<Home>Qualifying Exams><Analysis>Spring 2010 - Problem 5

Spring 2010 - Problem 5

density argument, measure theory

  1. For f∈L2(R)f \in L^2\p{\R} and a sequence {xn}n≥1⊆R\set{x_n}_{n\geq1} \subseteq \R which converges to zero, define

    fn(x)=f(x+xn).f_n\p{x} = f\p{x + x_n}.

    Show that {fn}n≥1\set{f_n}_{n\geq1} converges to ff in the L2L^2 sense.

  2. Let W⊆RW \subseteq \R be a Lebesgue measurable set of positive Lebesgue measure. Show that the set of differences

    W−W={x−y∣x,y∈W}W - W = \set{x - y \mid x, y \in W}

    contains an open neighborhood of the origin.
Solution.

  1. We first establish the result for compactly supported smooth functions. Let f∈Cc∞(R)f \in C_c^\infty\p{\R}. Let ε>0\epsilon > 0. Since ff has compact support, it is uniformly continuous on R\R, so there exists δ>0\delta > 0 such that if ∣x−y∣<δ\abs{x - y} < \delta, then ∣f(x)−f(y)∣<ε\abs{f\p{x} - f\p{y}} < \epsilon. Thus, if nn is large enough, ∣xn∣<δ\abs{x_n} < \delta. Let EE denote the support of ff, so in the worst case, the support of fn−ff_n - f has measure 2m(E)2m\p{E} when both functions have disjoint support, where mm denotes the Lebesgue measure. Thus,

    ∫R∣f(x+xn)−f(x)∣2 dx≤2m(E)ε2→ε→00,\int_\R \abs{f\p{x + x_n} - f\p{x}}^2 \,\diff{x} \leq 2m\p{E} \epsilon^2 \xrightarrow{\epsilon\to0} 0,

    so ∥fn−f∥L2→0\norm{f_n - f}_{L^2} \to 0 if f∈Cc∞(R)f \in C_c^\infty\p{\R}. For f∈L2(R)f \in L^2\p{\R}, we exploit the fact that Cc∞(R)C_c^\infty\p{\R} is dense in L2(R)L^2\p{\R} with respect to the L2L^2-norm.

    Let ε>0\epsilon > 0 and pick g∈Cc∞(R)g \in C_c^\infty\p{\R} such that ∥f−g∥L2<ε\norm{f - g}_{L^2} < \epsilon. Thus, if nn is large enough, ∥g(x+xn)−g(x)∥L2<ε\norm{g\p{x + x_n} - g\p{x}}_{L^2} < \epsilon also, which gives

    ∥f(x+xn)−f(x)∥L2≤∥f(x+xn)−g(x+xn)∥L2+∥f(x)−g(x)∥L2+∥g(x+xn)−g(x)∥L2<3ε.\begin{aligned} \norm{f\p{x + x_n} - f\p{x}}_{L^2} \leq \norm{f\p{x + x_n} - g\p{x + x_n}}_{L^2} + \norm{f\p{x} - g\p{x}}_{L^2} + \norm{g\p{x + x_n} - g\p{x}}_{L^2} < 3\epsilon. \end{aligned}

    Indeed, the first two terms are smaller than ε\epsilon by density and translation invariance, and the third term is smaller than ε\epsilon since nn is large. Letting ε→0\epsilon \to 0 completes the proof.

  2. Let f=χWf = \chi_W. By intersecting WW with a large enough ball centered at the origin, we may assume without loss of generality that 0<m(W)<∞0 < m\p{W} < \infty, and so f∈L2(R)f \in L^2\p{\R}. By (1),

    ∥f(x+y)−f(x)∥L22=∫(χW(x+y)−χW(x))2 dx=∫χW(x)2+χW(x+y)2−2χW(x+y)χW(x) dx=2m(W)−2∫χW(x)χW(x+y) dx\begin{aligned} \norm{f\p{x + y} - f\p{x}}_{L^2}^2 &= \int \p{\chi_W\p{x + y} - \chi_W\p{x}}^2 \,\diff{x} \\ &= \int \chi_W\p{x}^2 + \chi_W\p{x + y}^2 - 2\chi_W\p{x + y}\chi_W\p{x} \,\diff{x} \\ &= 2m\p{W} - 2 \int \chi_{W}\p{x}\chi_W\p{x + y} \,\diff{x} \end{aligned}

    tends to 00 as y→0y \to 0. In particular, there exists δ>0\delta > 0 so that if ∣y∣<δ\abs{y} < \delta, then the quantity is smaller than m(W)m\p{W}, and so

    2m(W)−2∫χW(x)χW(x+y) dx≤m(W)  ⟹  0<m(W)2≤∫χW(x)χW(x+y) dx.2m\p{W} - 2 \int \chi_{W}\p{x}\chi_W\p{x + y} \,\diff{x} \leq m\p{W} \implies 0 < \frac{m\p{W}}{2} \leq \int \chi_{W}\p{x}\chi_W\p{x + y} \,\diff{x}.

    In particular, if ∣y∣<δ\abs{y} < \delta, then there exists x∈Wx \in W such that x+y∈Wx + y \in W also, and so y=x+y−x∈W−Wy = x + y - x \in W - W, so W−WW - W contains an open neighborhood of 00.