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Spring 2011 - Problem 4

Fatou's lemma, Lp spaces

Let fn ⁣:[0,1]→[0,∞)\func{f_n}{\br{0,1}}{\pco{0,\infty}} be Borel functions with

sup⁡n∫01fn(x)log⁡(2+fn(x)) dx<∞.\sup_n \int_0^1 f_n\p{x} \log\p{2 + f_n\p{x}} \,\diff{x} < \infty.

Suppose fn→ff_n \to f Lebesgue almost everywhere. Show that f∈L1f \in L^1 and fn→ff_n \to f in the L1L^1 sense.

Hint: Consider gn(x)=max⁡ (fn(x),λ)g_n\p{x} = \max\,\p{f_n\p{x}, \lambda} for certain choices of λ\lambda.
Solution.

Set

C=sup⁡n∫01fn(x)log⁡(2+fn(x)) dx.C = \sup_n \int_0^1 f_n\p{x} \log\p{2 + f_n\p{x}} \,\diff{x}.

Since fn≥0f_n \geq 0, we see that log⁡(2+fn)≥log⁡2≥0\log\p{2 + f_n} \geq \log{2} \geq 0. Thus, we may apply Fatou's lemma to get

log⁡(2)∫01f(x)log⁡(2) dx≤∫01f(x)log⁡(2+f(x)) dx=∫01lim inf⁡n→∞ (fn(x)log⁡(2+fn(x))) dx(fn→f and continuity of log⁡x)≤lim inf⁡n→∞∫01fn(x)log⁡(2+fn(x)) dx≤C<∞,\begin{aligned} \log\p{2}\int_0^1 f\p{x}\log\p{2} \,\diff{x} &\leq \int_0^1 f\p{x}\log\p{2 + f\p{x}} \,\diff{x} \\ &= \int_0^1 \liminf_{n\to\infty}\, \p{f_n\p{x}\log\p{2 + f_n\p{x}}} \,\diff{x} && (f_n \to f \text{ and continuity of } \log{x}) \\ &\leq \liminf_{n\to\infty} \int_0^1 f_n\p{x} \log\p{2 + f_n\p{x}} \,\diff{x} \\ &\leq C \\ &< \infty, \end{aligned}

so f∈L1([0,1])f \in L^1\p{\br{0,1}}. To show convergence in L1L^1, first notice that for λ>0\lambda > 0,

∥fn−f∥L1≤∥fnχ{fn>λ}∥L1+∥fnχ{fn≤λ}−fχ{f≤λ}∥L1+∥fχ{f>λ}∥L1.\norm{f_n - f}_{L^1} \leq \norm{f_n\chi_{\set{f_n > \lambda}}}_{L^1} + \norm{f_n\chi_{\set{f_n \leq \lambda}} - f\chi_{\set{f \leq \lambda}}}_{L^1} + \norm{f\chi_{\set{f > \lambda}}}_{L^1}.

Let ε>0\epsilon > 0. For the first term, we have

∥fnχ{fn>λ}∥L1=∫{fn>λ}fn(x)⋅log⁡(2+fn(x))log⁡(2+fn(x)) dx≤1log⁡(2+λ)∫{fn>λ}fn(x)log⁡(2+fn(x)) dx≤Clog⁡(2+λ)→λ→∞0.\begin{aligned} \norm{f_n\chi_{\set{f_n > \lambda}}}_{L^1} &= \int_{\set{f_n > \lambda}} f_n\p{x} \cdot \frac{\log\p{2 + f_n\p{x}}}{\log\p{2 + f_n\p{x}}} \,\diff{x} \\ &\leq \frac{1}{\log\p{2 + \lambda}} \int_{\set{f_n > \lambda}} f_n\p{x} \log\p{2 + f_n\p{x}} \,\diff{x} \\ &\leq \frac{C}{\log\p{2 + \lambda}} \xrightarrow{\lambda\to\infty} 0. \end{aligned}

Thus, we may pick λ>0\lambda > 0 so that this term is smaller than ε\epsilon.

Next, notice that fnχ{fn≤λ}→fχ{f≤λ}f_n\chi_{\set{f_n \leq \lambda}} \to f\chi_{\set{f \leq \lambda}} almost everywhere. Indeed, if xx is such that fn(x)→f(x)f_n\p{x} \to f\p{x}, then there are three cases: if f(x)>λf\p{x} > \lambda, then for nn large enough, we must have fn(x)>λf_n\p{x} > \lambda as well, so fn(x)χ{fn≤λ}(x)=0=f(x)χ{f≤λ}f_n\p{x}\chi_{\set{f_n \leq \lambda}}\p{x} = 0 = f\p{x}\chi_{\set{f \leq \lambda}}. If f(x)=λf\p{x} = \lambda, then fn(x)χ{fn≤λ}(x)=λf_n\p{x}\chi_{\set{f_n \leq \lambda}}\p{x} = \lambda or fn(x)f_n\p{x}, both of which converge to λ\lambda. Finally, if f(x)<λf\p{x} < \lambda, then eventually fn(x)<λf_n\p{x} < \lambda, so fn(x)χ{fn≤λ}(x)=fn(x)→f(x)=f(x)χ{f≤λ}f_n\p{x}\chi_{\set{f_n \leq \lambda}}\p{x} = f_n\p{x} \to f\p{x} = f\p{x}\chi_{\set{f \leq \lambda}}. Moreover, fnχ{fn≤λ}≤λ∈L1([0,1])f_n\chi_{\set{f_n \leq \lambda}} \leq \lambda \in L^1\p{\br{0, 1}}, so by dominated convergence, the second term vanishes as n→∞n \to \infty.

Finally, observe that ∣f−fχ{f<n}∣→0\abs{f - f\chi_{\set{f < n}}} \to 0 almost everywhere and ∣f−fχ{f<n}∣≤2f∈L1([0,1])\abs{f - f\chi_{\set{f < n}}} \leq 2f \in L^1\p{\br{0,1}}, so by dominated convergence,

∣∫fχ{f≥n} dx∣≤∫∣f−fχ{f<n}∣ dx→n→∞0.\abs{\int f\chi_{\set{f \geq n}} \,\diff{x}} \leq \int \abs{f - f\chi_{\set{f < n}}} \,\diff{x} \xrightarrow{n\to\infty} 0.

Thus, making λ\lambda larger if necessary, we have ∥fχ{f>λ}∥L1<ε\norm{f\chi_{\set{f > \lambda}}}_{L^1} < \epsilon. Hence,

lim sup⁡n→∞ ∥fn−f∥L1≤2ε.\limsup_{n\to\infty}\, \norm{f_n - f}_{L^1} \leq 2\epsilon.

Sending ε→0\epsilon \to 0, we get fn→ff_n \to f in L1L^1.