# 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 $\hat{\alpha}$?

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What inner product are you putting on this space? –  Qiaochu Yuan Apr 30 '11 at 17:42
@Qiaochu: $\langle f, g \rangle_{H_{0}^{1}} = \langle f, g \rangle_{L^2} + \langle f', g' \rangle_{L^2}$ is the standard convention. –  t.b. Apr 30 '11 at 17:48
Why would you expect your function to converge? Rescaling the physical coordinates is given by a inverse scaling in Fourier 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). –  Willie Wong Jul 29 '11 at 22:45
@Willie: Thanks for your reading. This is my old question. Actually, it is from that question. It seems that one needs the Riemann-Lebesgue lemma finally. –  Jack Jul 30 '11 at 1:07
@Willie: +1. I didn't notice this since it's from the book(Hunter's Applied Analysis). –  Jack Jul 30 '11 at 3:36

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.

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If $\alpha\in L_2([0,1])$ then sequence $\alpha_n(x)$ converges weakly to $\int_0^1 \alpha(x)\,dx\$.

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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

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Sorry. I missed your edit. The first inner product corresponds to the $L^2$ norm. Clearly, convergence in $C$ implies convergence in $L^2$, so that $L^2$ is the weakeast type of convergence we might consider. However, $$\int _0^1\alpha _n(x)^2dx=\int _0^1\alpha (nx)^2dx=1/n\int _0^n\alpha (x)^2dx=\int _0^1\alpha (x)^2dx=\left\| \alpha \right\| _2$$ This shows that the only time this sequence converges (in any of the three topologies) is when $\alpha$ is the function that is identically $0$. –  Jonathan Gleason Apr 30 '11 at 18:42