Justify or give a counterexample to the following assertions:
If \(\{f_i\}\) is a sequence in \(L^2([0,1])\) converging weakly to \(f\) in \(L^2([0,1])\) then \(f_i^2\) converges weakly to \(f^2\) in \(L^1([0,1])\).
If \(\{f_i\}\) is a sequence in \(L^2([0,1])\) converging strongly to \(f\) in \(L^2([0,1])\), then \(f_i^2\) converges strongly to \(f^2\) in \(L^1([0,1])\).
Solution
1. Assertion (a) is false: the sequence\[ f_n(x)=\sin(2\pi n x) \]converges weakly to \(0\) in \(L^2([0,1])\).
For \(g\in L^2([0,1])\), finite measure gives \(g\in L^1([0,1])\). Hence the Riemann--Lebesgue lemma gives \[ \int_0^1 f_n(x)g(x)\,dx =\int_0^1 \sin(2\pi n x)g(x)\,dx \longrightarrow0. \] Thus \(f_n\rightharpoonup0\) in \(L^2([0,1])\).
2. The squares in step 1 do not converge weakly to \(0\) in \(L^1([0,1])\).
The constant function \(1\) belongs to \(L^\infty([0,1])=(L^1([0,1]))^*\), and \[ \int_0^1 f_n(x)^2\,dx =\int_0^1\sin^2(2\pi n x)\,dx =\frac12 \] for every \(n\). Since the weak limit asserted in part (a) would be \(0^2=0\), testing against \(1\) gives a contradiction.
3. Assertion (b) is true: if \(f_i\to f\) strongly in \(L^2([0,1])\), then\[ \norm{f_i^2-f^2}_{L^1}\longrightarrow0. \]
The sequence \((f_i)\) is bounded in \(L^2\), so there is \(M<\infty\) with \(\norm{f_i}_{L^2}\le M\) for every \(i\). By the Cauchy--Schwarz inequality in $L^2$, \[ \begin{aligned} \norm{f_i^2-f^2}_{L^1} &=\int_0^1 \abs{f_i-f}\,\abs{f_i+f}\,dx\\ &\le \norm{f_i-f}_{L^2}\norm{f_i+f}_{L^2}\\ &\le \qty{M+\norm{f}_{L^2}}\norm{f_i-f}_{L^2} \longrightarrow0. \end{aligned} \]
4. Q.E.D.
Steps 1 and 2 give the counterexample required for part (a), and step 3 proves part (b).