Section 4

Singular Integrals. Weierstrass' Approximation Theorem

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Problem 2.128

If the functions

p1(t), p2(t), …, pn(t), … p_{1}(t), \ p_{2}(t), \ \ldots , \ p_{n}(t), \ \ldots

are continuous on the interval [a,b]\left[a,b\right] and if they satisfy the conditions

pn(t)≥0,∫abpn(t) dt=1,n=1,2,3,…, p_{n}(t) \geq 0, \qquad \int _{a}^{b} p_{n}(t) \, dt = 1, \qquad n = 1, 2, 3, \ldots ,

then the terms of the sequence

∫abp1(t)f(t) dt,∫abp2(t)f(t) dt, …, ∫abpn(t)f(t) dt, … \int _{a}^{b} p_{1}(t) f(t) \, dt, \qquad \int _{a}^{b} p_{2}(t) f(t) \, dt, \ \ldots , \ \int _{a}^{b} p_{n}(t) f(t) \, dt, \ \ldots

are between the minimum and the maximum of the continuous function f(t)f(t). (Cf. Problems 65, 79, 83 of Part I.)

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Problem 2.129

Let xx be a fixed point of the interval [a,b]\left[a,b\right] considered in Problem 128. In order that

lim⁡n→∞∫abpn(t)f(t) dt=f(x) \lim _{n \to \infty } \int _{a}^{b} p_{n}(t) f(t) \, dt = f(x)

holds for all functions f(t)f(t) continuous on [a,b]\left[a,b\right] it is necessary and sufficient that

lim⁡n→∞(∫ax−εpn(t) dt+∫x+εbpn(t) dt)=0 \lim _{n \to \infty } \left(\int _{a}^{x-\varepsilon } p_{n}(t) \, dt + \int _{x+\varepsilon }^{b} p_{n}(t) \, dt\right) = 0

for all positive values of ε\varepsilon for which a<x−ε<x+ε<ba < x-\varepsilon < x+\varepsilon < b (if x=ax=a or x=bx=b the first or second integral resp. under the limit sign has to be omitted.) (Cf. Problems 66, 80, 84 of Part I.)

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Problem 2.130

We have

lim⁡ε→+0ε∫0∞e−εtf(t) dt=lim⁡t→∞f(t), \lim _{\varepsilon \to +0} \varepsilon \int _{0}^{\infty } e^{-\varepsilon t} f(t) \, dt = \lim _{t \to \infty } f(t),

provided that the integral on the left and the limit on the right hand side exist.

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Problem 2.131

If the integral

∫0∞tλf(t) dt \int _{0}^{\infty } t^{\lambda } f(t) \, dt

converges for λ=α\lambda = \alpha and for λ=β\lambda = \beta, α<β\alpha < \beta, it converges for α≤λ≤β\alpha \leq \lambda \leq \beta and it represents a continuous function of λ\lambda on that interval.

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Problem 2.132

We assume that

p1(x,t), p2(x,t), …, pn(x,t), … p_{1}(x,t), \ p_{2}(x,t), \ \ldots , \ p_{n}(x,t), \ \ldots

are continuous functions of xx and tt, a≤x,t≤ba \leq x, t \leq b, and that for each nn

pn(x,t)≥0,∫abpn(x,t) dt=1. p_{n}(x,t) \geq 0, \qquad \int _{a}^{b} p_{n}(x,t) \, dt = 1 .

Let f(t)f(t) denote a continuous function. The functions

fn(x)=∫abpn(x,t)f(t) dt,n=1,2,3,… f_{n}(x) = \int _{a}^{b} p_{n}(x,t) f(t) \, dt, \qquad n = 1, 2, 3, \ldots

lie between the minimum and the maximum of f(t)f(t) on [a,b]\left[a,b\right] for any xx, a≤x≤ba \leq x \leq b; i.e. min⁡a≤x≤bf(x)≤fn(x)≤max⁡a≤x≤bf(x)\min_{a \leq x \leq b} f(x) \leq f_{n}(x) \leq \max_{a \leq x \leq b} f(x). Furthermore

lim⁡n→∞fn(x)=f(x)fora<x<b, \lim _{n \to \infty } f_{n}(x) = f(x) \quad \text{for} \quad a < x < b,

provided

lim⁡n→∞(∫ax−εpn(x,t) dt+∫x+εbpn(x,t) dt)=0 \lim _{n \to \infty } \left(\int _{a}^{x-\varepsilon } p_{n}(x,t) \, dt + \int _{x+\varepsilon }^{b} p_{n}(x,t) \, dt\right) = 0

uniformly for a+ε≤x≤b−εa+\varepsilon \leq x \leq b-\varepsilon, ε\varepsilon fixed and positive. The convergence is uniform on any closed subinterval of (a,b)(a,b).

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Problem 2.133

Let f(x)f(x) denote a continuous function on [0,1]\left[0,1\right]. The convergence of

lim⁡n→∞12⋅32⋅54⋯2n+12n∫01f(t)[1−(x−t)2]n dt=f(x) \lim _{n \to \infty } \frac{1}{2} \cdot \frac{3}{2} \cdot \frac{5}{4} \cdots \frac{2n+1}{2n} \int _{0}^{1} f(t) \left[1-(x-t)^{2}\right]^{n} \, dt = f(x)

is uniform for ε≤x≤1−ε\varepsilon \leq x \leq 1-\varepsilon, 0<ε<120 < \varepsilon < \frac{1}{2}, ε\varepsilon fixed.

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Problem 2.134

Let f(x)f(x) be a continuous periodic function with period 2π2\pi. Then

lim⁡n→∞12nπ∫02πf(t)(sin⁡nx−t2sin⁡x−t2)2 dt=f(x), \lim _{n \to \infty } \frac{1}{2n\pi } \int _{0}^{2\pi } f(t) \left(\frac{\sin n \dfrac {x-t}{2}}{\sin \dfrac {x-t}{2}}\right)^{2} \, dt = f(x),

the convergence is uniform for all xx.

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Problem 2.135

Every function defined and continuous on the finite interval [a,b]\left[a,b\right] can be approximated uniformly on [a,b]\left[a,b\right] by polynomials to any degree of accuracy. (Weierstrass' approximation theorem.)

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Problem 2.136

Every continuous function that is periodic with period 2π2\pi can be uniformly approximated by trigonometric polynomials (see §2 of Chapter 1, Part VI) to any assigned degree of accuracy. (Weierstrass' approximation theorem.)

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Problem 2.137

Let f(x)f(x) denote a function that is properly integrable over [a,b]\left[a,b\right] ([0,2π]\left[0,2\pi \right]). Two polynomials (trigonometric polynomials), p(x)p(x) and P(x)P(x), can then be found for any positive ε\varepsilon so that for a≤x≤ba \leq x \leq b (0≤x≤2π0 \leq x \leq 2\pi)

p(x)≤f(x)≤P(x) p(x) \leq f(x) \leq P(x)

and

∫abP(x) dx−∫abp(x) dx<ε(∫02πP(x) dx−∫02πp(x) dx<ε). \int _{a}^{b} P(x) \, dx - \int _{a}^{b} p(x) \, dx < \varepsilon \qquad \left(\int _{0}^{2\pi } P(x) \, dx - \int _{0}^{2\pi } p(x) \, dx < \varepsilon \right) .
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Problem 2.138

The nn-th moment of a function f(x)f(x) is given by

∫abf(t)tn dt. \int _{a}^{b} f(t) t^{n} \, dt .

If all the moments of a function that is defined and continuous on the finite interval [a,b]\left[a,b\right] vanish then the function vanishes identically.

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Problem 2.139

If all the moments

∫abf(t)tn dt,n=0,1,2,…, \int _{a}^{b} f(t) t^{n} \, dt, \qquad n = 0, 1, 2, \ldots ,

of a function that is properly integrable over the interval [a,b]\left[a,b\right] vanish then the function f(x)f(x) vanishes at every point of continuity.

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Problem 2.140

If the first nn moments vanish,

∫abf(x) dx=∫abf(x)x dx=∫abf(x)x2 dx=⋯=∫abf(x)xn−1 dx=0, \int _{a}^{b} f(x) \, dx = \int _{a}^{b} f(x) x \, dx = \int _{a}^{b} f(x) x^{2} \, dx = \cdots = \int _{a}^{b} f(x) x^{n-1} \, dx = 0,

of a function f(x)f(x) defined and continuous on the finite or infinite interval (a,b)(a,b) then the function changes sign (see §2 of Chapter 1, Part V) at least nn times in the interval (a,b)(a,b) unless it is identically 00.

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Problem 2.141

The 2n2n-th and (2n+1)(2n+1)-th trigonometric moment (Fourier constants, cf. §4 of Chapter 1, Part VI) of a function with period 2π2\pi are defined as

∫02πcos⁡nx f(x) dxand∫02πsin⁡nx f(x) dx. \int _{0}^{2\pi } \cos nx \, f(x) \, dx \qquad \text{and} \qquad \int _{0}^{2\pi } \sin nx \, f(x) \, dx .

If the first 2n+12n+1 trigonometric moments of a continuous function f(x)f(x) with period 2π2\pi vanish then f(x)f(x) changes sign at least 2n+22n+2 times in any interval of length >2π> 2\pi (see §2 of Chapter 1, Part V) unless f(x)f(x) is identically 00.

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Problem 2.142

Let the function φ(x)\varphi (x) be defined and continuous for x≥0x \geq 0. Suppose that the integral

J(k)=∫0∞e−kxφ(x) dx J(k) = \int _{0}^{\infty } e^{-kx} \varphi (x) \, dx

converges for k=k0k = k_{0} and that it vanishes for a sequence of kk's increasing in arithmetic progression:

J(k0)=J(k0+α)=J(k0+2α)=⋯=J(k0+nα)=⋯=0,α>0. J(k_{0}) = J(k_{0}+\alpha ) = J(k_{0}+2\alpha ) = \cdots = J(k_{0}+n\alpha ) = \cdots = 0, \qquad \alpha > 0 .

Then φ(x)\varphi (x) vanishes identically.

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Problem 2.143

The Γ\Gamma-function

Γ(s)=lim⁡n→∞nsn!s(s+1)⋯(s+n) \Gamma (s) = \lim _{n \to \infty } \frac{n^{s} n!}{s(s+1)\cdots (s+n)}

can be written as an integral (Problem 31). Use this fact to prove that Γ(s)\Gamma (s) does not have any zeroes.

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Remark.
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Γ(s+1)=sΓ(s)\Gamma (s+1) = s\Gamma (s) (Problem 142).

Problem 2.144

Work out the polynomials Kn(x)K_{n}(x), n=0,1,2,…n=0,1,2,\ldots for

f(x)=1,f(x)=x,f(x)=x2,f(x)=ex. f(x) = 1, \qquad f(x) = x, \qquad f(x) = x^{2}, \qquad f(x) = e^{x} .
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Problem 2.145

Let xx be any point on [0,1]\left[0,1\right] and

1=∑v=0n(nv)xv(1−x)n−v=ΣI+ΣII, 1 = \sum _{v=0}^{n} \binom {n}{v} x^{v}(1-x)^{n-v} = \Sigma ^{\mathrm{I}} + \Sigma ^{\mathrm{II}},

where ΣI\Sigma^{\mathrm{I}} refers to the subscripts for which ∣v−nx∣≤n3/4\left|v-nx\right| \leq n^{3/4} and ΣII\Sigma^{\mathrm{II}} to those for which ∣v−nx∣>n3/4\left|v-nx\right| > n^{3/4}, n≥1n \geq 1. Then

ΣII<14n−12. \Sigma ^{\mathrm{II}} < \frac{1}{4} n^{-\frac{1}{2}} .
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Problem 2.146

Let f(x)f(x) be continuous on [0,1]\left[0,1\right]. The polynomials Kn(x)K_{n}(x) converge uniformly to f(x)f(x) on [0,1]\left[0,1\right]. (New proof of Weierstrass' theorem, Problem 135.)

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