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Fall 2018 - Problem 1

construction, Lp spaces

Let $\set{f_n}_n$ be a sequence of real-valued Lebesgue measurable functions on $\R$ and let $f$ be another such function. Assume that

$f_n \to f$ , Lebesgue almost everywhere,
$\int_\R \abs{x} \abs{f_n\p{x}} \,\diff{x} \leq 100$ , for all $n$ , and
$\int_\R \abs{f_n\p{x}}^2 \,\diff{x} \leq 100$ , for all $n$ >

Prove that $f_n \in L^1$ for all $n$ , that $f \in L^1$ , and that $\norm{f_n - f}_{L^1} \to 0$ . Also show that neither assumption (ii) nor assumption (iii) can be omitted while making these deductions.

Solution.

Let $g$ be a Lebesgue measurable function on $\R$ satisfying conditions (ii) and (iii). Then

\begin{aligned} \int_\R \abs{g\p{x}} \,\diff{x} &= \int_{\set{\abs{x} \leq 1}} \abs{g\p{x}} \,\diff{x} + \int_{\set{\abs{x} > 1}} \abs{g\p{x}} \,\diff{x} \\ &= \int_{\set{\abs{x} \leq 1} \cap \set{\abs{g\p{x}} \geq 1}} \abs{g\p{x}} \,\diff{x} + \int_{\set{\abs{x} \leq 1} \cap \set{\abs{g\p{x}} < 1}} \abs{g\p{x}} \,\diff{x} + \int_{\set{\abs{x} > 1}} \abs{g\p{x}} \,\diff{x} \\ &\leq \int_{\set{\abs{x} \leq 1} \cap \set{\abs{g\p{x}} \geq 1}} \abs{g\p{x}}^2 \,\diff{x} + \int_{\set{\abs{x} \leq 1}} \,\diff{x} + \int_{\set{\abs{x} > 1}} \abs{x} \abs{g\p{x}} \,\diff{x} \\ &\leq 200 + 2 \\ &< \infty, \end{aligned}

so $g \in L^1$ . Hence, $f_n \in L^1$ for all $n$ . By Fatou's lemma, we also see that

\begin{gathered} \int_\R \abs{x} \abs{f\p{x}} \,\diff{x} \leq \liminf_{n\to\infty} \int_\R \abs{x} \abs{f_n\p{x}} \,\diff{x} \leq 100 \\ \int_\R \abs{f\p{x}}^2 \,\diff{x} \leq \liminf_{n\to\infty} \int_\R \abs{f_n\p{x}}^2 \,\diff{x} \leq 100, \end{gathered}

so $f$ satisfies the same conditions, i.e., $f \in L^1$ as well. To prove $L^1$ convergence, fix $\epsilon > 0$ . Then observe that

\int_{\set{\abs{x} \geq R}} \abs{f_n\p{x}} \,\diff{x} \leq \int_{\set{\abs{x} \geq R}} \frac{\abs{x}\abs{f_n\p{x}}}{\abs{x}} \,\diff{x} \leq \frac{100}{R},

and this inequality is true with $f_n$ replaced with $f$ as well. Thus, we may pick $R > 0$ large enough so that $\int_{\set{\abs{x} \geq R}} \abs{f_n\p{x}} \,\diff{x} < \epsilon$ and $\int_{\set{\abs{x} \geq R}} \abs{f\p{x}} \,\diff{x} < \epsilon$ . On the finite measure set $B\p{0, R}$ , we may apply Egorov's theorem, since $f_n \to f$ almost everywhere. Hence, we get $E \subseteq B\p{0, R}$ such that $f_n \to f$ uniformly on $B\p{0, R} \setminus E$ and with $m\p{E} < \epsilon$ . Thus, if $n$ is large enough, then $\abs{f_n - f} < \frac{\epsilon}{2R}$ on $B\p{0, R} \setminus E$ , and so

\begin{aligned} \int_\R \abs{f_n - f} \,\diff{x} &\leq \int_{B\p{0,R}} \abs{f_n - f} \,\diff{x} + \int_{\set{\abs{x} \geq R}} \abs{f_n} \,\diff{x} + \int_{\set{\abs{x} \geq R}} \abs{f} \,\diff{x} \\ &\leq \int_E \abs{f_n - f} \,\diff{x} + \int_{B\p{0,R} \setminus E} \abs{f_n - f} \,\diff{x} + 2\epsilon \\ &\leq \norm{f_n - f}_{L^2} m\p{E}^{1/2} + 3\epsilon \\ &\leq \epsilon^{1/2}\p{\norm{f_n}_{L^2} + \norm{f}_{L^2}} + 3\epsilon \\ &\leq 20\epsilon^{1/2} + 3\epsilon, \end{aligned}

so $\norm{f_n - f}_{L^1} \to 0$ .

To see that (2) is necessary, consider $f_n\p{x} = f\p{x} = \abs{x}^{-1} \chi_{B\p{0,1}}$ . Then these satisfy (1) trivially and satisfy (2) since $\abs{x}\abs{f_n\p{x}} = \chi_{B\p{0,1}}$ , which certainly has integral bounded by $100$ . Moreover, it does not satisfy (3) since $f_n \notin L^2$ . However, $f_n \notin L^1$ , since $\abs{x}^{-1}$ is not integrable near the origin, so the conclusion fails.

Similarly, to see that (3) is necessary, let $f_n\p{x} = f\p{x} = \abs{x}^{-1} \chi_{\set{\abs{x} \geq 1}}$ . This it's clear that (1) and (3) are satisfied, but (2) is not satisfied since $\abs{x}\abs{f_n\p{x}} = \chi_{\set{\abs{x} \geq 1}}$ , which has infinite mass. As before, $f_n \notin L^1$ , so the conclusion fails.