Integration Theory / Mathematical Analysis (5B1479/MA429)

Solutions to homework assignment # 4

1.a. Use polar coordinates x = r cos θ, y = r sin θ and note that mentioned square is contained in the region \{0 ≤ r\sqrt{2}, 0 ≤ θ \frac{π}{2} \}, then f(x, y) = \frac{1}{r} and

∫_{[0,1]×[0,1]} f(x, y)dxdy ≤ ∫_{[0,\sqrt{2}]×[0, \frac{π}{2} ]} f · rdrdθ = ∫_{[0,\sqrt{2}]×[0, \frac{π}{2} ]} 1drdθ = \frac{\sqrt{2}π}{2} .

b.

∫_{0}^{1} ∫_{0}^{1} \frac{x − y}{(x + y)^{3}} dxdy = ∫_{0}^{1} ∫_{0}^{1} (\frac{1}{(x + y)^{2}} − \frac{2y}{(x + y)^{3}} ) dxdy = ∫_{0}^{1} (−\frac{1}{(x + y)^{1}} + \frac{y}{(x + y)^{2}} )|_{x=0}^{1} dy

= ∫_{0}^{1} (−\frac{1}{1 + y} + \frac{1}{y} + \frac{y}{(1 + y)^{2}} − \frac{y}{y^{2}} ) dy = ∫_{0}^{1} −\frac{1}{(1 + y)^{2}} dy = \frac{1}{1 + y} |_{y=0}^{1} = −\frac{1}{2} .

Likewise one shows that ∫_{0}^{1} ∫_{0}^{1} \frac{x−y}{(x+y)^{3}} dydx = \frac{1}{2}. Fubini theorem cannot be applied, since \frac{x−y}{(x+y)^{3}} is not Lebesgue integrable in the square [0, 1] × [0, 1]. One can see that by evaluating he iterated integral of its absolute value:

∫_{0}^{1} ∫_{0}^{1} |\frac{x − y}{(x + y)^{3}} | dxdy = ∫_{0}^{1} (∫_{0}^{y} −\frac{x − y}{(x + y)^{3}} dx + ∫_{y}^{1} \frac{x − y}{(x + y)^{3}} dx) dy

= ∫_{0}^{1} (− (−\frac{1}{(x + y)^{1}} + \frac{y}{(x + y)^{2}} )|_{x=0}^{y} + (−\frac{1}{(x + y)^{1}} + \frac{y}{(x + y)^{2}} )|_{x=y}^{1}) dy

= ∫_{0}^{1} (\frac{1}{2y} − \frac{1}{(1 + y)^{2}} ) dy = +∞ .

2.a. B_{1} is generated by open intervals, so B_{1} × B_{1} is generated by open rectangles, while B_{2} is generated by open discs. Any open rectangle is a countable union of open circles, and vice versa, therefore B_{1} × B_{1} = B_{2}. b. Consider a set E := A × {0} for some Lebesgue non-measurable subset A of the real line. Then E ⊂ ℝ × {0}, so it is a null set for the Lebesgue measure in ℝ^{2}, and hence E belongs to L_{2}. On the other hand section E^{0} = {x : (x, 0) ∈ E} = A ∉ L_{1}, so by the Lemma 2.15.1 set E does not belong to L_{1} × L_{1}.

One can also easily show, that L_{1} × L_{1} ⊂ L_{2}. 3.a. Suppose that function h ∈ L^{r}(X, μ). Then by Hölder inequality for functions f = |h|^{p} and g = 1 and powers P = \frac{r}{p} and Q = \frac{r}{r−p} (clearly \frac{1}{P} + \frac{1}{Q} = 1) one has

∫ |h|^{p}dμ = ∫ fgμ ≤ (∫ f^{P} dμ)^{\frac{1}{P}} (∫ g^{Q}dμ)^{\frac{1}{Q}}

= (∫ |h|^{p \frac{r}{p}} dμ)^{\frac{1}{P}} (∫ 1dμ)^{\frac{1}{Q}} = (∫ |h|^{r}dμ)^{\frac{1}{P}} μ(X)^{\frac{1}{Q}} ≤ +∞ ,

since h ∈ L^{r} and μ(X) < ∞. Thus ∫ |h|^{p}dμ < ∞ and h ∈ L^{p}(X, μ).

One can give a proof not employing the Hölder inequality, but using obvious |f|^{p} ≤ 1 + |f|^{r} instead. b. Set X := [1, +∞), and let μ be the usual Lebesgue measure. It is easy to check that function h(x) = x^{−1/r} belongs to L^{p}[1, +∞) for any p > r, but not to L^{r}[1, +∞).

4. Suppose that ϕ is continuous and zero outside interval [−M, M]. Then it is uniformly continuous on [−M, M], meaning that lim_{ϵ→0} ω_{ϕ}(ϵ) = 0, where ω_{f}(ϵ) := sup_{|x−y|≤ϵ} |f(x) − f(y)| is its modulus of continuity. In this case

∥ϕ(x) − ϕ(x + ϵ)∥_{1} = ∫_{M}^{M+ϵ} ∥ϕ(x) − ϕ(x + ϵ)∥dx ≤ ∫_{M}^{M+ϵ} ω_{ϕ}(ϵ)dx = (2M + ϵ)ω_{ϕ}(ϵ) ⟶^{ϵ→0} 0,

so lim_{ϵ→0} ∥ϕ(x) − ϕ(x + ϵ)∥_{1} = 0.

Now take arbitrary function f ∈ L^{1}(ℝ). Fix some δ > 0. Since lim_{M→∞} ∫_{−∞}^{−M} + ∫_{M}^{∞} |f|dx = 0, we

can pick such M that ∫_{−∞}^{−M} + ∫_{M}^{∞} |f|dx < δ/4. Now by the corollary of Lusin’s theorem one can find a

continuous function ϕ, zero outside [−M, M], such that ∫_{−M}^{M} |f − ϕ|dx < δ/4. Then

∥f − ϕ∥_{1} = ∫_{ℝ} |f − ϕ|dx = ∫_{−M}^{M} |f − ϕ|dx + ∫_{−∞}^{−M} + ∫_{M}^{∞} |f|dx < δ/2.

Hence we can write

lim sup_{ϵ→0} ∥f(x) − f(x + ϵ)∥_{1} ≤ lim sup_{ϵ→0} (∥f(x) − ϕ(x)∥_{1} + ∥ϕ(x) − ϕ(x + ϵ)∥_{1} + ∥ϕ(x + ϵ) − f(x + ϵ)∥_{1})
≤ ∥f(x) − ϕ(x)∥_{1} + lim sup_{ϵ→0} ∥ϕ(x) − ϕ(x + ϵ)∥_{1} + ∥f(x) − ϕ(x)∥_{1} ≤ δ.

Since δ was arbitrary, this implies that lim_{ϵ→0} ∥f(x) − f(x + ϵ)∥_{1} = 0. 5.a. Using in (∗) Fubini theorem for non-negative functions, we get

∫_{ℝ} |f ∗ g(x)|dx = ∫_{ℝ} |∫_{ℝ} f(x − y)g(y)dy| dx ≤ ∫_{ℝ} ∫_{ℝ} |f(x − y)g(y)| dydx =^{∗} ∫_{ℝ} ∫_{ℝ} |f(x − y)g(y)| dydx
= ∫_{ℝ} ∫_{ℝ} |f(x − y)||g(y)|dxdy = ∫_{ℝ} (∫_{ℝ} |f(x − y)|dx) |g(y)|dy
= ∫_{ℝ} ∥f∥_{1}|g(y)|dy = ∥f∥_{1} ∫_{ℝ} |g(y)|dy = ∥f∥_{1}∥g∥_{1} < ∞.

Thus f ∗ g belongs to L^{1}(ℝ). b. Changing the variable y = x − u we get (whenever the first integral makes sense)

f ∗ g(x) = ∫_{−∞}^{+∞} f(x − y)g(y)dy = ∫_{+∞}^{−∞} f(x − (x − u))g(x − u)d(x − u)
= − ∫_{+∞}^{−∞} f(u)g(x − u)du = ∫_{−∞}^{+∞} f(u)g(x − u)du = g ∗ f(x).

c. Since g is zero outside [a, b] and continuous, it belongs to L^{1} and thus convolution makes sense by part a. Denoting by ω_{g}(ϵ) its modulus of continuity we can write

|g ∗ f(x + ϵ) − g ∗ f(x)| = |∫_{ℝ} g(x + ϵ − y)f(y) + g(x − y)f(y)dy|
≤ ∫_{ℝ} |g(x + ϵ − y)f(y) + g(x − y)f(y)| dy = ∫_{ℝ} |g(x + ϵ − y) − g(x − y)| · |f(y)|dy
≤ ∫_{ℝ} ω_{g}(ϵ) · |f(y)|dy = ω_{g}(ϵ) ∫_{ℝ} |f(y)|dy = ω_{g}(ϵ)∥f∥_{1} ⟶^{ϵ→0} 0,

and f ∗ g = g ∗ f is continuous.