Consider the sequence of functions \(g_n(x)=[\sin(nx)]^2\) on \([0,2\pi]\). Define each of the following notions of convergence and determine whether the sequence converges in that sense; if so, determine the limit:
pointwise convergence;
strong convergence in \(L^1\);
weak convergence in \(L^1\).
Solution
Recall the identity \[ g_n(x)=\sin^2(nx)=\frac12-\frac12\cos(2nx). \]
1. The sequence does not converge pointwise on \([0,2\pi]\).
At \(x=\pi/2\), one has \[ g_n(\pi/2)=\sin^2(n\pi/2), \] which alternates between \(1\) for odd \(n\) and \(0\) for even \(n\). Hence \(g_n(\pi/2)\) has no limit, so \((g_n)\) does not converge pointwise on the whole interval.
2. The sequence does not converge strongly in \(L^1([0,2\pi])\).
Strong convergence in \(L^1\) means convergence in the norm \[ \|h\|_1=\int_0^{2\pi}|h(x)|\,dx. \] If \(g_n\) converged strongly in \(L^1\) to some \(g\), then every subsequence would converge in measure to \(g\).
However, \(g_n\) cannot converge in measure. Indeed, using \[ g_n-\frac12=-\frac12\cos(2nx), \] we have for every \(n\) \[ m\left\{x\in[0,2\pi]:\left|g_n(x)-\frac12\right|>\frac14\right\} =m\{x:|\cos(2nx)|>1/2\}. \] Because \(x\mapsto 2nx\) traverses an integer number of full periods on \([0,2\pi]\), this measure is independent of \(n\) and equals \[ \frac{4\pi}{3}>0. \] Thus \(g_n\) does not converge in measure to \(1/2\).
On the other hand, part 3 below shows that every weak \(L^1\) limit would have to be \(1/2\); strong convergence would imply weak convergence to the same limit. Therefore strong \(L^1\) convergence is impossible.
3. The sequence converges weakly in \(L^1([0,2\pi])\) to the constant function \(1/2\).
Weak convergence in \(L^1\) means that \[ \int_0^{2\pi} g_n(x)\varphi(x)\,dx \longrightarrow \int_0^{2\pi} g(x)\varphi(x)\,dx \] for every bounded linear functional on \(L^1\); on this finite measure space these are represented by functions \(\varphi\in L^\infty([0,2\pi])\).
Fix \(\varphi\in L^\infty([0,2\pi])\). Since the interval has finite measure, \(\varphi\in L^1([0,2\pi])\). Using the trigonometric identity above, \[ \int_0^{2\pi}\left(g_n(x)-\frac12\right)\varphi(x)\,dx =-\frac12\int_0^{2\pi}\varphi(x)\cos(2nx)\,dx. \] By the Riemann–Lebesgue lemma, the last integral tends to \(0\). Hence \[ \int_0^{2\pi}g_n\varphi \longrightarrow \frac12\int_0^{2\pi}\varphi, \] which is exactly \[ g_n\rightharpoonup \frac12 \qquad\text{in }L^1([0,2\pi]). \]