Riesz's Sur certains systèmes singuliers d'équations intégrales

In what follows, the functions of bounded variation will play a leading role. We know the importance of this class of functions defined by M. Jordan, whose most remarkable properties become almost obvious after only one statement: that every real function of bounded variation is the difference of two bounded, never decreasing functions.

Consider the following problem: Given a function \( \text{A}(x) \), can we say whether or not there is function \( \alpha(x) \) of bounded variation of which it is the indefinite integral?

The analogous question for one or another class of functions always arises when it comes to determining a function to solve this problem, that it is easier to first calculate the original function. One knows the answer for certain classes: for example, through the results developed by several mathematicians in recent years, one can find the indefinite integrals of summable functions, of those whose square is summable, etc. Some time ago I stated a necessary and sufficient condition for functions of bounded variation^[1]; I return to the matter in this Memoir.

Let's try to develop a necessary condition first. Assume that \( \text{A}(x) \) is the indefinite integral of the real function \( \alpha(x) \) of bounded variation defined in the interval \( (a, b) \); moreover \( \alpha(x) \) can be continuous or not. Decompose the interval \( (a, b) \) into a finite number of subintervals \( (x_k, x_{k + 1}), \) \( k=0, 1, \cdots, m-1; x_0=a, x_m=b \), and consider the expression

\[ \begin{equation}{ {{\LARGE{\sum_{\large{k=1}}^{\large{m-1}}}} \left|{{\text{A}(x_{k+1}) - \text{A}(x_k)} \over {x_{k+1}-x_k}} - {{\text{A}(x_k) - \text{A}(x_{k-1})} \over {x_k-x_{k-1}}} \right|}.} \tag{1} \end{equation} \]

We have

\[ {{\text{A}(x_{k+1}) - \text{A}(x_k)} \over {x_{k+1}-x_k}} = { \displaystyle {\int_{x_k}^{x_{k+1}} \alpha(x)dx } \over {x_{k+1}-x_k}} = \beta_k; \]

where \( \beta_k \) designates a fixed quantity between the lower and upper limits of \( \alpha(x) \) in the interval \( (x_k, x_{k + 1}) \). It follows that the value of the sum (1) can not exceed the total variation of \( \alpha(x) \). That is to say it is always true that when \( \text{A}(x) \) has the required property, the sum of (1) does not exceed some finite bound, independent of the number of subintervals and the manner of decomposition. So this is a necessary condition; we will see that this condition is sufficient at the same time.

Suppose the condition is satisfied, with \( G \) the upper bound. We seek to determine a function \( \alpha(x) \) of bounded variation whose indefinite integral is \( \text{A}(x) \).

In reality, the function \( \text{A}(x) \) has derivatives on the left and on the right. As for our purpose, it suffices to show that it admits any of the four derived numbers, and that they are functions of bounded variation. For every function of bounded variation is Riemann integrable, and when a derived number is integrable, its original function is the indefinite integral^[2].

Now let \( h \) be a positive quantity less than \( \large{{b-a} \over 3} \); this condition being fulfilled, we easily deduce that for any \( x \) and \( |h'| < h \), the ratio \( \large{{\text{A}(x+h') - \text{A}(x)} \over h'} \) cannot surpass in absolute value the larger of the quantities

\[ G + {\left| {\text{A}(a+h) - \text{A}(a)} \over h \right|}, \hskip{20pt} G + {\left| {\text{A}(b-h) - \text{A}(b)} \over h \right|}. \]

We conclude that the function \( \text{A}(x) \) has bounded derived numbers. Let \( \alpha(x) \) be a derived number, for example the upper right derived number; function \( \alpha(x) \) is thus determined for \( a \leq x < b \), and we complete its definition by assuming \( \alpha(b) \) equal to the upper left derived number at \( b \).

The function \( \alpha(x) \) is of bounded variation; its total variation does not exceed \( G \). Indeed, decomposing the interval \( (a, b) \) in some way into a finite number of subintervals

\[ (x_k,x_{k+1}) \hskip{20pt} k=0,1, \cdots, m-1; \;\; x_0 =a, \; x_m=b \]

we will have to prove that the value of the sum

\[ \begin{equation}{\sum_{k=0}^{m-1} |\alpha(x_{k+1}) - \alpha(x_k)|} \tag{2} \end{equation} \]

does not exceed \( G \). Or, for an approximate value of this sum differing as little we want, we can replace \( \alpha(x_k) \) by \( \large{{\text{A}(x_k + h_k) - \text{A}(x_k)} \over h_k} \), the amounts \( h_k \) being positive for \( k < m \), negative for \( k = m \), and chosen in a suitable manner; at the same time, we can assume the intervals \( (x_k, x_k + h_k) \) to be quite small without encroaching upon each other. However, it is clear that the sum (2) as amended does not exceed \( G \); we conclude that the sum (2) itself, however large the approximation, does not exceed \( G \). This fact does not depend on the manner of partitioning the interval \( (a, b) \); consequently, the function \( \alpha(x) \) is of bounded variation and its total variation does not exceed \( G \). We can also conclude, applying the already mentioned theorem, that the given function \( \text{A}(x) \) is the indefinite integral of \( \alpha(x) \).

In the above, we have assumed that the functions \( \text{A}(x), a(x) \) are real. Our findings extend to the general case of complex-valued functions. To realize that, we must break down the functions into their real and imaginary parts; the real and imaginary parts of \( \text{A}(x) \) always separately satisfy our condition with respect to the bounds \( G1, G2 \) of function \( \text{A}(x) \) and satisfy it. \( G1 + G2 \) gives a suitable bound. On the other hand, \( \text{A}(x) \) itself fulfills our condition compared to a bound \( G \), its real and imaginary parts also being bound by \( G \).

To state our result:

For a function \( \text{A}(x) \) defined in the interval \( (a,b) \), real or not, to be the indefinite integral of a function of bounded variation, it is necessary and sufficient that the value of the sum (1) does not exceed some finite bound, independent of the number of subintervals and how we decompose the interval \( (a,b) \).

More specifically speaking, the function \( \text{A}(x) \) is the indefinite integral of a function whose total variation is equal to the smallest upper bound. As for the actual functions, this follows immediately from the above considerations; one could also adapt these considerations to the general case. Yet the statement follows also, in all its generality, from developments in section II.

Consider the Stieltjes integral

\[ \begin{equation}{\int_a^b f(x)d\alpha(x).} \tag{3} \end{equation} \]

We understand this integral, when it exists, to be the limit of the sum

\[ \begin{equation}{ \sum_k f(\xi_k)[\alpha(x_{k+1}) - \alpha(x_k)],} \tag{4} \end{equation} \]

corresponding to a decomposition of the interval \( (a, b) \) into a finite number of subintervals; \( \xi_k \) denotes any element of the interval \( (x_k, x_{k+1}) \). The passage to the limit is subject only to the sole condition that the length of the subintervals tends uniformly to zero.

Suppose \( f(x) \) is continuous and \( \alpha(x) \) is of bounded variation. Then the integral (3) exists and in this case

\[ \begin{equation}{{\left|\int_a^b f(x)d\alpha(x)\right|} \leq \text{ maximum of } |f(x)| \times \text{ total variation of } \alpha(x).} \tag{5} \end{equation} \]

Indeed, if the integral (3) exists, the inequality (5) immediately follows due to an inequality analogous to that for (4), which is otherwise obvious. But in the conditions laid down, the integral (3) does exist because by (4), the difference in the two sums corresponds to two decompositions which cannot in absolute value exceed the quantity

\[ |\varepsilon| \times \text{total variation of } \alpha(x); \]

we will denote by \( \varepsilon \) the largest oscillation of \( f(x) \) in the intervals whose length does not exceed twice that of the largest interval in our two decompositions. The function \( f(x) \) is assumed to be continuous; let \( \varepsilon \) approach zero as one passes from (4) to the limit (3). We conclude that the integral (3) exists.

By applying a well known transformation dating back to Abel, we see without difficulty that in the conditions laid down, the integral

\[ \begin{equation}{\int_a^b \alpha(x)df(x)} \tag{6} \end{equation} \]

also exists; at the same time, this establishes the relation^[3]

\[ \begin{equation}{{\int_a^b f(x)d\alpha(x) + \int_a^b \alpha(x)df(x)} = {f(b)\alpha(b) - f(a)\alpha(a)}.}\tag{7} \end{equation} \]

Now introduce the function

\[ \text{A}(x) = \int_a^x \alpha(x)dx + \text{A}(a), \]

the indefinite integral of \( \alpha(x) \). The sum

\[ \Large{\sum_{\large{k}}} \normalsize{{{[f(x_{k+1}) - f(x_k)][\text{A}(x_{k+1}) - \text{A}(x_k)]} \over {x_{k+1}-x_k}}}, \]

corresponding to a certain division of \( (a, b) \) and the sum corresponding to the same division

\[ \sum_k [f(x_{k+1}) - f(x_k)] \; \alpha(\xi_k) \]

which serves to define the integral (6), differ in each interval by an absolute value no more than

\[ \text{maximum of } |f(x_{k+1}) - f(x_k)| \times \text{ total variation of } \alpha(x). \]

Therefore, these two sums tend to the same limit (6), which we can denote by

\[ \int_a^b {{df \; dA} \over {dx}}. \]

By virtue of (7) we have

\[ \begin{equation}{{\int_a^b f(x)d\alpha(x) = {f(b)\alpha(b) - f(a)\alpha(a) - \int_a^b {{{df \; dA(x)} \over {dx}}}}.}}\tag{8} \end{equation} \]

Some comments will be useful. Consider first if we can change the function \( \alpha(x) \), without at the same time changing the value ​​of the integral (3). What you see immediately is that if we add a constant to \( \alpha(x) \), the value of (3) does not change. Thanks to the continuity of \( f(x) \), we also see easily that if one assigns any new values to \( \alpha(x) \) at a finite number of points \( x \) in the interior of \( (a, b) \), the change will not affect the value of our integral. But there's more: The set of points where \( \alpha(x) \) is modified can become countably infinite without this change causing a change in value of ​​(3). The following theorem specifies all permitted changes: For the integral (3) to equal zero for all continuous functions \( f(x) \), it is necessary and sufficient that the function of bounded variation \( \alpha(x) \) be constant over the entire interval \( (a, b) \) except possibly for a countable set of points in \( (a, b) \).

Indeed, suppose the function \( \alpha(x) \) to be such that all the integrals (3) are all zero. In (7) successively let \( f (x) \equiv 1 \); then \( f(x) = x \) for \( a \leq x \leq \xi, f(x) = \xi \) for \( x > \xi \). This results in

\[ \alpha(a) = \alpha(b), \hskip{16pt} {\int_a^x \alpha(x) \; dx} = {\alpha(a) \; x;} \]

and one concludes (since functions of bounded variation are continuous except, at most, for a countable set) that \( \alpha(x) = \alpha(a) = \alpha(b) \). On the other hand, to see that our condition is sufficient, suppose that \( \alpha(a) = \alpha(b) = \alpha(x) \), except possibly for a countable set; in this case we have

\[ \text{A}(x) = {\int_a^x \alpha(x) \; dx} = {\alpha(a) \; x}, \]

and from identity (8) it follows immediately that

\[ {\int_a^b f(x) \; d\alpha(x)} = 0. \]

Here is a remarkable result of this theorem: For all the points of discontinuity of \( \alpha(x) \) interior to the interval \( (a,b) \) we may replace \( \alpha(x) \) by \( \alpha(x-0) \) or \( \alpha(x+0) \), or by any of the values \( \theta \; \alpha (x-0) + (1-\theta) \alpha (x+0) \; [0 < \theta <1 ] \), without changing the value (3) at the same time. And this method does not increase the total variation of \( \alpha(x) \); on the contrary it always decreases \( \alpha(x) \) using \( \theta \; \alpha (x-0) + (1-\theta) \alpha (x+0) \) for any value \( 0 \leq \theta \leq 1 \).

For functions \( \alpha(x) \) such that at any interior point of \( (a,b) \), \( \alpha(x) \) coincides with one of the values \( \theta \; \alpha (x-0) + (1-\theta) \alpha (x+0) \), \( \theta \) ranging from \( 0 \) to \( 1 \), the total variation of \( \alpha(x) \) cannot be reduced without one or the other values of (3) changing. Following inequality (5), we can add this remark: As for these functions \( \alpha(x) \) one can choose \( f(x) \) such that

\[ {\left|\int_a^b f(x) \; d\alpha(x)\right|} > \text{ maximum of } |f(x)| \times [- \varepsilon + \text{ total variation of } \alpha(x)]; \]

where \( \varepsilon \) is a positive quantity as small as we want. Indeed, by definition, one can approximate the total variation of \( \alpha(x) \) within \( \varepsilon / 2 \) of the sum

\[ \sum_{k=0}^{m-1} |\alpha(x_{k+1}) - \alpha(x_k)| \hskip{16pt} (x_0=a <x_1 \cdots x_m = b); \]

one has only to select the proper points of division \( x_k \); under present conditions, we may choose these points to be limited to points of continuity of \( \alpha(x) \). Marking off around these division points relatively small intervals \( (y, z) \) in which the oscillation of \( \alpha(x) \) does not exceed the amount \( \Large \varepsilon \over \large{{6(m-1)}} \) in absolute value, set^[4]

\[ f(x) = \overline{\text{sign}} \; [\alpha(z) - \alpha(y)]; \]

to accomplish the above definition of a continuous function \( f(x) \), it is supposed linear in the small intervals designated.

Consider the collection \( \Omega \) of all continuous functions \( f(x) \) on the interval \( (a, b) \), real or not. For the class \( \Omega \), we define the concept of limit function by the assumption of uniform convergence. A functional operation \( \text{A}[f(x)] \) which associates with each element of \( \Omega \) a matching number, real or not, will be called continuous if when \( f(x) \) is the limit function of \( f_i(x) \), then \( \text{A}[f_i] \) tends to \( \text{A}[f] \). Any distributive and continuous operation will be called linear.

For example, when \( a(x) \) denotes any function integrable on \( (a, b) \), the law

\[ \begin{equation}{\text{A}[f(x)] = \int_a^b a(x)f(x) \; dx} \tag{9} \end{equation} \]

defines a linear operation. On the other hand, the linear operation \( \text{A}[f(x)] = f(x_0) \), where \( x_0 \) designates a specific point in \( (a, b) \), cannot be put in the form (9). M. Hadamard had demonstrated the remarkable fact that every linear operation is the limit of a sequence of operations of the form (9); in a more precise way, Hadamard's theorem asserts the existence of a sequence of continuous functions \( {a_n(x)} \) such that^[5]

\[ \text{A}[f(x)] = {\displaystyle \lim_{n \rightarrow \infty}}{\int_a^b a_n(x)f(x) \; dx}. \]

We will show that any linear operation is expressed by a Stieltjes integral. Conversely, by inequality (5): if \( \alpha(x) \) denotes a specific function of bounded variation, then integral (3) defines a linear operation.

First we will establish the following lemma: if \( \text{A}[f(x)] \) designates any linear operation, then associated with it is a positive number \( \text{M}_{\small{\text{A}}} \) such that for every continuous function f(x)

\[ \begin{equation}{|\text{A}[f(x)]| \leq \text{M}_{\small{\text{A}}} \times \text{ maximum of } |f(x)|.} \tag{10} \end{equation} \]

Indeed, otherwise there exists an infinite sequence of continuous functions \( f_k(x) \) such that

\[ |f_k(x)| \leq 1, \hskip{16pt} |\text{A}[f_k(x)]| > k^2. \]

Now, the series

\[ \Large{\sum_{\large{k=1}}} \normalsize{{{f_k(x)} \over {k^2}} \; \overline{\text{sign}}[\text{A}[f_k(x)]} \]

converges uniformly to a continuous function \( f^*(x) \); and the operation \( \text{A}[f] \) being assumed continuous, we would have

\[ \text{A}[f^*(x)] = \Large{\sum_{\large{k=1}}^{\large{\infty}}} \normalsize{{|\text{A}[{f_k(x)]|} \over {k^2}}}. \]

But the series on the right obviously diverges, so our lemma is established.

After these preliminaries, given the linear operation \( A[f(x)] \), we define the function \( \text{A}(x) \) by the formula \( \text{A}(\xi) = \text{A}([f(x, \xi)]) \), where we denote by \( f(x, \xi) \) the function equal to \( x \) for \( a \leq x \leq \xi \) and to \( \xi \) for \( x > \xi \). This being done, decompose the interval \( (a, b) \) into a finite number of subintervals \( (x_k, x_{k+1}) (k = 0, 1, \cdots, m-1; x_0 = a, x_m = b) \) and define \( m - 1 \) continuous functions \( f_k(x) (k=1, \cdots, m-1) \) by putting \( f_k(x) = 0 \) for \( x \leq x_{k-1} \) and \( x \geq x_{k+1} \),

\[ f_k(x_k) = {\overline{\text{sign}}\left[{{\text{A}(x_{k+1}) - \text{A}(x_k)} \over {x_{k+1}-x_k}} - {{\text{A}(x_k) - \text{A}(x_{k-1})} \over {x_k-x_{k-1}}} \right]}, \]

and \( f_k(x) \) linear in the intervals \( (x_{k-1}, x_k), (x_k, x_{k+1}) \). Put

\[ f(x) = {\sum_{k=1}^{m-1} f_k(x)}; \]

we obviously have \( |f(x)| \leq 1 \). Apply inequality (10) to \( f(x) \); it follows that

\[ {{\LARGE{\sum_{\large{k=1}}^{\large{m-1}}}} \left|{{\text{A}(x_{k+1}) - \text{A}(x_k)} \over {x_{k+1}-x_k}} - {{\text{A}(x_k) - \text{A}(x_{k-1})} \over {x_k-x_{k-1}}} \right|} = \text{A}[f(x)] \leq \text{M}_{\small{\text{A}}}. \]

We conclude that there are functions of bounded variation of which \( \text{A}(x) \) is the indefinite integral; we will define one for interior points \( x \) of \( (a, b) \) by putting

\[ \alpha(x) = {\displaystyle \lim_{h \rightarrow 0}} \; {{\text{A}(x+h) - {\text{A}(x-h)} \over 2h}}; \]

and on the points \( a \) and \( b \), put \( \alpha(b) = 0, \alpha(a) = \text{A}[u(x)], u(x) \) denoting the function with constant value 1.

The function \( \alpha(x) \) being of bounded variation, the integral (3) exists for any continuous function \( f(x) \). In particular, when the function \( f(x) \) consists of a finite number of linear pieces, formula (8) tells us immediately that the integral (3) gives \( \text{A}[f(x)] \). Noting that every continuous function is a limit of such functions, and relying on inequalities (5) and (10), we conclude that the same fact holds for every continuous function \( f(x) \). This gives the following theorem:

Given the linear operation \( \text{A}[f(x)] \), one can determine the function of bounded variation \( \alpha(x) \) such that for every continuous function \( f(x) \) we have \[ \text{A}[f(x)] = \int_a^b f(x) \; d\alpha(x). \]

We can add a few remarks about the total variation of \( \alpha(x) \). This total variation equals the upper limit of

\[ \left|\int_a^b f(x) \; d\alpha(x)\right|, \]

for all functions \( f(x) \) with \( |f(x)| \leq 1 \). Indeed, inequality (5) tells us that the total variation of \( \alpha(x) \) is not less than this upper limit; secondly, it can not exceed it because according to the remarks at the end of section II, the function \( f(x) \) can be chosen such that \( |f(x)| \leq 1 \), and that at the same time in formula (5), the sign is applied to \( \varepsilon \), where \( \varepsilon \) is a positive amount as small as we want.

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^ 1. Sur les opérations fonctionnelles linéaires (Comptes rendus, 29 November 1909). I there briefly announced the results developed in sections I to III of this Memorandum; the content of sections IV to VIII was summarized in a second note: Sur certaines systémes d'équations fonctionnelles et l'approximation des fonctions continues (Comptes rendus, 14 March 1910).

^ 2. See H. Lebesgue, Leçons sur l'intégration et la recherche des fonctions primitives, p. 81.

^ 3. Cf. the memoir of Stieltjes, Recherches sur les fractions continues (Annales de Toulouse, 8, 1894).

^ 4. Here and below, we set \( \text{sign} \; z = e^{\varphi \sqrt{-1}}, \; \overline{\text{sign}} \; z = e^{-\varphi \sqrt{-1}} \), where \( z = r e^{\varphi \sqrt{-1}} \), \( r > 0; \) \( \text{sign} \; 0 = \overline{\text{sign}} \; 0 = 0 \).

^ 5. Sur les opérations fonctionnelles (Comptes rendus, 9 February 1903). Cf. also M. Fréchet, Sur les opérations linéaires (Transactions American Math. Soc., 5, 1904, p. 493-499); J. Hadamard, Leçons sur le calcul des variations, collected by M. Fréchet, Paris, 1910.

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Translated by Mike Bertrand (May 22, 2015)