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Fall 2015 - Problem 2

Lp spaces

Let f∈Lp(R)f \in L^p\p{\R}, 1<p<∞1 < p < \infty, and let a∈Ra \in \R be such that a>1−1pa > 1 - \frac{1}{p}. Show that the series

∑n=1∞∫nn+n−a∣f(x+y)∣ dy\sum_{n=1}^\infty \int_n^{n + n^{-a}} \abs{f\p{x + y}} \,\diff{y}

converges for almost all x∈Rx \in \R.
Solution.

Let N∈NN \in \N. We will show that

∫NN+1∑n=1∞∫nn+n−a∣f(x+y)∣ dy dx<∞,\int_N^{N+1} \sum_{n=1}^\infty \int_n^{n+n^{-a}} \abs{f\p{x + y}} \,\diff{y} \,\diff{x} < \infty,

which proves the claim. Since ∣f(x+y)∣\abs{f\p{x + y}} is non-negative, we may apply Fubini-Tonelli (viewing summation as integration with respect to the counting measure), which yields

∫NN+1∑n=1∞∫nn+n−a∣f(x+y)∣ dy dx=∫NN+1∑n=1∞∫Rχ[n,n+n−a](y)∣f(x+y)∣ dy dx=∫NN+1∑n=1∞∫Rχ[n,n+n−a](y−x)∣f(y)∣ dy dx(y↦x+y)=∫R∣f(y)∣∫NN+1∑n=1∞χ[y−(n+n−a),y−n](x) dx dy.\begin{aligned} \int_N^{N+1} \sum_{n=1}^\infty \int_n^{n+n^{-a}} \abs{f\p{x + y}} \,\diff{y} \,\diff{x} &= \int_N^{N+1} \sum_{n=1}^\infty \int_\R \chi_{\br{n, n+n^{-a}}}\p{y} \abs{f\p{x + y}} \,\diff{y} \,\diff{x} \\ &= \int_N^{N+1} \sum_{n=1}^\infty \int_\R \chi_{\br{n, n+n^{-a}}}\p{y - x} \abs{f\p{y}} \,\diff{y} \,\diff{x} && \p{y \mapsto x + y} \\ &= \int_\R \abs{f\p{y}} \int_N^{N+1} \sum_{n=1}^\infty \chi_{\br{y-\p{n+n^{-a}}, y-n}}\p{x} \,\diff{x} \,\diff{y}. \end{aligned}

Notice that in the inner integral, χ[y−(n+n−a),y−n](x)\chi_{\br{y-\p{n+n^{-a}}, y-n}}\p{x} will vanish whenever y−n<Ny - n < N or y−(n+n−a)>N+1y - \p{n + n^{-a}} > N + 1. Thus, the only terms that remain satisfy n≤y−Nn \leq y - N and n+n−a≥y−(N+1)n + n^{-a} \geq y - \p{N + 1}. In particular, if n∈[(y−N)−1,y−N]n \in \br{\p{y - N} - 1, y - N}, then both inequalities are satisfied. Thus, there are at most two non-vanishing terms, so

∫NN+1∑n=1∞χ[y−(n+n−a),y−n](x) dx≤∑(y−N)−1≤n≤y−Nmin⁡ {1,n−a}≤2min⁡ {1,∣y−N∣−a}.\begin{aligned} \int_N^{N+1} \sum_{n=1}^\infty \chi_{\br{y-\p{n+n^{-a}}, y-n}}\p{x} \,\diff{x} &\leq \sum_{\p{y - N} - 1 \leq n \leq y - N} \min\,\set{1, n^{-a}} \\ &\leq 2\min\,\set{1, \abs{y - N}^{-a}}. \end{aligned}

Hence, our original expression is bounded by

2∫R∣f(y)∣min⁡ {1,∣y−N∣−a} dy=2∥f∥Lp∥min⁡ {1,∣y−N∣−a}∥Lq,\begin{aligned} 2\int_\R \abs{f\p{y}} \min\,\set{1, \abs{y - N}^{-a}} \,\diff{y} &= 2\norm{f}_{L^p} \norm{\min\,\set{1, \abs{y - N}^{-a}}}_{L^q}, \end{aligned}

where 1p+1q=1\frac{1}{p} + \frac{1}{q} = 1. Observe that by assumption, aq>1aq > 1, so ∣x∣−aq\abs{x}^{-aq} is integrable away from 00. Hence,

∥min⁡ {1,∣y−N∣−a}∥Lqq≤∫B(N,1) dy+∫B(N,1)c∣y−N∣−aq dy<∞.\norm{\min\,\set{1, \abs{y - N}^{-a}}}_{L^q}^q \leq \int_{B\p{N,1}} \,\diff{y} + \int_{B\p{N,1}^\comp} \abs{y - N}^{-aq} \,\diff{y} < \infty.

Thus, the series is finite for almost every x∈[N,N+1]x \in \br{N, N + 1}, and because N∈NN \in \N was arbitrary, it follows that the series is finite almost everywhere.