Let $u : \mathbb{R} \to [-1, 1]$ be defined by $u(y) = \cos(2\pi y)$.Set $u_\epsilon(x) = u\left(\frac{x}{\epsilon}\right)$ for $x \in (a,> b)$, where $a, b \in \mathbb{R}$ and $a < b$. Show that $ u_\epsilon \rightharpoonup 0 $ in $ L^2(a, b) $ as $ \epsilon \to 0 $.
SolutionWith $u(y)$, set $y=\frac{x}{\epsilon}$,
The $L^2$ norm for $u_\epsilon$ is given by
$$||{u_\epsilon}||_{L^2(a,b)} = \left( \int_a^b \left| \cos\left( 2\pi \frac{x}{\epsilon} \right) \right|^2 \, dx \right)^{1/2}.$$
Since $ \left| \cos \left( 2\pi \frac{x}{\epsilon} \right) \right|^2 $ simplifies to $ \cos^2\left( 2\pi \frac{x}{\epsilon} \right) $, we use the trigonometric identity:
$$\cos^2\left( 2\pi \frac{x}{\epsilon} \right) = \frac{1}{2} \left( 1 + \cos\left( 4\pi \frac{x}{\epsilon} \right) \right).$$
Now we solve the integral:
$$\int_a^b \cos^2\left( 2\pi \frac{x}{\epsilon} \right) \, dx = \int_a^b \frac{1}{2} \left( 1 + \cos\left( 4\pi \frac{x}{\epsilon} \right) \right) \, dx.$$
Splitting this into two integrals:
$$\int_a^b \frac{1}{2} \, dx + \int_a^b \cos\left( 4\pi \frac{x}{\epsilon} \right) \, dx.$$
And we obtain
$$\int_a^b \frac{1}{2} \, dx = \frac{b - a}{2}.$$
The second integral,
$$\int_a^b \cos\left( 4\pi \frac{x}{\epsilon} \right) \, dx,$$
oscillates rapidly as $ \epsilon \to 0 $. Since the oscillations average out over the interval $ (a, b) $, the integral tends to zero. So if we take the limit, we obtain
$$\lim_{\epsilon\to0}\int_a^b \cos\left( 4\pi \frac{x}{\epsilon} \right) \, dx=0,$$
So we have only the first integral, and hence the norm on the open interval $(a,b)$ is:
$$||{u_\epsilon}|| = \sqrt{ \frac{b - a}{2}} = C \ \ \ \forall b>a.$$
But this shows that $u_\epsilon$ converges to some constant on the open interval (a,b) in the $L^2$ norm, and not zero.
Any ideas what went wrong here?
Thanks