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Fall 2012 - Problem 5

Egorov's theorem, Lp spaces

Let fn ⁣:R3→R\func{f_n}{\R^3}{\R} be a sequence of functions such that sup⁡n∥fn∥L2<∞\sup_n\norm{f_n}_{L^2} < \infty. Show that if fnf_n converges almost everywhere to a function f ⁣:R3→R\func{f}{\R^3}{\R}, then

∫R3∣∣fn∣2−∣fn−f∣2−∣f∣2∣ dx→0.\int_{\R^3} \abs{\abs{f_n}^2 - \abs{f_n - f}^2 - \abs{f}^2} \,\diff{x} \to 0.
Solution.

First, observe that

∣∣fn∣2−∣fn−f∣2−∣f∣2∣=∣∣fn∣2−(fn2+f2−2ffn)−∣f∣2∣=∣2ffn−2f2∣=2∣f∣∣f−fn∣.\begin{aligned} \abs{\abs{f_n}^2 - \abs{f_n - f}^2 - \abs{f}^2} &= \abs{\abs{f_n}^2 - \p{f_n^2 + f^2 - 2ff_n} - \abs{f}^2} \\ &= \abs{2ff_n - 2f^2} \\ &= 2\abs{f}\abs{f - f_n}. \end{aligned}

Let M=sup⁡n∥fn∥L2M = \sup_n \norm{f_n}_{L^2}, and so by Fatou's lemma,

∫∣f∣2≤lim inf⁡n→∞∫∣fn∣2≤M2<∞\int \abs{f}^2 \leq \liminf_{n\to\infty} \int \abs{f_n}^2 \leq M^2 < \infty

by assumption. In other words, f∈L2(R3)f \in L^2\p{\R^3}. We need to analyze

∫R3∣∣fn∣2−∣fn−f∣2−∣f∣2∣ dx=∫R32∣f∣∣f−fn∣ dx.\int_{\R^3} \abs{\abs{f_n}^2 - \abs{f_n - f}^2 - \abs{f}^2} \,\diff{x} = \int_{\R^3} 2\abs{f}\abs{f - f_n} \,\diff{x}.

Let ε>0\epsilon > 0. Since f∈L2(R3)f \in L^2\p{\R^3}, there exists R>0R > 0 large enough so that

∫B(0,R)c∣f∣2<ε.\int_{B\p{0,R}^\comp} \abs{f}^2 < \epsilon.

Also, observe that E↦∫E∣f∣2 dxE \mapsto \int_E \abs{f}^2 \,\diff{x} is a measure absolutely continuous with respect to the Lebesgue measure. Hence, there exists δ>0\delta > 0 such that if m(E)<δm\p{E} < \delta, then ∫E∣f∣2 dx<ε\int_E \abs{f}^2 \,\diff{x} < \epsilon.

On the other hand, since B(0,R)B\p{0, R} is a set of finite measure, Egorov's theorem gives us a set E⊆B(0,R)E \subseteq B\p{0, R} such that fn→ff_n \to f uniformly on B(0,R)∖EB\p{0, R} \setminus E and m(E)<δm\p{E} < \delta, where mm denotes the Lebesgue measure. Thus, on B(0,R)∖EB\p{0, R} \setminus E, there exists N∈NN \in \N such that ∣fn−f∣<ε\abs{f_n - f} < \epsilon uniformly, and so by Cauchy-Schwarz,

∫R32∣f∣∣f−fn∣ dx=∫B(0,R)∖E2∣f∣∣f−fn∣ dx+∫E2∣f∣∣f−fn∣ dx+∫B(0,R)c2∣f∣∣f−fn∣ dx≤2∥f∥L2∥(f−fn)χB(0,R)∖E∥L2+2∥fχE∥L2∥f−fn∥L2+2∥fχB(0,R)c∥L2∥f−fn∥L2≤2Mε1/2+4Mε1/2+4Mε1/2=10Mε1/2.\begin{aligned} \int_{\R^3} 2\abs{f}\abs{f - f_n} \,\diff{x} &= \int_{B\p{0, R} \setminus E} 2\abs{f}\abs{f - f_n} \,\diff{x} + \int_E 2\abs{f}\abs{f - f_n} \,\diff{x} + \int_{B\p{0, R}^\comp} 2\abs{f}\abs{f - f_n} \,\diff{x} \\ &\leq 2 \norm{f}_{L^2} \norm{\p{f - f_n}\chi_{B\p{0,R} \setminus E}}_{L^2} + 2 \norm{f\chi_E}_{L^2} \norm{f - f_n}_{L^2} + 2\norm{f\chi_{B\p{0,R}^\comp}}_{L^2} \norm{f - f_n}_{L^2} \\ &\leq 2M\epsilon^{1/2} + 4M\epsilon^{1/2} + 4M\epsilon^{1/2} \\ &= 10M\epsilon^{1/2}. \end{aligned}

Sending ε→0\epsilon \to 0, we get the claim.