve the problem. This is your homework, me telling you how to solve it will mean that you might fail your test, drop out of school, etc. I can't handle that kind of guilt.

Why don't you solve it, and we'll check it.

5. But I'm not actually asking you to solve me the problem completely, a hint could be a help.

6. Originally Posted by Connected
But I'm not actually asking you to solve me the problem completely, a hint could be a help.
I gave you a big hint. If $\displaystyle f_n\to f$ in $\displaystyle C[0,1]$ then $\displaystyle f$ must be continuous, but a quick check shows that $\displaystyle f_n\to\mathbf{1}_{\mathbb{Q}}$ which is the indicator function on the rationals.