Let \(f_j,f\in L^2(\RR^n)\) and suppose \[ \norm{f_j-f}_2\longrightarrow0. \]
If \(\Omega\subset\RR^n\) has finite measure, prove that \[ \int_\Omega \widehat f_j(\xi)\,d\xi\longrightarrow\int_\Omega \widehat f(\xi)\,d\xi. \]
If \(\abs{\Omega}=\infty\), must the same conclusion hold? Prove your answer or give a counterexample.
Solution
1. Finite-measure sets.
By the Plancherel theorem, \[ \norm{\widehat f_j-\widehat f}_2=\norm{f_j-f}_2\longrightarrow0. \] Since \(\abs{\Omega}<\infty\), the Cauchy--Schwarz inequality in $L^2$ gives \[ \abs{\int_\Omega(\widehat f_j-\widehat f)(\xi)\,d\xi} \le \abs{\Omega}^{1/2}\,\norm{\widehat f_j-\widehat f}_2\longrightarrow0. \] This proves part (a).
2. Infinite-measure sets: the conclusion can fail.
Assume \(\abs{\Omega}=\infty\). Choose measurable sets \(E_j\subset\Omega\) with \[ 0<\abs{E_j}<\infty, \qquad \abs{E_j}\longrightarrow\infty. \] Define \[ g_j(\xi)=\frac{\mathbf1_{E_j}(\xi)}{\abs{E_j}}. \] Then \[ \norm{g_j}_2^2=\frac1{\abs{E_j}}\longrightarrow0, \qquad \int_\Omega g_j(\xi)\,d\xi=1. \] By the Plancherel theorem, for each \(j\) there exists \(f_j\in L^2(\RR^n)\) with Fourier transform \(g_j\). Taking \(f=0\), we have \[ \norm{f_j-f}_2\to0, \] but \[ \int_\Omega \widehat f_j=1 \qquad\text{for every }j, \] while \(\int_\Omega\widehat f=0\). Thus the conclusion in part (a) need not hold when \(\abs{\Omega}=\infty\).
3. Q.E.D.
Step 1 proves part (a), and step 2 supplies the counterexample required for part (b).