Topics on Measurable Functions of Real Variables

Abstract

Let f be a real-valued function defined and bounded in some interval $[a,b]\subset \mathbb {R}$.

Appendices

Problems and Complements

1c Functions of Bounded Variations

1.1. :

The continuous function on [0, 1]

$$\begin{aligned} f(x)=\left\{ \begin{array}{ll} {\displaystyle x^2\cos \frac{\pi }{x}}\quad &{}\text { for }\> x\in (0,1]\\ 0&{} \text { for }\> x=0 \end{array} \right. \end{aligned}$$

is of bounded variation in [0, 1].

1.2. :

Let f be the continuous function defined in [0, 1] by

$$\begin{aligned} \begin{array}{cl} f(0)=0,\quad f\left( \frac{1}{2n+1}\right) =0,\quad f\left( \frac{1}{2n}\right) =\frac{1}{2n}\quad &{}\text { for all }\>n\in \mathbb {N}\\ f\>\text { is affine on the intervals }\> \left[ \frac{1}{m+1},\frac{1}{m}\right] \quad &{}\text { for all }\> m\in \mathbb {N}. \end{array} \end{aligned}$$

Such an f is not of bounded variation in [0, 1].

1.3. :

Prove Propositions 1.1–1.3.

1.4. :

Let $\{f_n\}$ be a sequence of functions in [a, b] converging pointwise in [a, b] to f. Then $\mathcal {V}_f[a,b]\le \liminf \mathcal {V}_{f_n}[a,b]$, and strict inequality may occur as shown by the sequence

$$\begin{aligned} f_n(x)=\left\{ \begin{array}{ll} 0\quad &{}\text { for }\> x=0\\ \frac{1}{2}\quad &{}\text { for }\> x\in (0,\frac{1}{n}]\\ 0\quad &{}\text { for }\> x\in (\frac{1}{n},1]. \end{array} \right. \end{aligned}$$

1.5. :

Let f be continuous and of bounded variation in [a, b]. Then the functions $x\rightarrow \mathcal {V}_f[a,x]$, $\mathcal {V}_f^+[a,x]$, $\mathcal {V}_f^-[a,x]$ are continuous in [a, b].

1.6. :

Prove that the distribution function $f_*(\cdot )$ of a measurable function f on a measure space $\{X,\mathcal {A},\mu \}$, as defined by (15.1c) of Chap. 4, is of bounded variation in every interval $[a,b]\subset \mathbb {R}$.

1.1 1.1c The Function of the Jumps

Let f be of bounded variation in [a, b] and regard it as defined in $\mathbb {R}$, by extending f to be equal to f(a) on $(-\infty ,a]$ and equal to f(b) in $[b,\infty )$. Prove that the limits

$$\begin{aligned} f(x^{\pm })\buildrel \mathrm { def}\over {=}\lim _{h\rightarrow 0} f(x\pm h)\quad \text { for }\> h>0 \end{aligned}$$

exist for all $x\in [a,b]$. The function of the jumps of f is defined by

$$\begin{aligned} J_f(x)=\mathop {\textstyle {\sum }}\limits _{a\le c_j\le x} \big [f(c_j^+)-f(c_j^-)\big ]. \end{aligned}$$

The difference $f-J_f$ is continuous in [a, b]. Also $J_f$ is of bounded variation in [a, b] and

$$\begin{aligned} \mathcal {V}_f[a,b]=\mathcal {V}_{f-J_f}[a,b]+\mathcal {V}_{J_f}[a,b]. \end{aligned}$$

Therefore a function f of bounded variation in [a, b] can be decomposed into the continuous function $f-J_f$ and $J_f$. The latter bears the possible discontinuities of f in [a, b].

1.7. :

Construct a non-decreasing function in [0, 1] which is discontinuous at all the rational points of [0, 1].

1.2 1.2c The Space BV[a, b]

Let $[a,b]\subset \mathbb {R}$ be a finite interval and denote by BV[a, b] the collection of all functions $f:[a,b]\rightarrow \mathbb {R}$ of bounded variation in [a, b]. One verifies that BV[a, b] is a linear vector space. Also setting

$$\begin{aligned} d(f,g)=|f(a)-g(a)|+\mathcal {V}_{f-g}[a,b]\quad \text { for }\> f,g\in BV[a,b] \end{aligned}$$

(1.1c)

defines a distance in BV[a, b] by which $\{BV[a,b];d\}$ is a metric space.

For any two functions $f,g\in BV[a,b]$

$$\begin{aligned} \sup _{[a,b]}|f-g|\le |f(a)-g(a)|+\mathcal {V}_{f-g}[a,b]. \end{aligned}$$

(1.2c)

Fig. 1

Cauchy sequence in the sup-norm and not in the BV-norm

Therefore a Cauchy sequence in BV[a, b] is also Cauchy in the sup-norm. The converse is false as illustrated in Fig. 1c. The sequence $\{f_n\}$ generated as in Fig. 1c is not a Cauchy sequence in the topology of BV[0, 1], while it is a Cauchy sequence in the topology of C[0, 1].

1.2.1 1.2.1.c Completeness of BV[a, b]

Let $\{f_n\}$ be a Cauchy sequence in BV[a, b]. There exists $f\in BV[a,b]$ such that $\{f_n\}\rightarrow f$ in the topology of BV[a, b].

The proof is in two steps. First one uses (1.2c) to identify the limit f. Then one proves that such an f is actually in BV[a, b] by using that $\{f_n\}$ is Cauchy in BV[a, b]. As a consequence BV[a, b] is a complete metric space.

2c Dini Derivatives

2.1. :

Compute $D^+f(0)$ and $D_+f(0)$ for the function in (1.1).

2.2. :

Let f have a maximum at some $c\in (a,b)$. Then $D^-f(c)\ge 0$.

2.3. :

Let f be continuous in [a, b]. If $D^+f\ge 0$ in [a, b], then f is non-decreasing in [a, b]. The assumption that f be continuous cannot be removed.

Hint: Fix $[\alpha ,\beta ]\subset (a,b)$, and by continuity extend the function f to be equal to $f(\beta )$ for $x\ge \beta$ and to be equal to $f(\alpha )$ for $x\le \alpha$. Having fixed $\varepsilon >0$, the assumption implies that for each $x\in [\alpha ,\beta ]$ there exists

$$\begin{aligned} 0<h_x=h(x,\varepsilon )<\textstyle {\frac{1}{2}}(b-\beta ) \end{aligned}$$

such that

$$\begin{aligned} f(x)<f(x+h_x)+ h\varepsilon . \end{aligned}$$

By continuity, such inequality continues to hold for all $y\in (x-\delta _x,x+\delta _x)$ for some $\delta _x>0$, which without loss of generality we may assume not to exceed $\frac{1}{4} h_x$. The collection of these open intervals covers $[\alpha ,\beta ]$. From this select a finite sub-collection, say

$$\begin{aligned} (x_j-\delta _j,x_j+\delta _j)\quad \text { for }\>j=1,\dots ,n. \end{aligned}$$

To these are associated positive numbers $h_j<\frac{1}{2}(\beta -b)$ such that

$$\begin{aligned} f(y)<f(y+h_j)+ h_j\varepsilon \quad \text { for all }\> y\in (x_j-\delta _j,x_j+\delta _j). \end{aligned}$$

Starting with $y=\alpha$ and iterating this inequality $m\le n$ times gives

$$\begin{aligned} f(\alpha )<f\Big (\alpha +\mathop {\textstyle {\sum }}\limits _{j=1}^m h_j\Big )+ \mathop {\textstyle {\sum }}\limits _{j=1}^mh_j\varepsilon . \end{aligned}$$

By construction, there exists $m\le n$ such that

$$\begin{aligned} \alpha +\mathop {\textstyle {\sum }}\limits _{j=1}^m h_j\ge \beta \quad \text { and } \quad \mathop {\textstyle {\sum }}\limits _{j=1}^m h_j \le b-a. \end{aligned}$$

Therefore

$$\begin{aligned} f(\alpha )<f(\beta )+\varepsilon (b-a) \quad \text { for all }\>\varepsilon >0. \end{aligned}$$

2.4. :

Let $f\in BV[a,b]$ and let $J_f$ the function of the jumps of f introduced in § 1.1c. Prove that $J_f^\prime =0$ a.e. in [a, b].

Hint: Assume that f is non-decreasing, so that $J_f$ is non decreasing. Denoting by $\{c_n\}$ the sequence of the jumps of f

$$\begin{aligned} J_f(b)-J_f(a) =\mathop {\textstyle {\sum }}\limits _{a<c_j\le b}\big [f(c_j^+)-f(c_j^-)\big ]. \end{aligned}$$

(2.1c)

Fix $m\in \mathbb {N}$ and let $\varepsilon >0$ be so small that the intervals

$$\begin{aligned} (\alpha _j,\beta _j]=\Big (c_j-\frac{\varepsilon }{2m}\,,\, c_j+\frac{\varepsilon }{2m}\Big ]\mathop {\textstyle {\bigcap }}\limits (a,b] \quad \text { for }\> j=1,\dots ,m \end{aligned}$$

are disjoint. The complement of their union is the finite union of disjoint intervals $[\alpha _j^\prime ,\beta _j^\prime )$. For $t>0$ assume that $\mu ([DJ_f>t])>0$, and choose

$$\begin{aligned} \varepsilon <\textstyle {\frac{1}{2}}\mu ([DJ_f>t]). \end{aligned}$$

For such a choice estimate

$$\begin{aligned} \mu \Big ([DJ_f>t]\mathop {\textstyle {\bigcap }}\limits \mathop {\textstyle {\bigcup }}\limits _{j=1}^{m^\prime } [\alpha _j^\prime ,\beta _j^\prime ]\big )>\frac{1}{2}\mu ([DJ_f>t])>0 \end{aligned}$$

for a positive integer $m^\prime$. By Proposition 2.2

$$\begin{aligned} J_f(\beta _j^\prime )-J_f(\alpha _j^\prime )\ge t\mu \big (\big [DJ_f>t\big ]\cap \big [\alpha _j^\prime ,\beta _j^\prime \big ]\big ). \end{aligned}$$

Therefore

$$\begin{aligned} \begin{aligned} J_f(b)-J_f(a)&=\mathop {\textstyle {\sum }}\limits _{j=1}^m\big [J_f(\beta _j)-J_f(\alpha _j)\big ] +\mathop {\textstyle {\sum }}\limits _{j=1}^{m+1} \big [J_f(\beta _j^\prime )-J_f(\alpha _j^\prime )\big ]\\&\ge \mathop {\textstyle {\sum }}\limits _{j=1}^m \big [f(c_j^+)-f(c_j^-)\big ] +\textstyle {\frac{1}{2}}\mu ([DJ_f>t]). \end{aligned} \end{aligned}$$

Letting $m\rightarrow \infty$ and taking into account (2.1c) implies $\mu ([DJ_f>t])=0$ for all $t>0$. For a different approach use Theorem 4.1.

1.1 2.1c A Continuous, Nowhere Differentiable Function ([167])

For a real number x, denote by $\{x\}$ the distance from x to its nearest integer and set

$$\begin{aligned} f(x)=\mathop {\textstyle {\sum }}\limits _{n=0}^\infty \>\frac{\{10^n x\}}{10^n}. \end{aligned}$$

(2.2c)

Each term of the series is continuous. Moreover the series is uniformly convergent being majorized by the geometric series ${\sum } 10^{-n}$. Therefore f is continuous. Since $f(x)=f(x+j)$ for every integer j and all $x\in \mathbb {R}$, it suffices to consider $x\in [0,1)$. Any such x has a decimal expansion of the form $x=0.a_1a_2\dots a_n\dots$, where $a_i$ are integers from 0 to 9. By excluding the case when $a_i=9$ for all i larger than some m, such a representation is unique.

For $n\in \mathbb {N}$ fixed compute

$$\begin{aligned} \begin{array}{ll} \{10^n\,x\}=0.a_{n+1}a_{n+2}\dots &{}\text { if }\> 0.a_{n+1}a_{n+2}\dots \le \frac{1}{2}\\ \{10^n\,x\}=1- 0.a_{n+1}a_{n+2}\dots &{}\text { if }\> 0.a_{n+1}a_{n+2}\dots >\frac{1}{2}. \end{array} \end{aligned}$$

Having fixed $x\in [0,1)$ choose increments

$$\begin{aligned} h_m=\left\{ \begin{array}{ll} -10^{-m}\qquad &{}\text { if either } a_m=4 \text { or } a_m=9\\ +10^{-m}\qquad &{}\text { otherwise}. \end{array} \right. \end{aligned}$$

Then form the difference quotients of f at x

$$\begin{aligned} \frac{f(x+h_m)-f(x)}{h_m}=10^m\mathop {\textstyle {\sum }}\limits _{n=0}^\infty \,\pm \frac{\{10^n(x\pm 10^{-m})\}-\{10^n\,x\}}{10^n}. \end{aligned}$$

The numerators of the terms of this last series, all vanish for $n\ge m$, whereas for $n=0,1,\dots ,(m-1)$ they are equal to $\pm 10^{n-m}$. Therefore the difference quotient reduces to the sum of m terms each of the form $\pm 1$. Such a sum is an integer, positive or negative, which has the same parity of m. Thus the limit as $h_m\rightarrow 0$ of the difference ratios does not exists.

Remark 2.1c

The function in (2.2c) is not of bounded variation in any interval $[a,b]\subset \mathbb {R}$.

1.2 2.2c An Application of the Baire Category Theorem

The existence of a continuous and nowhere differentiable function, can be established indirectly, by a category-type argument. The Dini derivatives $D_+f$ and $D^+f$ introduced in (2.1), are called the right Dini numbers.

Proposition 2.1c

(Banach [12]) There exists a real-valued, continuous function in [0, 1] such that its Dini’s numbers $|D_+f|$ and $|D^+f|$ are infinity at every point of (0, 1).

Proof

For $n\in \mathbb {N}$ let $E_n$ denote the collection of all functions $f\in C[0,1]$, for which there exists at least one point $t\in [0,1-\frac{1}{n}]$ for which

$$\begin{aligned} \Big |\frac{f(t+h)-f(t)}{h}\Big |\le n \quad \text { for all }\>h\in (0,1-t). \end{aligned}$$

Each $E_n$ is closed and nowhere dense in C[0, 1]. Both statements are meant with respect to the topology of the uniform convergence in C[0, 1]. To prove that $E_n$ is nowhere dense in C[0, 1] observe that any continuous function in [0, 1] can be approximated in the sup-norm by continuous functions with polygonal graph of arbitrarily large Lipschitz constant. Then the complement $C[0,1]-\cup E_n$ is non-empty.$\blacksquare$

4c Differentiating Series of Monotone Functions

4.1. :

Let $\{f_n\}$ be a sequence of functions of bounded variation in [a, b] such that the series $\mathop {\textstyle {\sum }}\limits f_n(x)$ and $\mathop {\textstyle {\sum }}\limits \mathcal {V}_{f_n}[a,x]$ are both convergent in [a, b]. Then the sum f of the first series is of bounded variation in [a, b] and the derivative can be computed term by term, a.e. in [a, b].

 

5c Absolutely Continuous Functions

5.1. :

Let f be absolutely continuous in [a, b]. Then f is Lipschitz continuous in [a, b] if and only $f^\prime$ is a.e. bounded in [a, b].

5.2. :

The function

$$\begin{aligned} f(x)=\left\{ \begin{array}{ll} {\displaystyle x^{1+\varepsilon }\sin \frac{1}{x}}\quad &{}\text { for }\>x\in (0,1]\\ 0&{}\text { for }\>x=0 \end{array} \right. \end{aligned}$$

is absolutely continuous in [0, 1] for all $\varepsilon >0$.

1.1 5.1c The Cantor Ternary Function ([23])

Set $f(0)=0$ and $f(1)=1$. Divide the interval [0, 1] into 3 equal subintervals and on the central interval $[\frac{1}{3},\frac{2}{3}]$ set $f=\frac{1}{2}$, i.e., f is defined to be the average of its values at the extremes of the parent interval [0, 1]. Next divide the interval $[0,\frac{1}{3}]$ into 3 equal subintervals, and on the central interval $[\frac{1}{3^2},\frac{2}{3^2}]$ set $f=\frac{1}{4}$, i.e., f is defined to be the average of its values at the extremes of the parent interval $[0,\frac{1}{3}]$. Likewise divide the interval $[\frac{2}{3},1]$ into 3 equal subintervals, and on the central interval $[\frac{7}{3^2},\frac{8}{3^2}]$ set $f=\frac{3}{4}$, i.e., f is defined to be the average of its values at the extremes of the parent interval $[\frac{2}{3},1]$.

Proceeding in this fashion we define f in the whole [0, 1], by successive averages. By construction f is non-constant, non-decreasing, and continuous in [0, 1]. Since it is constant on each of the intervals making up the complement of the Cantor set $\mathcal {C}$, its derivative vanishes in [0, 1] except on $\mathcal {C}$. Thus $f^\prime =0$ a.e. on [0, 1].

1.1.1 5.1.1.c Another Construction of the Cantor Ternary Function

The same function can be defined by an alternate procedure that uses the ternary expansion of the elements of the Cantor set. For $x\in \mathcal {C}$ let $\{\epsilon _j\}$ be sequence, with entries only 0 or 1, corresponding to the ternary expansion of x, as in (2.1) of Chap. 1. Then define

$$\begin{aligned} f(x)=f\Big (\mathop {\textstyle {\sum }}\limits _{j=1}^\infty \, \frac{2}{3^j}\epsilon _i{x,j}\Big )\buildrel \mathrm { def}\over {=} \mathop {\textstyle {\sum }}\limits _{j=1}^\infty \,\frac{1}{2^j}\epsilon _{x,j}. \end{aligned}$$

(5.1c)

Let $(\alpha _n,\beta _n)$ be an interval removed in the $n^{\text {th}}$ step of the construction of the Cantor set. The extremes $\alpha _n$ and $\beta _n$ belong to $\mathcal {C}$ and their ternary expansion is described in 2.2 of the Complements of Chap. 1. From the form of such expansion compute

$$\begin{aligned} f(\alpha _n)-f(\beta _n)=\frac{1}{2^n}-\mathop {\textstyle {\sum }}\limits _{j=n+1}^\infty \,\frac{1}{2^j}=0. \end{aligned}$$

If $(\alpha _n,\beta _n)$ is an interval in $[0,1]-\mathcal {C}$ we set

$$\begin{aligned} f(x)=f(\alpha _n)\qquad \text { for all } \> x\in [\alpha _n,\beta _n]. \end{aligned}$$

In such a way f is defined in the whole [0, 1] and $f^\prime =0$ a.e. in [0, 1]. The right-hand side of (5.1c) is the decimal representation of the number in [0, 1] whose binary representation is the sequence $\epsilon _x$. As x ranges over $\mathcal {C}$, the sequences $\epsilon _x$, with only entries 0 or 1, range over all such sequences. Therefore f maps $\mathcal {C}$ onto [0, 1]. To show that f is continuous, observe that f is monotone and finite. Therefore its possible discontinuity points are discrete jumps. If f were not continuous then it would not be a surjection over [0, 1].

The Cantor ternary function is continuous, of bounded variation, but not absolutely continuous. This can be established indirectly by means of Corollary 5.2. Give a direct proof.

1.2 5.2c A Continuous Strictly Monotone Function with a.e. Zero Derivative

The Cantor ternary function is piecewise constant on the complement of the Cantor set. This accounts for $f^\prime =0$ a.e. in [0, 1]. We next exhibit a continuous strictly increasing function in [0, 1], whose derivative vanishes a.e. in [0, 1].

Let $t\in (0,1)$ be fixed and define $f_o(x)=x$ and

$$\begin{aligned} f_1(x)=\left\{ \begin{array}{ll} (1+t)x\quad &{}\text { for }\> 0\le x\le \frac{1}{2}\\ (1-t)x+t\quad &{}\text { for }\> \frac{1}{2}\le x\le 1. \end{array} \right. \end{aligned}$$

The function $f_1$ is constructed by dividing [0, 1] into two equal subintervals, by setting $f_1=f_o$ at the end points of [0, 1], by setting

$$\begin{aligned} f_1\Big (\frac{1}{2}\Big )=\frac{1-t}{2}f_o(0)+ \frac{1+t}{2}f_o(1)=\frac{1+t}{2} \end{aligned}$$

and by defining $f_1$ to be affine in the intervals $[0,\frac{1}{2}]$ and $[\frac{1}{2},1]$.

This procedure permits one to construct an increasing sequence $\{f_n\}$ of strictly increasing functions in [0, 1]. Precisely if $f_n$ has been defined, it must be affine in each of the subintervals

$$\begin{aligned} \Big [\frac{j}{2^n}\,,\,\frac{j+1}{2^n}\Big ] \qquad j=0,1,\dots ,2^n-1. \end{aligned}$$

Subdivide each of these into two equal subintervals, and define $f_{n+1}$ to be affine on each of these with values at the end points, given by

$$\begin{aligned} f_{n+1}\Big (\frac{j}{2^n}\Big )&=f_n\Big (\frac{j}{2^n}\Big )\\ f_{n+1}\Big (\frac{j+1}{2^n}\Big )&=f_n\Big (\frac{j+1}{2^n}\Big )\\ f_{n+1}\Big (\frac{2j+1}{2^{n+1}}\Big )&= \frac{1-t}{2} f_n\Big (\frac{j}{2^n}\Big ) +\frac{1+t}{2}f_n\Big (\frac{j+1}{2^n}\Big ). \end{aligned}$$

By construction $\{f_n\}$ is increasing and

$$\begin{aligned} f_m\Big (\frac{j}{2^n}\Big )=f_n\Big (\frac{j}{2^n}\Big ) \quad \text { for all }\> m\ge n,\quad j=0,1,\dots ,2^n-1. \end{aligned}$$

(5.2c)

The limit function f is non-decreasing. We show next that it is continuous and strictly increasing in [0, 1]. Every fixed $x\in [0,1]$ is included into a sequence of nested and shrinking intervals $[\alpha _n,\beta _n]$ of the type

$$\begin{aligned} \alpha _n=\frac{m_{n,x}}{2^n}\quad \beta _n= \frac{m_{n,x}+1}{2^n}\quad \text { for some }\> m_{n,x}\in \mathbb {N}\cup \{0\}. \end{aligned}$$

By the construction of $f_{n+1}$, if the parent interval of $[\alpha _n,\beta _n]$ is $[\alpha _n,\beta _{n-1}]$

$$\begin{aligned} f_{n+1}(\beta _n)-f_{n+1}(\alpha _n)=\frac{1+t}{2} [f_n(\beta _{n-1})-f_n(\alpha _n)]. \end{aligned}$$

Likewise if the parent interval of $[\alpha _n,\beta _n]$ is $[\alpha _{n-1},\beta _n]$

$$\begin{aligned} f_{n+1}(\beta _n)-f_{n+1}(\alpha _n)=\frac{1-t}{2} [f_n(\beta _n)-f_n(\alpha _{n-1})]. \end{aligned}$$

Therefore by (5.2c) either

$$\begin{aligned} f_{n+1}(\beta _n)-f_{n+1}(\alpha _n)=\frac{1+t}{2} [f_n(\beta _{n-1})-f_n(\alpha _{n-1})] \qquad {(5.2\mathrm{c})_+} \end{aligned}$$

or

$$\begin{aligned} f_{n+1}(\beta _n)-f_{n+1}(\alpha _n)=\frac{1-t}{2} [f_n(\beta _{n-1})-f_n(\alpha _{n-1})].\qquad {(5.2\mathrm{c})_-} \end{aligned}$$

From this by iteration

$$\begin{aligned} f_{n+1}(\beta _n)-f_{n+1}(\alpha _n) =\mathop {\textstyle {\prod }}\limits _{i=1}^n\,\frac{1+\varepsilon _i t}{2} \quad \text { where }\quad \varepsilon _i=\pm 1. \end{aligned}$$

Since for all $m\ge n+1$

$$\begin{aligned} f_m(\beta _n)=f_{n+1}(\beta _n)\quad \text { and }\quad f_m(\alpha _n)=f_{n+1}(\alpha _n) \end{aligned}$$

the previous equality implies

$$\begin{aligned} f(\beta _n)-f(\alpha _n)=\mathop {\textstyle {\prod }}\limits _{i=1}^n\,\frac{1+\varepsilon _i t}{2} \quad \text { where }\quad \varepsilon _i=\pm 1. \end{aligned}$$

For each fixed n the right-hand side is strictly positive. Thus $f(\beta _n)>f(\alpha _n)$, i.e., f is strictly monotone. On the other hand (5.2c)${}_{\pm }$ imply also

$$\begin{aligned} f(\beta _n)-f(\alpha _n)\le \Big (\frac{1+t}{2}\Big )^n\longrightarrow 0 \quad \text { as }\quad n\rightarrow \infty . \end{aligned}$$

Thus f is continuous in [0, 1]. Still from (5.2c)${}_{\pm }$ we compute

$$\begin{aligned} \frac{f(\beta _n)-f(\alpha _n)}{\beta _n-\alpha _n}= \mathop {\textstyle {\prod }}\limits _{i=1}^n\,(1+\varepsilon _i t). \end{aligned}$$

As $n\rightarrow \infty$ the right-hand side either converges to zero, or diverges to infinity or the limit does not exists. However, since f is monotone it is a.e. differentiable. Therefore the limit exists for a.e. $x\in [0,1]$ and is zero. By Corollary 5.2 such a function is not absolutely continuous. Give a direct proof.

5.3. :

The function f constructed in § 14 of Chap. 3 is not absolutely continuous.

5.4. :

Let $\mu$ be a Radon measure on $\mathbb {R}$ defined on the same $\sigma$-algebra of the Lebesgue measurable sets in $\mathbb {R}$, and absolutely continuous with respect to the Lebesgue measure on $\mathbb {R}$. Then set

$$\begin{aligned} f(x)=\left\{ \begin{array}{ll} +\mu ([\alpha ,x])\quad &{}\text { for }\> x\in [\alpha ,\infty )\\ -\mu ([x,\alpha ])\quad &{}\text { for }\> x\in (-\infty ,\alpha ]. \end{array} \right. \end{aligned}$$

The function f is locally absolutely continuous, i.e., its restriction to any bounded interval is absolutely continuous. The function f can be used to generate the Lebesgue-Stieltjes measure $\mu _f$. The measure $\mu _f$ coincides with $\mu$ on the Lebesgue measurable sets.

1.3 5.3c Absolute Continuity of the Distribution Function of a Measurable Function

For a measurable function f on a measure space $\{X,\mathcal {A},\mu \}$, let $\mu _f$ and $f_*$ be respectively, the distribution measure and the distribution function of f, as defined (9.1) and (15.1c of the Complements of Chap. 4).

Proposition 5.1c

The distribution function $f_*$ is absolutely continuous, in every closed subinterval of $\mathbb {R}$, if and only if the distribution measure $\mu _f$ is absolutely continuous with respect to the Lebesgue measure on $\mathbb {R}$. In such a case

$$\begin{aligned} f_*(t)=\int _{-\infty }^t \varphi (s)ds \end{aligned}$$

where $\varphi$ is the Radon-Nikodým derivative of $\mu$ with respect to the Lebesgue measure ds.

5.5. :

Give an example of $\{X,\mathcal {A},\mu \}$ and f for which $f_*$ is continuous and not absolutely continuous.

7c Derivatives of Integrals

7.1. :

Construct a measurable set $E\subset (-1,1)$ such that $d_E^{\,\prime }(0)=\frac{1}{2}$. For $n\in \mathbb {N}$ set⁵

$$\begin{aligned} I_n=\Big [\frac{1}{2^n},\frac{1}{2^{n-1}}\Big ) \quad \text { so that }\quad (0,1)=\mathop {\textstyle {\bigcup }}\limits _{n\in \mathbb {N}} I_n. \end{aligned}$$

Divide each $I_n$ into $2^n$ equal subintervals, each of length $2^{-2n}$ and retain only those $\frac{1}{2}$-closed intervals of even parity, to obtain sets

$$\begin{aligned} E_n=\mathop {\textstyle {\bigcup }}\limits _{j=0}^{2^{n-1}-1} \Big [\frac{2^n+2j}{2^{2n}}\,,\, \frac{2^n+2j+1}{2^{2n}}\Big ),\quad \text { and }\quad E^+=\mathop {\textstyle {\bigcup }}\limits E_n. \end{aligned}$$

Verify that $\mu (E_n)=\frac{1}{2}\mu (I_n)$ and $\mu (E^+)=\frac{1}{2}$. having fixed $h\in (0,1)$ there exists $n_h, i_h\in \mathbb {N}$, such that

$$\begin{aligned} h\in \Big [\frac{1}{2^{n_h}},\frac{1}{2^{n_h-1}}\Big ]\quad \text { and }\quad h\in \Big [\frac{2^{n_h}+i_h}{2^{2n_h}}\,,\, \frac{2^{n_h}+(i_h+1)}{2^{2n_h}}\Big ]. \end{aligned}$$

The first of these implies that $h=O(2^{-{n_h}})$. Next compute

$$\begin{aligned} (0,h]\cap E^+=\Big (\mathop {\textstyle {\bigcup }}\limits _{\ell =n_h+1}^\infty E_{\ell }\Big )\mathop {\textstyle {\bigcup }}\limits \Big (\mathop {\textstyle {\bigcup }}\limits _{j=0}^{2^{n_h-1}-1}(0,h]\mathop {\textstyle {\bigcap }}\limits \Big (\frac{2^{n_h}+2j}{2^{2n_n}}\,,\, \frac{2^{n_h}+2j+1}{2^{2n_h}}\Big )\Big ). \end{aligned}$$

From this

$$\begin{aligned} \mu \big [(0,h]\cap E\big ]={\textstyle \frac{1}{2}}h+ O(2^{-{2n_h}})\quad \Longrightarrow \quad \mu \big [(0,h]\cap E^+\big ]={\textstyle \frac{1}{2}}h+ O(h^2). \end{aligned}$$

Thus

$$\begin{aligned} \lim _{h\rightarrow 0}\frac{1}{h} \int _0^h \chi _{E^+} dx=\frac{1}{2}. \end{aligned}$$

Let $E^-$ be the symmetric of $E^+$ in $(-1,0]$ and set $E=E^-\cup E^+$.

7.2. :

Let f be absolutely continuous in [a, b]. Then the function $x\rightarrow \mathcal {V}_f[a,x]$ is also absolutely continuous in [a, b]. Moreover

$$\begin{aligned} \mathcal {V}_f[a,x]=\int _a^x|f^\prime (t)|dt\quad \text { for all }\> x\in [a,b]. \end{aligned}$$

7.3. :

Let f be of bounded variation in [a, b]. The singular part of f is the function ([90])

$$\begin{aligned} \sigma _f(x)=f(x)-f(a)-\int _a^xf^\prime (t)dt. \end{aligned}$$

(7.1c)

The singular part of f is of bounded variation and $\sigma ^\prime =0$ a.e. in [a, b]. It has the same singularities as f, and $(f-\sigma )$ is absolutely continuous.

Thus every function f of bounded variation in [a, b], can be decomposed into the sum of an absolutely continuous function in [a, b] and a singular function. Compare the $\sigma _f$ with the functions of the jumps $J_f$ given in § 1.1c of the Complements.

7.4. :

Let f be Lebesgue integrable in the interval [a, b] and let F denote a primitive of f. Then for every absolutely continuous function g defined in [a, b]

$$\begin{aligned} \int _a^b fgdx=F(b)g(b)-F(a)g(a)- \int _a^b Fg^\prime dx. \end{aligned}$$

7.5. :

Let $f,g:[a,b]\rightarrow \mathbb {R}$ be absolutely continuous. Then

$$\begin{aligned} \int _a^b fg^\prime dx+\int _a^b f^\prime gdx= f(a)g(a)- f(b)g(b). \end{aligned}$$

7.6. :

Let $h:[a,b]\rightarrow [c,d]$ be absolutely continuous, increasing and such that $h(a)=c$ and $h(b)=d$. Then for every nonnegative, Lebesgue measurable function $f:[c,d]\rightarrow \mathbb {R}$, the composition f(h) is measurable and

$$\begin{aligned} \int _c^d f(s)ds=\int _a^b f\big (h(t)\big ) h^\prime (t)dt. \end{aligned}$$

This is established sequentially for f the characteristic function of an interval, the characteristic function of an open set, the characteristic function of a measurable set, for a simple function.

Proposition 7.1c

Let f be absolutely continuous in [a, b]. Then for every measurable $E\subset [a,b]$ of measure zero, $f(E)\subset \mathbb {R}$ is a set of measure zero.

Proof

Combine Proposition 7.2 with and Vitali’s absolute continuity of the integral (Theorem 11.1 of Chap. 4).$\blacksquare$

The converse of Proposition 7.1c is in false. The characteristic function of the rationals maps any set into a set of measure zero. Such a function however is not continuous.

7.7. :

Prove by a counterexample that the converse of Proposition 7.1c is false, even if f is assumed to be continuous. Hint: The function f in (1.1) is continous in [0, 1] and not BV[0, 1]. To show that it maps measurable sets of measure zero, into sets of measure zero, observe that $f\in AC[\varepsilon ,1]$ for all $\varepsilon >0$.

Proposition 7.2c

Let $f:[a,b]\rightarrow \mathbb {R}$ be continuous and monotone and such that for every measurable $E\subset [a,b]$ of measure zero, $f(E)\subset \mathbb {R}$ is a set of measure zero. Then f is absolutely continuous in [a, b].

Proof

Assuming f non-decreasing the function $h(x)=x+f(x)$ is strictly increasing. The set function $\nu (E)=\mu (h(E))$ for all Lebesgue measurable sets in [a, b] is a measure on the same $\sigma$-algebra, satisfying $\nu \ll \mu$. Apply the Radon-Nykodým theorem.$\blacksquare$

1.1 7.1c Characterizing BV[a, b] Functions

Denote by $C_o^1[a,b]$ the collection of continuously differentiable functions $\varphi$ of compact support in [a, b].

Proposition 7.3c

Let $f\in BV[a,b]$. Then

$$\begin{aligned} \sup _{\genfrac{}{}{0.0pt}{}{\varphi \in C_o^1(\mathbb {R})}{|\varphi |\le 1}} \int _a^b f\varphi ^\prime dx \le \mathcal {V}_f[a,b]. \end{aligned}$$

(7.2c)

Proof

One may assume that f is monotone increasing and nonnegative. There exists an increasing sequence of simple functions such that $\{f_n\}\rightarrow f$ everywhere in [a, b] (Proposition 3.1 of Chap. 4). By the monotonicity of f, the construction of $\{f_n\}$ identifies a partition

$$\begin{aligned} P=\{a=x_o<x_1<\cdots <x_n=b\} \end{aligned}$$

(7.3c)

of [a, b] such that

$$\begin{aligned} f_n(x) =f(x_{j-1}) \quad \text { in the interval }\> [x_{j-1},x_j]\quad \text { for } \> j=1,\dots ,n. \end{aligned}$$

For $0<\delta \ll 1$ and n fixed construct the Lipschitz continuous functions

$$\begin{aligned} f_{n,\delta }(x)=\mathop {\textstyle {\sum }}\limits _{j=1}^n \Big \{f(x_{j-1})\chi _{[x_{j-1},x_j]} +[f(x_j)-f(x_{j-1})] \chi _{[x_j-\delta ,x_j]} \frac{x-x_j +\delta }{\delta }\Big \}. \end{aligned}$$

One verifies that

$$\begin{aligned} \int _a^b f\varphi ^\prime dx =\lim _{n\rightarrow \infty }\lim _{\delta \rightarrow 0} \int _a^b f_{n,\delta }\varphi ^\prime dx. \end{aligned}$$

Since $f_{n,\delta }$ are absolutely continuous in [a, b], integration by parts is justified. Hence for every $\varphi \in C_o^1[a,b]$,

$$\begin{aligned} \Big |\int _a^b f_{n,\delta }\varphi ^\prime dx\Big |=\Big | \mathop {\textstyle {\sum }}\limits _{j=1}^n\int _{x_{j-1}}^{x_j}f_{n,\delta } \varphi ^\prime dx\Big |\le \mathcal {V}_f[a,b]. \end{aligned}$$

$\blacksquare$

The converse of (7.2c) is in general false. For example $\chi _{\{\frac{1}{2}\}}\in BV[0,1]$ with variation 2, whereas the left-hand side of (7.2c) is zero. The latter only requires that f be integrable, whereas the notion of bounded variation requires that f be defined at every point of [a, b]. However if f is integrable in [a, b] then it is unambigously defined by (11.1), or (11.1)${}_{N=1}^\prime$, everywhere in [a, b] except for a set of measure zero. Given f integrable in [a, b] introduce partitions $P_f$ as in (7.3c), where however $x_j$ are differentiability points of f. Then define the essential variation of f in (a, b) as

$$\begin{aligned} {\text {ess}}-\mathcal {V}_f(a,b)=\sup _{P_f}\mathop {\textstyle {\sum }}\limits _{j=1}^n|f(x_{j})-f(x_{j-1})|. \end{aligned}$$

Proposition 7.4c

Let f be integrable in [a, b]. Then

$$\begin{aligned} {\text {ess}}-\mathcal {V}_f(a,b)\le \sup _{\genfrac{}{}{0.0pt}{}{\varphi \in C_o^1[a,b]}{|\varphi |\le 1}} \int _a^b f\varphi ^\prime dx. \end{aligned}$$

(7.4c)

Proof

Assume the right-hand side of (7.4c) is finite, and fix a partition $P_f$. Without loss of generality may assume that a and b are points of $P_f$, and construct the polygonal of vertices $\big (x_j, f(x_j)\big )$. Assume momentarily that such a polygonal changes its monotonicity at each of its veritice, i.e.,

$$\begin{aligned} \begin{aligned}&\text {if }\> f(x_{j-1})< f(x_j)\>\text { then }\> f(x_j)> f(x_{j+1});\\&\text {if }\> f(x_{j-1})> f(x_j)\>\text { then }\> f(x_j)< f(x_{j+1}). \end{aligned} \end{aligned}$$

(7.5c)

Assume $f(a)<f(x_1)$ and set

$$\begin{aligned} \varphi _\delta (x)=\left\{ \begin{array}{ll} {\displaystyle \frac{a-x}{\delta }} \>&{}\text { for }\> a\le x<a+\delta ,\\ -1\>&{}\text { for }\> a+\delta \le x\le x_1-\delta ,\\ {\displaystyle \frac{x-x_1+\delta }{\delta }-1}\>&{}\text { for }\> x_1-\delta \le x<x_1.\\ \end{array}\right. \end{aligned}$$

Then, for $j=1,\dots ,n$, define $\varphi$ recursively as

$$\begin{aligned} \begin{array}{ll} {\varphi _\delta (x)=\left\{ \begin{array}{ll} {\displaystyle \frac{x_{j-1}-x}{\delta }}\>&{}\text { for }\> x_{j-1}\le x<x_{j-1}+\delta ,\\ -1\>&{}\text { for }\> x_{j-1}+\delta \le x\le x_j-\delta ,\\ {\displaystyle \frac{x-x_j+\delta }{\delta }-1}\>&{}\text { for }\> x_j-\delta \le x<x_j, \end{array}\right. } \quad &{}\text { if }\> f(x_{j-1})<f(x_j);\\ {}\\ {}\\ {\varphi (x)_\delta =\left\{ \begin{array}{ll} {\displaystyle \frac{x-x_{j-1}}{\delta }}\>&{}\text { for }\> x_{j-1}\le x<x_{j-1}+\delta ,\\ 1\>&{}\text { for }\> x_{j-1}+\delta \le x\le x_j-\delta ,\\ {\displaystyle \frac{x_j-\delta -x}{\delta }+1}\>&{}\text { for }\> x_j-\delta \le x<x_j, \end{array}\right. } \quad&\text { if }\> f(x_{j-1})>f(x_j). \end{array} \end{aligned}$$

Assuming momentarily that such a choice is admissible, compute

$$\begin{aligned} \sup _{\genfrac{}{}{0.0pt}{}{\varphi \in C_o^1[a,b]}{|\varphi |\le 1}}\int _a^b f\varphi ^\prime dx&\ge \limsup _{\delta \rightarrow 0}\Big |\int _a^b f\varphi _\delta ^\prime dx\Big |\\&=\limsup _{\delta \rightarrow 0}\Big |\frac{\text { sign }{\varphi _\delta ^\prime }}{\delta } \int _a^{a+\delta } fdx\\&\qquad \qquad + \mathop {\textstyle {\sum }}\limits _{j=1}^{n-1} \frac{\text { sign }{\varphi _\delta ^\prime }}{\delta }\int _{x_j-\delta }^{x_j+\delta } fdx +\frac{\text { sign }{\varphi _\delta ^\prime }}{\delta }\int _{b-\delta }^b fdx\Big |\\&\ge \mathop {\textstyle {\sum }}\limits _{j=1}^n |f(x_j)-f(x_{j-1})|, \end{aligned}$$

since $x_j$ are differentiability points of f.$\blacksquare$

Complete the proof by the following steps:

i. :

Prove that in (7.4c) the supremum can be taken over all Lipschitz continuous functions $\varphi \in C_o[a,b]$.

ii. :

Remove the assumptions and that $P_f$ satisfies (7.5c), to establish the inequality for all partitions $P_f$ of [a, b]. $\blacksquare$

Corollary 7.1c

An integrable function f in [a, b] is of essentially bounded variation in [a, b] if and only if the right-hand side of (7.4c) is finite.

1.2 7.2c Functions of Bounded Variation in N Dimensions [55]

The characterization of Corollary 7.1c suggests a notion bounded variations in several dimensions. A function f locally integrable in $\mathbb {R}^N$ is of bounded variation in a Lebesgue measurable set $E\subset \mathbb {R}^N$ if there exists a constant C such that

$$\begin{aligned} \Big |\int _Ef{\text {div}}{\varvec{\varphi }}dx\Big |\le C \end{aligned}$$

(7.6c)

for all vector valued functions

$$\begin{aligned} {\varvec{\varphi }}=(\varphi _1,\dots ,\varphi _N)\in [C_o^1(\mathbb {R}^N)]^N \quad \text { such that }\> |{\varvec{\varphi }}|\le 1. \end{aligned}$$

The smallest constant C for which (7.6c) holds is the variation of f in E and is denoted by $\Vert Df\Vert (E)$.

1.2.1 7.2.1c Perimeter of a Set

Let E be a bounded measurable set in $\mathbb {R}^N$ with smooth boundary ${\partial }{E}$. Prove that $\chi _E\in BV(\mathbb {R}^N)$ and that $\mathcal {V}_{\chi _E}(\mathbb {R}^N)=\mathcal {H}^{N-1}({\partial }{E})$, where $\mathcal {H}^{N-1}$ is the $(N-1)$-dimensional Hausdorff measure of sets in $\mathbb {R}^N$.

If ${\partial }{E}$ is not smooth, (7.6c) suggests defining the “measure of ${\partial }{E}$” or, roughly speaking, the perimeter of E, by

$$\begin{aligned} {\text {Per}}(E)=\sup _{{\varvec{\varphi }}\in [C_o^1]^N,\,|{\varvec{\varphi }}|\le 1} \int _{\mathbb {R}^N} \chi _E{\text {div}}{\varvec{\varphi }}dx, \end{aligned}$$

(7.7c)

provided the right-hand side is finite. Sets $E\subset \mathbb {R}^N$ for which ${\text {Per}}(E)<\infty$ are called of finite perimeter. They correspond, roughly speaking, to sets for which the Gauss-Green theorem holds.

13c Convex Functions

13.1. :

Give an example of a bounded, discontinuous, convex function in [a, b]. Give an example of a convex function unbounded in (a, b).

13.2. :

A continuous function f in (a, b) is convex if and only if

$$\begin{aligned} f\Big (\frac{x+y}{2}\Big )\le \frac{f(x)+f(y)}{2}\qquad \text { for all }\> x,y\in (a,b). \end{aligned}$$

13.3. :

Let f be convex, non-decreasing and non-constant in $(0,\infty )$. Then $f(x)\rightarrow \infty$ as $x\rightarrow \infty$.

13.4. :

Let f be convex in $[0,\infty )$. Then the limit of $x^{-1}f(x)$ as $x\rightarrow \infty$, exists finite or infinite.

13.5. :

Let $\{f_n\}$ be a sequence of convex functions in (a, b) converging to some real-valued function f. Then the convergence is uniform within any closed subinterval of (a, b). The conclusion is false if f is permitted to take values in $\mathbb {R}^*$, as shown by the sequence $\{x^n\}$ for $x\in (0,2)$.

13.6. :

Let $f\in C^2(a,b)$. Then f is convex in (a, b) if and only if $f^{\prime \prime }(x)\ge 0$ for each $x\in (a,b)$. Proof: Having fixed $x<y$ it suffices to prove that

$$\begin{aligned}{}[0,1]\ni t\rightarrow \varphi (t) = f(tx+(1-t)y)-t f(x)-(1-t) f(y) \end{aligned}$$

(13.1c)

is nonpositive in [0, 1]. Such a function vanishes at the end points of [0, 1] and its extrema are minima since

$$\begin{aligned} \varphi ^{\prime \prime }(t)=(x-y)f^{\prime \prime }(tx+(1-t)y)\le 0. \end{aligned}$$

Proposition 13.1c

A continuous function f in (a, b) is convex if and only if either one of the two one-sided derivatives $D_\pm f$ is non-decreasing.

Proof

Assume for example that $D_+f$ is non-decreasing. If the function $\varphi$ in (13.1c) has a positive maximum $\varphi (t_o)>0$, at some $t_o\in (0,1)$, then

$$\begin{aligned} D_+\varphi (t_o)=(x-y)D_+f(t_ox+(1-t_o)y)+f(y)-f(x)\le 0. \end{aligned}$$

Therefore, since $D_+f$ is non-decreasing, $D_+\varphi (t)$ is non-positive in $[0,t_o]$. Thus $\varphi$ is non-increasing in $[0,t_o]$ and $\varphi (t_o)\le 0$.$\blacksquare$

13.7. :

The function $f(x)=|x|^p$ is convex for $p\ge 1$ and concave for $p\in (0,1)$.

1.1 13.8c Convex Functions in $\mathbb {R}^N$

Let E be a convex subset of $\mathbb {R}^N$. A function $f:E\rightarrow \mathbb {R}^*$ is convex if for every pair of points x and y in E and every $t\in [0,1]$

$$\begin{aligned} f(t x+(1-t)y)\le tf(x)+(1-t)f(y). \end{aligned}$$

The $(N+1)$-dimensional set

$$\begin{aligned} \mathcal {G}_f=\left\{ (x,x_{N+1})\in \mathbb {R}^{N+1}\bigm | x\in E,\quad x_{N+1}\ge f(x)\right\} \end{aligned}$$

is the epigraph of f. The function f is convex if and only if its epigraph is convex.

13.9. :

Let $E\subset \mathbb {R}^N$ be open and convex. A function $f\in C^2(E)$ is convex if and only if ${\mathop {\textstyle {\sum }}\limits }_{i,j=1}^N f_{x_ix_j}\xi _i\xi _j\ge 0$ for all $\xi \in \mathbb {R}^N$.

Hint: Fix $B_\rho (x)\subset E$ and $\xi$ in the unit sphere of $\mathbb {R}^N$. The function $(-\rho ,\rho )\ni t\rightarrow \varphi (t)=f(x+t\xi )$ is convex.

13.10. :

Construct a non convex function $f\in C^2(\mathbb {R}^2)$ such that $f_{xx}$ and $f_{yy}$ are both nonnegative.

13.11. :

Let $E\subset \mathbb {R}^N$ be open and convex, and let f be convex and real-valued in E. Then f is continuous on E. Moreover for every $x\in E$ there exist the left and right directional derivatives

$$\begin{aligned} D^\pm _{\mathbf {u}}f(x)= D^\pm _t f(x+t\mathbf {u})\bigm |_{t=0} \qquad \text { for all }\>|\mathbf {u}|=1. \end{aligned}$$

Moreover $D^-_{\mathbf {u}}f\le D^+_{\mathbf {u}}f$. In particular for each $x\in E$, there exist the left and right derivatives $D^\pm _{x_j}f$, along the coordinate axes and $D^-_{x_j}f\le D^+_{x_j}f$.

13.12. :

Let $E\subset \mathbb {R}^N$ be convex. A function f defined in E is convex if and only if $f(x)=\sup \pi (x)$, where $\pi \le f$ is affine.

13.13. :

Let $f:\mathbb {R}^N\rightarrow \mathbb {R}$ be convex. There exist a positive number k such that $\liminf _{|x|\rightarrow \infty }|x|^{-1}f(x)\ge -k$.

1.2 13.14c The Legendre Transform ([92])

The Legendre transform $f^*$ of a convex function $f:\mathbb {R}^N\rightarrow \mathbb {R}^*$ is defined by

$$\begin{aligned} f^*(x)=\sup _{y\in \mathbb {R}^N}\{x\cdot y-f(y)\}. \end{aligned}$$

(13.2c)

Proposition 13.2c

$f^*$ is convex in $\mathbb {R}^N$ and $f^{**}=f$.

Proof

The convexity of f follows from 13.12. From the definition (13.2c), $f(y)+f^*(x)\ge y\cdot x$, for all $x,y\in \mathbb {R}^N$. Therefore

$$\begin{aligned} f(y)\ge \sup _{x\in \mathbb {R}^N}\{y\cdot x-f^*(x)\}= f^{**}(y). \end{aligned}$$

Also, still from (13.2c)

$$\begin{aligned} f^{**}(x)&=\sup _{y\in \mathbb {R}^N}\big \{x\cdot y- \sup _{z\in \mathbb {R}^N}\{y\cdot z-f(z)\}\big \}\\&=\sup _{y\in \mathbb {R}^N}\inf _{z\in \mathbb {R}^N}\{y\cdot (x-z)+f(z)\}. \end{aligned}$$

Since f is convex, for a fixed $x\in \mathbb {R}^N$, there exists a vector $\mathbf {m}$ such that

$$\begin{aligned} f(z)-f(x)\ge \mathbf {m}\cdot (z-x)\qquad \text { for all }\> z\in \mathbb {R}^N. \end{aligned}$$

Combining these inequalities yields

$$\begin{aligned} f^{**}(x)\ge f(x)+\sup _{y\in \mathbb {R}^N}\inf _{z\in \mathbb {R}^N} (z-x)\cdot (\mathbf {m}-y)=f(x). \end{aligned}$$

$\blacksquare$

1.3 13.15c Finiteness and Coercivity

The Legendre transform $f^*$, as defined by (13.2c), could be infinite even if f is finite in $\mathbb {R}^N$. For example in $\mathbb {R}$

$$\begin{aligned} |x|^*=\left\{ \begin{array}{ll} 0\quad &{}\text { if }\> |x|\le 1\\ \infty \quad &{}\text { if }\> |x|>1. \end{array} \right. \end{aligned}$$

A convex function $f:\mathbb {R}^N\rightarrow \mathbb {R}$ is coercive at infinity if

$$\begin{aligned} \lim _{|x|\rightarrow \infty }\frac{f(x)}{|x|}=\infty . \end{aligned}$$

Proposition 13.3c

If f is coercive at infinity, then $f^*$ is finite in $\mathbb {R}^N$. If f is finite, then $f^*$ is coercive at infinity.

Proof

Assume f is coercive at infinity. If the $\sup$ in (13.2c) is achieved for $y=0$ the assertion is obvious. Otherwise

$$\begin{aligned} f^*(x)=\sup _{y\in \mathbb {R}^N-\{0\}}|y| \Big \{x\cdot \frac{y}{|y|}-\frac{f(y)}{|y|}\Big \}. \end{aligned}$$

Therefore the supremum is achieved for some finite y and $f^*(x)$ is finite.

To prove the converse statement, fix $\lambda >0$ and write

$$\begin{aligned} f^*(x)&=\sup _{y\in \mathbb {R}^N}\{x\cdot y-f(y)\}\ge \{x\cdot y-f(y)\}\bigm |_{y=\lambda x/|x|}\\&=\lambda |x|-f\big (\lambda \frac{x}{|x|}\big )\ge \lambda |x|-\sup _{|\mathbf {u}|=\lambda }|f(\mathbf {u})|. \end{aligned}$$

Therefore, since $x\in \mathbb {R}^N-\{0\}$ is arbitrary

$$\begin{aligned} \lim _{|x|\rightarrow \infty }\frac{f^*(x)}{|x|}\ge \lambda \qquad \text { for all }\>\lambda >0. \end{aligned}$$

$\blacksquare$

1.4 13.16c The Young’s Inequality

Prove that the Legendre transform of the convex function

$$\begin{aligned} \mathbb {R}\ni a\rightarrow f(a)=\frac{1}{p} |a|^p\quad \text { for }\> 1< p<\infty \end{aligned}$$

is

$$\begin{aligned} \mathbb {R}\ni b\rightarrow f^*(b)=\frac{1}{q} |b|^q\quad \text { for }\> 1<q<\infty \quad \text { and }\>\frac{1}{p}+\frac{1}{q}=1. \end{aligned}$$

Then the definition (13.2c) of the Legendre transform implies the Young’s inequality

$$\begin{aligned} |ab|\le \frac{1}{p}|q|^p+\frac{1}{q}|b|^q\quad \text { for all }\> a,b\in \mathbb {R}. \end{aligned}$$

(13.3c)

The inequality continues to holds for the limiting case $p=1$ and $q=\infty$. For a different proof see Proposition 2.1 of Chap. 6.

14c Jensen’s Inequality

Proposition 14.1c

(Hölder [75]) Let $\{\alpha _i\}$ be a sequence of nonnegative numbers such that $\sum \alpha _i=1$, and let $\{\xi _i\}$ be a sequence in $\mathbb {R}$. Then

$$\begin{aligned} \exp \left( \mathop {\textstyle {\sum }}\limits \,\alpha _i\xi _i\right) \le \mathop {\textstyle {\sum }}\limits \, \alpha _i\exp (\xi _i). \end{aligned}$$

(14.1c)

Proof

Apply (14.1) to $e^x$ with $\eta =\xi _j$ and $\alpha =\mathop {\textstyle {\sum }}\limits \alpha _i\xi _i$, to get

$$\begin{aligned} \exp \left( \mathop {\textstyle {\sum }}\limits \,\alpha _i\xi _i\right) + m_j\left( \xi _j- \mathop {\textstyle {\sum }}\limits \,\alpha _i\xi _i\right) \le \exp (\xi _j). \end{aligned}$$

Multiply by $\alpha _j$ and add over j.$\blacksquare$

Corollary 14.1c

Let $\{\alpha _i\}$ be a sequence of nonnegative numbers such that $\sum \alpha _i=1$, and let $\{\xi _i\}$ be a sequence of positive numbers. Then

$$\begin{aligned} \mathop {\textstyle {\prod }}\limits \,\xi _i^{\alpha _i}\le \mathop {\textstyle {\sum }}\limits \,\alpha _i\xi _i. \end{aligned}$$

(14.2c)

1.1 14.1c The Inequality of the Geometric and Arithmetic Mean

In the case where $\alpha _i=0$ for $i>n$ and $\alpha _i=1/n$ for $i=1,2,\dots ,n$, inequality (14.2c) reduces to the inequality between the geometric and arithmetic mean of n positive numbers⁶

$$\begin{aligned} (\xi _1\xi _2\cdots \xi _n)^{1/n}\le \frac{\xi _1+\xi _2+\cdots +\xi _n}{n}. \end{aligned}$$

(14.3c)

1.2 14.2c Integrals and Their Reciprocals

Proposition 14.2c

Let E be a measurable set of finite measure and let $f:E\rightarrow \mathbb {R}^+$ be measurable. Then

$$\begin{aligned} \frac{1}{\displaystyle \left( \frac{1}{\mu (E)}\int _Ef d\mu \right) ^p}\le \frac{1}{\mu (E)} \int _E\frac{1}{f^p} d\mu ,\quad \text { for all }\> p>0. \end{aligned}$$

(14.4c)

Proof

Assume first that f is integrable and that $f\ge \varepsilon$, and apply Jensen’s inequality with $\varphi (t)=t^{-p}$. $\blacksquare$

15c Extending Continuous Functions

15.1. :

Let f be convex in a closed interval [a, b] and assume that $D_+f(a)$ and $D_-f(b)$ are both finite. There exists a convex function $\widetilde{f}$ defined in $\mathbb {R}$ such that $f=\widetilde{f}$ in [a, b].

16c The Weierstrass Approximation Theorem

16.1. :

Let $E\subset \mathbb {R}^N$ be open, bounded and with smooth boundary ${\partial }{E}$. Let $f\in C^1({\bar{E}})$ vanish on ${\partial }{E}$, and let $P_j$ denote the jth Stieltjes polynomial relative to f . Then

$$\begin{aligned} \lim _{j\rightarrow \infty } \frac{\partial P_j}{\partial x_i}=\frac{\partial f}{\partial x_i} \qquad j=1,\dots ,N\quad \text { in }\> E. \end{aligned}$$

16.2. :

Let $E\subset \mathbb {R}^N$ be bounded and open, and fet $f:{\bar{E}}\rightarrow \mathbb {R}$ and be Lipschitz continuous in ${\bar{E}}$. with Lipschitz constant L. Then the Stieltjes polynomials $P_j$ relative to f are equi-Lipschitz continuous in E, with the same constant L.

16.3. :

A continuous function $f:[0,1]\rightarrow \mathbb {R}$, can be approximated by the Bernstein polynomials $B_j$, relative to f

$$\begin{aligned} B_j(x)=\mathop {\textstyle {\sum }}\limits _{i=1}^j\,\Big ( \begin{array}{c} j\\ i \end{array} \Big ) f\Big (\frac{i}{j}\Big ) x^i(1-x)^{j-i}. \end{aligned}$$

State and prove a N-dimensional version of such an approximation ([98]).

16.4. :

Let f be uniformly continuous on a bounded, open set $E\subset \mathbb {R}^N$ and denote by $\mathcal {P}_n$ the set of all polynomials of degree n in the coordinate variables. Then

$$\begin{aligned} \int _Efp_ndx=0\>\text { for all } p_n\in \mathcal {P}_n \text { and all } n\in \mathbb {N}\>\Longrightarrow \> f=0. \end{aligned}$$

17c The Stone-Weierstrass Theorem

17.1. :

The Stone-Weierstrass theorem fails for complex valued functions.

Let D be the closed, unit disc in the complex plane $\mathbb {C}$ and denote by $C(D;\mathbb {C})$ the linear space of all the continuous complex valued functions defined in D endowed with the topology generated by the metric in (17.1).

Consider also the subset $\mathcal {H}(D)$ of $C(D;\mathbb {C})$, consisting of all holomorphic functions defined in D. One verifies that $\mathcal {H}(D)$ is an algebra. Moreover uniform limits of holomorphic functions in D are holomorphic ([24], Chap. V, Théoréme 1, page 145). Thus $\mathcal {H}(D)$ is closed under the metric in (17.1). The algebra $\mathcal {H}(D)$ is called the disc algebra.

Such an algebra separates points since it contains the holomorphic function $f(z)=z$. Moreover $\mathcal {H}(D)$ contains the constants. However $\mathcal {H}(D)\ne C(D;\mathbb {C})$. Indeed the function $f(z)=\overline{z}$ is continuous but not holomorphic in D.

17.2. :

Let $f:\mathbb {R}\rightarrow \mathbb {R}$ be continuous and $2\pi$-periodic. For every $\varepsilon >0$ there exists a function of the type

$$\begin{aligned} \varphi (x)= a_o+\mathop {\textstyle {\sum }}\limits _{n=1}^m (b_n\cos {nx}+ c_n\sin {nx}) \end{aligned}$$

such that $\sup _{\mathbb {R}}|f-\varphi |\le \varepsilon$ (Hint: Use Stone’s Theorem).

19c A General Version of the Ascoli-Arzelà Theorem

The proof of Theorem 19.1 uses only the separability of $\mathbb {R}^N$ and the metric structure of $\mathbb {R}$. Thus it can be extended into any abstract framework with these two properties.

Let $\{f_n\}$ be a countable collection of continuous functions from a separable topological space $\{X;\mathcal {U}\}$ into a metric space $\{Y;d_Y\}$. The functions $f_n$ are equibounded at x if the closure in $\{Y;d_Y\}$ of the set $\{f_n(x)\}$ is compact.

The functions $f_n$ are equi-continuous at a point $x\in X$ if for every $\varepsilon >0$, there exists an open set $\mathcal {O}\in \mathcal {U}$ containing x and such that

$$\begin{aligned} d_Y(f_n(x),f_n(y))\le \varepsilon \quad \text { for all }\> y\in \mathcal {O}\quad \text { and all }\> n\in \mathbb {N}. \end{aligned}$$

Theorem 19.1c

Let $\{f_n\}$ be a sequence of continuous functions from a separable space $\{X;\mathcal {U}\}$ into a metric space $\{Y;d_Y\}$. Assume that the functions $f_n$ are equibounded and equi-continuous at each $x\in X$. Then, there exists a subsequence $\{f_{n^\prime }\}\subset \{f_n\}$ and a continuous function $f:X\rightarrow Y$ such that $\{f_{n^\prime }\}\rightarrow f$ pointwise in X. Moreover the convergence is uniform on compact subsets of X.

State and prove an analog of Proposition 19.1.