<Home>Qualifying Exams><Analysis>Fall 2016 - Problem 1

Fall 2016 - Problem 1

Lp spaces

We consider the space L1(μ)L^1\p{\mu} of integrable functions on a measure space (X,M,μ)\p{X, \mathcal{M}, \mu}. For g∈L1(μ)g \in L^1\p{\mu}, let

∥g∥L1=∫∣g(x)∣ dμ\norm{g}_{L^1} = \int \abs{g\p{x}} \,\diff\mu

be the corresponding L1L^1-norm. Suppose that ff and fnf_n for n∈Nn \in \N are functions in L1(μ)L^1\p{\mu} such that

  1. fn(x)→f(x)f_n\p{x} \to f\p{x} for μ\mu-almost every x∈Xx \in X and
  2. ∥fn∥L1→∥f∥L1\norm{f_n}_{L^1} \to \norm{f}_{L^1}.

Show that then ∥fn−f∥L1→0\norm{f_n - f}_{L^1} \to 0.
Solution.

Observe that

∣fn−f∣≤∣fn∣+∣f∣  ⟹  ∣fn∣+∣f∣−∣fn−f∣≥0\abs{f_n - f} \leq \abs{f_n} + \abs{f} \implies \abs{f_n} + \abs{f} - \abs{f_n - f} \geq 0

by the triangle inequality. Hence, we may apply Fatou's lemma to obtain

∫lim inf⁡n→∞ (∣fn∣+∣f∣−∣fn−f∣) dμ≤lim inf⁡n→∞ (∫∣fn∣ dμ+∫∣f∣ dμ−∫∣fn−f∣ dμ)=lim inf⁡n→∞ (∥fn∥L1+∥f∥L1−∥fn−f∥L1)=2∥f∥L1−lim sup⁡n→∞ ∥fn−f∥L1.\begin{aligned} \int \liminf_{n\to\infty}\,\p{\abs{f_n} + \abs{f} - \abs{f_n - f}} \,\diff\mu &\leq \liminf_{n\to\infty}\,\p{\int \abs{f_n} \,\diff\mu + \int \abs{f} \,\diff\mu - \int \abs{f_n - f} \,\diff\mu} \\ &= \liminf_{n\to\infty}\,\p{\norm{f_n}_{L^1} + \norm{f}_{L^1} - \norm{f_n - f}_{L^1}} \\ &= 2\norm{f}_{L^1} - \limsup_{n\to\infty}\,\norm{f_n - f}_{L^1}. \end{aligned}

Notice that because fn→ff_n \to f almost everywhere, the left-hand side becomes

∫lim inf⁡n→∞ (∣fn∣+∣f∣−∣fn−f∣) dμ=∫2∣f∣ dμ=2∥f∥L1.\int \liminf_{n\to\infty}\,\p{\abs{f_n} + \abs{f} - \abs{f_n - f}} \,\diff\mu = \int 2\abs{f} \,\diff\mu = 2\norm{f}_{L^1}.

Rearranging the first inequality, we get

lim sup⁡n→∞ ∥fn−f∥L1≤0,\limsup_{n\to\infty}\,\norm{f_n - f}_{L^1} \leq 0,

so ∥fn−f∥L1→0\norm{f_n - f}_{L^1} \to 0.