Some Properties of Real Functions

We define

and

Is $G(x)$ integrable?

Let the function $f(x)$ (cf. also Problem 169 of this Part, Problem 240 of Part VIII) be given by

Show that $f(x)$ is continuous for every irrational $x$, discontinuous for every rational $x$ and properly integrable over any interval.

The two functions $f(x)$ and $\varphi (x)$ are properly integrable over $\left[a, b\right]$. Subdivide the interval:

If $\max (x_{v}-x_{v-1}) \to 0$ (the subinterval of maximal length converges to $0$) we obtain the relation

Suppose that $f(x)$ is properly integrable over $\left[a, b\right]$ and $\varphi (x)$ properly integrable over $\left[a, b+d\right]$, $d > 0$. Then

Let $f(x)$ denote a properly integrable function on $\left[a, b\right]$. There exist to every positive number $\varepsilon$ two step-functions, $\psi (x)$ and $\Psi (x)$, such that for the entire interval $\left[a, b\right]$

and

It is even possible to choose $\psi (x)$ and $\Psi (x)$ so that their points of discontinuity are equidistant.

(continued). If $f(x)$ is of bounded variation $\psi (x)$ and $\Psi (x)$ may be chosen so that the total variation of neither exceeds the total variation of $f(x)$.

We define

Then we have ($n$ integer) the limit relation

for any properly integrable function $f(x)$ on $\left[0, 1\right]$.

Let $f(x)$ be properly integrable over $\left[a, b\right]$. Then we can prove that

(continued). Yet

The function $\left(\dfrac {1}{x} - \left[\dfrac {1}{x}\right]\right) x^{\alpha }$ is properly integrable over $\left[0, 1\right]$ for $\alpha \geq 0$.

If $f(x)$ is properly integrable over $\left[a, b\right]$ the points of continuity of $f(x)$ are everywhere dense on this interval.

The function $f(x)$ is properly integrable over $\left[a, b\right]$. The equation

is satisfied if and only if $f(\xi ) = 0$ at every point of continuity $\xi$ of $f(x)$, $a < \xi < b$.

Assume that the function $y = f(x)$ is properly integrable over $\left[a, b\right]$, that $m \leq f(x) \leq M$ on this interval and that the function $\varphi (y)$ is continuous on $\left[m, M\right]$. Then also $\varphi \left[f(x)\right]$ is properly integrable over $\left[a, b\right]$.

If $f(x)$ and $\varphi (y)$ are properly integrable $\varphi \left[f(x)\right]$ need not be properly integrable (see Problems 98, 99).

If $f(x)$ is a monotone function on the interval $(0, 1]$ and if $\int_{0}^{1} x^{a} f(x) \, dx$ exists then

If $f(x)$ is monotone on the interval $[1, \infty )$ and if $\int_{1}^{\infty } x^{a} f(x) \, dx$ exists then

Determine those pairs of real values $\alpha , \beta$ for which the integral

is convergent.

Construct a function $f(x)$ that takes positive values and is bounded and integrable in any finite subinterval of $[0, \infty )$ and such that

converges for $\alpha = 1$ but diverges for any real value of $\alpha$ different from $1$.

Construct a function $f(x)$ that takes positive values and is bounded and integrable in any finite subinterval of $[0, \infty )$ and such that

converges for $\alpha = 1$ but diverges for any real value of $\alpha$ different from $1$.

The functions

are properly integrable over any finite interval and they satisfy the following conditions:

On every finite interval we have $\lim_{n \to \infty } f_{n}(x) = f(x)$ uniformly in $x$.

There is a function $F(x)$ such that $\left|f_{n}(x)\right| \leq F(x)$ and $\int_{-\infty }^{\infty } F(x) \, dx$ exists.

Then the limit and the integral can be interchanged:

Prove Problem 58 using Problem 31 of Part VI.

If the Dirichlet series (see §5 of Chapter 1, Part VIII)

converges for $s = \sigma$, $\sigma > 0$ we obtain for $s > \sigma$

where

Suppose that $f(x)$ is properly integrable over every finite interval and that $\int_{-\infty }^{+\infty } \left|f(x)\right| \, dx$ exists. Then we have

There are rational functions $R(x)$ such that for an arbitrary function $f(x)$ of the real variable $x$

provided that the integral on the right hand side exists. Show that this property belongs to those, and only to those rational functions that are of the form

where $\varepsilon = +1$ or $-1$, $\alpha , \alpha_{1}, \alpha_{2}, \ldots , \alpha_{n}$ are real, and $p_{1}, p_{2}, \ldots , p_{n}$ are positive numbers; $n \geq 0$. (The case $n=0$ must be interpreted as meaning $R(x) = \varepsilon (x-\alpha )$.)

There are rational functions $R(x)$ such that for an arbitrary function $f(x)$ of the real variable $x$

provided that the integral on the right hand side exists. Show that this property belongs to those, and only to those rational functions that are of the form

where $\varepsilon = +1$ or $-1$, $\alpha , \alpha_{1}, \alpha_{2}, \ldots , \alpha_{n}$ are real, and $p_{1}, p_{2}, \ldots , p_{n}$ are positive numbers; $n \geq 0$. (The case $n=0$ must be interpreted as meaning $R(x) = \varepsilon (x-\alpha )$.)

Are there actually functions of three variables?

More precisely: Is it possible to write every real function $f(x,y,z)$ of three variables with the help of two functions $\varphi (x,y)$ and $\psi (u,z)$ in the form of

Discuss the question:

(1) if $f(x,y,z)$, $\varphi (x,y)$, $\psi (u,z)$ are defined for all real values of the variables,

(2) if these functions are defined for all real values of the variables and are continuous.

Setting $x+y = S(x,y)$, $xy = P(x,y)$ we can write

in this formula $xy+yz+zx$ is composed of four functions of two variables ``boxed'' in each other. Prove that it is impossible to express $f(x,y,z)$ using only three functions of two variables boxed in one another if these functions are defined for all pairs of real values and if they are arbitrarily often differentiable.

Setting $x+y = S(x,y)$, $xy = P(x,y)$ we can write

in this formula $xy+yz+zx$ is composed of four functions of two variables ``boxed'' in each other. Prove that it is impossible to express $f(x,y,z)$ using only three functions of two variables boxed in one another if these functions are defined for all pairs of real values and if they are arbitrarily often differentiable.

Let the function $f(x)$ have a continuous derivative on $(a,b)$. Decide whether it is possible to find for each point $\xi$ of this interval two points $x_{1}, x_{2}$, $x_{1} < \xi < x_{2}$, such that

Assume that the function $f(x)$ is differentiable on $\left[a,b\right]$, but not a constant and that $f(a) = f(b) = 0$. Then there exists at least one point $\xi$ on $(a,b)$ for which

The function $f(x)$ is twice differentiable. Then there exists a point $\xi$ in $(x_{0}-r, x_{0}+r)$ where

When is the mean value of a function in an interval a simple mean of those two values that the function takes at the endpoints of the interval?

Assume that $f(x)$ is defined, bounded and integrable in the interval $\left[a,b\right]$. Introduce the abbreviations

and determine the most general function $f(x)$ satisfying

for all $u$ and $v$ subject to the condition

When is the mean value of a function in an interval a simple mean of those two values that the function takes at the endpoints of the interval?

Assume that $f(x)$ is defined, bounded and integrable in the interval $\left[a,b\right]$. Introduce the abbreviations

and determine the most general function $f(x)$ satisfying

for all $u$ and $v$ subject to the condition

(continued). Analogous question for

assume that $f(x) > 0$ for $a \leq x \leq b$.

(continued). Analogous question for

assume that $f(x) > 0$ for $a \leq x \leq b$.

(continued). Analogous question for

assume that $f(x) > 0$ for $a \leq x \leq b$.

(continued). Analogous question for

assume that $f(x) > 0$ for $a \leq x \leq b$.

The numbers $p_{0}, p_{1}, p_{2}, \ldots , p_{n}, \ldots$ are non-negative, at least two of the $p_{v}$'s do not vanish. Then the logarithm of the series

is a function of $x$, convex on every interval where the series converges.

A bounded convex function (see the definition of convexity given above, at the start of Chapter 2) is everywhere continuous and it is even everywhere differentiable from the left and from the right.

Suppose that the real function $f(x)$ is defined on a finite or infinite interval and that it has there a continuous derivative $f'(x)$. Consider the points of intersection of all the horizontal tangents of the curve $y=f(x)$ with the $y$-axis, i.e. the set $M$ of all points

Prove that the set $M$ cannot fill out an entire interval. (This proposition admits a farreaching generalization.)

If a monotone sequence of continuous functions converges on a closed interval to a continuous function it converges uniformly. (Theorem of Dini.)

Prove the following counterpart to Problem 126: If a sequence of monotone (continuous or discontinuous) functions converges on a closed interval to a continuous function it converges uniformly.

If the functions

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

then the terms of the sequence

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

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

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

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

We have

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

If the integral

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

We assume that

are continuous functions of $x$ and $t$, $a \leq x, t \leq b$, and that for each $n$

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

lie between the minimum and the maximum of $f(t)$ on $\left[a,b\right]$ for any $x$, $a \leq x \leq b$; i.e. $\min_{a \leq x \leq b} f(x) \leq f_{n}(x) \leq \max_{a \leq x \leq b} f(x)$. Furthermore

provided

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

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

is uniform for $\varepsilon \leq x \leq 1-\varepsilon$, $0 < \varepsilon < \frac{1}{2}$, $\varepsilon$ fixed.

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

the convergence is uniform for all $x$.

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

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

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

and

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

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

If all the moments

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

If the first $n$ moments vanish,

of a function $f(x)$ defined and continuous on the finite or infinite interval $(a,b)$ then the function changes sign (see §2 of Chapter 1, Part V) at least $n$ times in the interval $(a,b)$ unless it is identically $0$.

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

If the first $2n+1$ trigonometric moments of a continuous function $f(x)$ with period $2\pi$ vanish then $f(x)$ changes sign at least $2n+2$ times in any interval of length $> 2\pi$ (see §2 of Chapter 1, Part V) unless $f(x)$ is identically $0$.

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

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

Then $\varphi (x)$ vanishes identically.

The $\Gamma$-function

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

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

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

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

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