Convergence of a sequence of periodic functions Motivated by the homogenization theory which studies the effects of high-frequency oscillations in the coefficients upon solutions of PDE, I am thinking about the following question.  
Let the periodic function$$\alpha(x+1)=\alpha(x),\quad\alpha(x)>0,\quad x\in{\bf R}$$
and the sequence $$\alpha_n=\alpha(nx)\quad n\in{\bf N}$$ Consider the Hilbert space $$H^1_0([0,1]):=\{u:[0,1]\to{\bf R}\,|\,u,u'\in L^2([0,1]), u(0)=u(1)=0\}.$$
Here is my question:

What kind of convergence can one
  expect for the sequence
  $(\alpha_n(x))_{n=1}^{\infty}$, and
  what is the corresponding limit?

Edit: According to Qiaochu's comment, I assume TWO different inner products here:
$$\langle u,v\rangle_1=\int_{0}^1uvdx$$
and
$$\langle u,v\rangle_2=\int_{0}^1uvdx+\int_{0}^1u'v'dx$$
For what topology can one expect the convergence of the above sequence?

Edit: If one defines
$$\hat{\alpha} = \frac{1}{\int_0^1\frac{1}{\alpha(x)}dx}$$ can one expect some relationship between $(\alpha_n)$ and $\hat{\alpha}$?
 A: If $\alpha\in L_2([0,1])$ then sequence $\alpha_n(x)$ converges weakly to $\int_0^1 \alpha(x)\,dx\ $.
A: Let's first figure out what the limit would be.  There are essentially two types of convergence that we might care about:  convergence in $H_0^1\left( [0,1]\right)$ and convergence in $C\left( [0,1]\right)$.  I have ordered them from strongest to weakest.  (As the interval is not changing, from now on, I'll just write the first letter for these spaces).
If the sequence converges in $H_0^1$, then it is a bounded sequence in $H_0^1$.  Thus, by the Sobolev Embedding Theorem (see, e.g., Brezis, Functional Analysis, Sobolev Spaces, and Partial Differential Equations), pg. 212-213), there is a subsequecne $\alpha _{m_n}$ that converges in $C$.  You can use this argument to show that every subsequence of $\alpha _n$ has in turn a subsequence that converges in $C$, so that the original sequence converges in $C$.  Thus, convergence in $H_0^1$ is stronger than convergence in $C$, so we might as well only concern ourselves with convergence in $C$.
Now, suppose $\alpha _n$ converges in $C$.  For each rational $x\in [0,1]$, $n$ will eventually be large enough $nx\in \mathbb{Z}$, so that $\alpha _n(x)=\alpha (nx)=\alpha (0)=0$.  Thus, the limiting function must be $0$ on the rationals, and hence by continuity, must be identically $0$.  Thus, if $\alpha _n$ is to convergence in either $H_0^1$ or $C$, it must convergen to the function that is identically $0$.
This clearly implies that this sequence will not converge for a lot of $\alpha$.  For example, $\alpha (x)=\sin (\pi x)$.  This certainly does not converge to $0$, and as this is the only possibility, it must not converge at all.
I hope I didn't screw any of that up.  Please point out any mistakes I have made.  In any case, correct or not, I hope what I have said has been useful.
-Jonny Gleason
A: Why would you expect your functions to converge? Rescaling the physical coordinates is given by a inverse scaling in frequency space: $\alpha_n$ has the same $L^2$ as $\alpha$, but the $\dot{H}^1$ norm grows unboundedly. So you have no convergence in any sense related to $H^1_0$, and you get weak convergence in $L^2$ (since any bounded sequence in a Hilbert space has a weakly converging subsequence by Banach-Alaoglu).
Furthermore, unless $\alpha$ is constant, the weak convergence in $L^2$ cannot be upgraded to strong convergence, since for any fixed $n$, you have $$\lim_{m\to\infty} \|\alpha_m-\alpha_n\|_{L^2} = 2\|\alpha - \bar{\alpha}\|_{L^2}$$ where $\bar\alpha$ is the mean of $\alpha$, by Andrew's observation.

As a side remark, while what you wrote for the definition of $H^1_0$ is intuitively acceptable, it is not technically correct as the definition. (Note that $L^2$ "functions" are equivalent classes, and it doesn't really make sense to require that an $L^2$ function vanishes on a measure zero set.) $H^1_0$ is better defined as the the completion of $C^\infty_0$ under the $H^1$ norm.
