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Spring 2015 - Problem 2

Lp spaces

Let $f \in L^2_{\mathrm{loc}}\p{\R^n}$ , $g \in L^3_{\mathrm{loc}}\p{\R^n}$ . Assume that for all real $r \geq 1$ , we have

\int_{\set{r \leq \abs{x} \leq 2r}} \abs{f\p{x}}^2 \,\diff{x} \leq r^a, \quad \int_{\set{r \leq \abs{x} \leq 2r}} \abs{g\p{x}}^3 \,\diff{x} \leq r^b.

Here $a, b \in \R$ are such that $3a + 2b + n < 0$ . Show that $fg \in L^1\p{\R^n}$ .

Solution.

Observe that $\frac{1}{2} + \frac{1}{3} + \frac{1}{6} = 1$ . Hence, by Hölder's inequality,

\begin{aligned} \int_{B\p{0,1}} \abs{f\p{x}g\p{x}} \,\diff{x} &\leq \norm{f\chi_{B\p{0,1}}}_{L^2} \norm{g\chi_{B\p{0,1}}}_{L^3} \norm{\chi_{B\p{0,1}}}_{L^6} \\ &\leq \norm{f\chi_{B\p{0,1}}}_{L^2} \norm{g\chi_{B\p{0,1}}}_{L^3} m\p{B\p{0,1}}^{1/6} \\ &< \infty, \end{aligned}

since $f \in L^2_{\mathrm{loc}}\p{\R^n}$ and $g \in L^3_{\mathrm{loc}}\p{\R^n}$ . Hence, it remains to consider the integral on $B\p{0, 1}^\comp$ . First, recall that $m\p{B\p{0, r}} = Cr^n$ for some constant $C > 0$ depending only on the dimension $n$ . Thus, applying Hölder again,

\begin{aligned} \int_{B\p{0,1}^\comp} \abs{f\p{x}g\p{x}} \,\diff{x} &= \sum_{k=0}^\infty \int_{\set{2^k \leq \abs{x} \leq 2^{k+1}}} \abs{f\p{x}g\p{x}} \,\diff{x} \\ &\leq \sum_{k=0}^\infty \p{\int_{\set{2^k \leq \abs{x} \leq 2^{k+1}}} \abs{f\p{x}}^2 \,\diff{x}}^{1/2} \p{\int_{\set{2^k \leq \abs{x} \leq 2^{k+1}}} \abs{f\p{x}}^3 \,\diff{x}}^{1/3} \p{\int_{\set{2^k \leq \abs{x} \leq 2^{k+1}}} \diff{x}}^{1/6} \\ &\leq \sum_{k=0}^\infty 2^{ka/2} 2^{kb/3} \p{m\p{B\p{0, 2^{k+1}}} - m\p{B\p{0, 2^k}}}^{1/6} \\ &= C^{1/6}\p{2^{n} - 1} \sum_{k=0}^\infty 2^{ka/2} 2^{kb/3} 2^{kn/6} \\ &= C^{1/6}\p{2^{n} - 1} \sum_{k=0}^\infty \p{2^{\p{3a+2b+n}/6}}^k \\ &< \infty, \end{aligned}

since $3a + 2b + n < 0$ by assumption. Thus, $fg \in L^1\p{\R^n}$ , as required.