In this post we will study integration in abstract spaces.
Let \(E\) be a normed linear space and \(K\) the closed interval \([a,\,b]\) of the real number line. We consider an operator \(x = x(t),\) which need not be linear and maps \(K\) into \(E.\) In the following, we will call such an operator an abstract function on the interval \([a,\,b].\)
For these functions, we shall define and deduce properties of the fundamental operations of analysis.
Definition of the Integral
We consider a partition \(A = A[t_0 ,\, t_1 ,\, \cdots ,\, t_n ]\) of the closed interval \([a,\,b]\) into closed subintervals \([t_i ,\, t_{i+1}]\) with \[a = t_0 < t_1 < \cdots < t_n = b.\] The partition \(B = B[t_0 ' ,\, t_1 ' ,\, \cdots ,\, t_m ' ]\) is called a refinement of \(A = A[t_0 ,\, t_1 ,\, \cdots ,\, t_n ],\) if every interval \([t_i ' ,\, t_{i+1} ']\) is contained in an interval \([t_i ,\, t_{i+1}].\) In the partition \(B,\) therefore, every interval of \(A\) is again partitioned. If every interval \([t_i ,\, t_{i+1}]\) of the partition \(A\) has a length which does not exceed a positive number \(\tau ,\) \(t_{i+1} - t_i \le \tau ,\) then \(A\) is called \(\tau\)-partition of the interval \([a,\,b]\) and is designated by \(A_\tau .\)
We call \[S(A,\,x(t)) = \sum_{i=0}^{n-1} x(t_i )(t_{i+1} - t_i ) \tag{1}\] an integral sum of the abstract function \(x(t)\) with \(x\in E\) and \(t\in [a,\,b]\) for the partition \(A = A[t_0 ,\, t_1 ,\, \cdots ,\, t_n ].\)
Theorem 1. Let \(x(t)\) be a continuous function in a complete space \(E\) with \(t\in [a,\,b]\) and \(A_{r_n}\) a sequence of partitions of the interval \([0,\,1]\) such that \(\tau_n \to 0\) as \(n \to \infty .\) If we now form the integral sums \(S(A_{\tau_n} ,\, x(t)),\) then they tend to a limit \(S\) as \(n\to\infty .\) The limit \(S\) does not depend on the choice of the \(A_{\tau_n} .\)
The proof of the theorem is based on the following three lemmas.
Lemma 1. Every continuous function \(x(t)\in E\) with \(t\in [a,\,b]\) is uniformly continuous, i.e., for every positive number \(\epsilon\) we can determine a number \(\tau\) such that for \(t ' ,\) \(t ' ' \in [a,\,b],\) \[\lVert x(t ' ) - x(t ' ' ) \rVert < \epsilon , \tag{2}\] whenever \(\lvert t ' - t ' ' \rvert < \tau.\)
Proof
We use \(\varphi (t,\, t ' )\) to designate the function which is defined on the square \(a \le t \le b,\) \(a \le t ' \le b,\) by \[\varphi (t,\, t ' ) = \lVert x(t) - x(t ' ) \rVert .\] The function \(\varphi\) is continuous on this square, therefore uniformly continuous. Consequently, we can determine for every \(\epsilon > 0,\) a number \(\tau\) such that \[\lvert \varphi ( t,\, t ' ' ) - \varphi ( t ' ,\, t ' ' ) \rvert < \epsilon \,\, \text{for} \,\, \lvert t - t ' \rvert < \tau \tag{3}\] is fulfilled.
Furthermore, for \(t ' = t,\) \(\varphi (t,\,t) = \lVert x(t) - x(t) \rVert =0.\) Hence according to (3) \[\varphi (t,\,t ' ) = \lvert \varphi (t,\,t ' ) - \varphi (t ,\,t) \rvert < \epsilon \,\, \text{for} \,\, \lvert t-t ' \rvert < \tau\] or \[\lVert x(t ' ) -x(t) \rVert < \epsilon \,\, \text{for} \,\, \lvert t-t ' \rvert < \tau . \tag*{\(\blacksquare\)}\]
Lemma 2. If the partition \(W\) of the closed interval \([a,\,b]\) is a refinement of the \(\tau\)-partition \(V = V_\tau\) then \[\lVert S(V,\,x(t) - S(W,\,x(t)) \rVert \le \epsilon (b-a).\tag{4}\]
Proof
If the point \(t ' \) belongs to the interval \([t_i ,\, t_{i+1}]\) of the \(\tau\)-partition \(V_\tau ,\) then \[\lvert t ' - t_i \rvert \le t_{i+1} - t_i \le \tau .\] Hence, on account of (2), \[\lVert x(t ' ) - x(t_i ) \rVert \le \epsilon .\tag{5}\] Let \(n,\) \(m\) be the number of intervals \([t_i ,\, t_{i+1}],\) \(t_j ' ,\, t_{j+1} ' ]\) of \(V,\) \(W,\) respectively. Since \(W\) is a refinement of \(V,\) each of the points \(t_\ell ,\) \(\ell = 1,\) \(2,\) \(\cdots,\) \(n\) coincides with one of the points \(t_j ' : t_{k_{\ell}} ' = t_{\ell} ;\) We have, therefore, the ordering \[0 = k_0 < k_1 < \cdots < k_n = m ,\,\, m \ge n.\] Therefore, the interval \([t_\ell ,\, t_{\ell +1}]\) of \(V\) is partitioned into \(k_{\ell +1} - k_{\ell}\) intervals \([t_j ' ,\, t_{j+1} ']\) of partition \(W,\) where \(j = k_{\ell} ,\) \(k_\ell +1 ,\) \(\cdots ,\) \(k_{\ell +1} -1 .\) According to (5), we have for every \(j\) \[\lVert x(t_j ' ) - x(t_i ) \rVert \le \epsilon .\] Therefore \[\begin{align} S(V,\,x(t)) &= \sum_{\ell=0}^{n-1} x(t_\ell )(t_{\ell +1} - t_\ell ) \\[6pt] &= \sum_{\ell =0}^{n-1} x(t_\ell ) \sum_{j=k_\ell}^{k_{\ell +1} -1} (t_{j+1} ' - t_j ' ),\\[8pt] S(W,\,x(t)) &= \sum_{j=0}^{m-1} x(t_j ' )(t_{j+1} ' - t_j ' ) \\[6pt] &= \sum_{\ell =0}^{n-1} \sum_{j=k_\ell}^{k_{\ell +1} -1} x(t_j ' ) (t_{j+1} ' - t_j ' ),\\[8pt] \lVert S(V,\,x(t)) - S(W,\,x(t)) \rVert &=\left\lVert \sum_{\ell =0}^{n-1} \sum_{j=k_\ell}^{k_{\ell +1} -1} [x(t_\ell ) - x(t_j ' )] (t_{j+1} ' - t_j ' ) \right\rVert \\[6pt] &\le \sum_{\ell =0}^{n-1} \sum_{j=k_\ell}^{k_{\ell +1} -1} \lVert x(t_\ell ) - x(t_j ' ) \rVert (t_{j+1} ' - t_j ' ) \\[6pt] &\le \sum_{\ell =0}^{n-1} \sum_{j=k_\ell}^{k_{\ell +1} -1} \epsilon (t_{j+1} ' - t_j ' ) \\[6pt] &= \epsilon (b-a) .\tag*{\(\blacksquare\)} \end{align}\]
Lemma 3. Let \(V_\tau\) and \(V_{\tau '}\) be arbitrary \(\tau ,\) \(\tau ' \) partitions, respectively, of the closed interval \([a,\,b].\) Then \[\lVert S(V_\tau ,\, x(t)) - S(V_{\tau '} ,\, x(t)) \rVert \le (\epsilon_\tau + \epsilon_{\tau '})(b-a) .\tag{6}\]
Proof
We can always select a partition \(W\) of the interval \([a,\,b]\) which is a refinement of both \(V_\tau\) and \(V_{\tau '}.\) Then, according to (4), \[\lVert S(V_\tau ,\, x(t)) - S(W,\, x(t)) \rVert \le \epsilon_\tau (b-a),\\[6pt] \lVert S(V_{\tau '} ,\, x(t)) - S(W,\, x(t)) \rVert \le \epsilon_{\tau '} (b-a),\] from which \[\lVert S(V_\tau ,\, x(t)) - S(V_{\tau '} ,\, x(t)) \rVert \le (\epsilon_\tau + \epsilon_{\tau '} ) (b-a).\tag*{\(\blacksquare\)}\]
Now we prove the theorem 1. For a given \(\epsilon ,\) we consider a sequence of \(\tau_n\)-partitions \(V_{\tau_n}\) of the interval \([a,\,b],\) where \(\tau_n \to 0\) as \(n\to\infty .\) Because of Lemma 3, the inequality \[\lVert S(V_{\tau_n} ,\, x(t)) - S(V_{\tau_{n+1}} ,\, x(t)) \lVert \le (\epsilon_{\tau_n} + \epsilon_{\tau_{n+1}} )(b-a)\] holds for the corresponding integral sums \(S(V_{\tau_n} ,\, x(t)).\)
Since it holds for every \(\epsilon,\) the sequence \(S(V_{\tau_n},\, x(t))\) is a fundamental sequence and because \(S(V_{\tau_n} ,\, x(t)) \in E\) and \(E\) is complete, this sequence has a limit \(S\) in \(E.\)
Now let \(V_{\tau_n}\) be another sequence of partitions of the interval \([a,\,b]\) where \(\tau ' \to 0.\) Using the preceding argument as a basis, \(S(V_{\tau_n '},\,x(t))\) converges to a limit \(S_1\) as \(n\to\infty.\) We shall prove that \(S = S_1 .\)
To do this, we consider the sequence \(V_{\tau_1} ,\) \(V_{\tau_1 '},\) \(V_{\tau_2} ,\) \(V_{\tau_2 '} ,\) \(\cdots .\) The integral sums \(S(V_{\tau_n} ,\, x(t)),\) \(S(V_{\tau_n '} ,\, x(t)) ,\) respectively, corresponding to these sequences, form a convergent sequences. If their limits are equal to \(S_2 ,\) then they must be equal to the limits \(S\) and \(S_1 .\) Consequently, \(S = S_1 = S_2 ,\) and the theorem is proved.
We call \(S\) the integral of \(x(t)\) over \([a,\,b]\) and designate it by \[\int_a^b x(t) dt.\]
Properties of the Integral
Based on the definition of the integral, the following properties are evident.
[1] Integration is additive: \[\int_a^b [x(t) + y(t)] dt = \int_a^b x(t) dt + \int_a^b y(t)dt.\] [2] If \(\lambda\) is a fixed scalar then \[\int_a^b \lambda x(t) dt = \lambda \int_a^b x(t) dt.\] [3] We also have \[\left\lVert \int_a^b x(t) dt \right\rVert \le \int_a^b \lVert x(t) \rVert dt .\tag{7}\] For, \[\begin{align} \lVert S(V_\tau ,\, x(t)) \rVert &= \left\lVert \sum_{i=0}^{n-1} x(t_i )(t_{i+1} - t_i ) \right\rVert \\[6pt] &\le \sum_{i=0}^{n-1} \lVert x(t) \rVert (t_{i+1} - t_i ) \\[6pt] &= S(V_\tau ,\, \lVert x(t) \rVert ). \tag{8} \end{align}\] As \(\tau \to 0,\) the integral sums tend to the integrals \[\int_a^b x(t) dt \,\,\,\text{and}\,\,\, \int_a^b \lVert x(t) \rVert dt,\] and the inequality (8) goes over to (7).
[4] If there is defined a right-sided(or left-sided) multiplication for the elements \(x\in E\) with elements \(y\in E_1 ,\) then \[\int_a^b x(t) y dt = \int_a^b x(t) dt \,y.\tag{9}\] For, \[\begin{align} S(V_\tau ,\, x(t)y ) &= \sum_{i=0}^{n-1} x(t_i ) y(t_{i+1} - t_i ) \\[6pt] &=\left(\sum_{i=0}^{n-1} x(t_i )(t_{i+1} - t_i ) \right) y \\[6pt] &= S(V_\tau ,\, x(t)) y \tag{10} \end{align}\] holds for a \(\tau\)-partition \(V_\tau (t_0 ,\, t_1 ,\, \cdots ,\, t_n ).\) As \(\tau \to 0,\) equation (10) goes over to (9).
Analogously, for left-sided multiplication, \[\int_a^b yx(t)dt = y\int_a^b x(t)dt \tag{9\( ' \)}\] holds.
Example 1. If \(A\) is a linear operator mapping \(E\) into \(E_1 ,\) and \(x(t)\in E ,\) then \[\int_a^b Ax(t)dt = A \int_a^b x(t)dt.\]
Example 2. If \(A = A(t)\) is an operator from \((E \to E_1 ),\) which depends continuously on \(t,\) and if \(x\in E,\) then \[\int_a^b A(t) x dt = \left( \int_a^b A(t) dt \right) x.\]
In the first example, the operator \(A\) is a constant left-sided factor and in the second example, \(x\) is a constant right-sided factor.
If, in particular, \(f\) is a linear functional from \(\overline{E} ,\) then \[f\left( \int_a^b x(t) dt \right) = \int_a^b f(x(t)) dt,\\[6pt] \int_a^b f(t) (x) dt = \left( \int_a^b f(t) dt \right) (x).\tag{10\( ' \)}\]
[5] If the function \(x=x(t),\) \(x\in E,\) \(t\in [a,\,b]\) has a continuous derivative with respect to \(t,\) \(x ' (t) = \frac{d}{dt} x(t) ,\) then \[\int_a^b x ' (t) dt = x(b) - x(a) .\tag{11}\] For, according to (10'), we obtain, for an arbitrary linear functional \(L,\) \[L \left( \int_a^b x ' (t) dt \right) = \int_a^b L [x ' (t)] dt = \int_a^b \frac{d}{dt} \left\{ L[x(t)] \right\} dt.\] \([x(t)]\) is, however, a function of \(t\) whose range consists of numbers. Therefore: \[\int_a^b \left\{ L [x(t)] \right\} ' dt = L[x(b)] - L[x(a)].\] Hence, \[L \left( \int_a^b x ' (t) dt \right) = L [x(b) - x(a)].\tag{12}\] The equation (12) holds for an arbitrary linear functional \(L \in \overline{E}.\) With this, (11) is proved.
Applications to Differential Equations
We consider the differential equation \[\frac{dx}{dt} = f(t,\,x).\tag{13}\] \(x = x(t)\) and \(f(t,\,x)\) are elements form a normed space \(E;\) let \(t\in [a,\,b].\) We assume that \(f(t,\,x)\) is continuous in \(t\) and that, with respect to \(x,\) the Lipschitz condition \[\lVert f(t,\,x_1 ) - f(t,\,x_2 ) \lVert \le M \lVert x_1 - x_2 \rVert \tag{14}\] is satisfied.
We designate by \(C^E [a,\,b]\) the space of continuous functions \(x(t)\) with \(t\in [a,\,b]\) and \(x(t)\in E.\) We introduce a norm in \(C^E [a,\,b]\) by \[\lVert x \rVert_C = \max_{a\le t\le b} \lVert x(t) \rVert .\] Besides \(E,\) \(C^E [a,\,b]\) is a complete space. We prove this assertion in the same way as that for the special case \(E=\mathbb{R}\) and \(C^E [a,\,b] = C[0,\,1].\)
Besides the equality (13), we consider \[x(t) = x_0 + \int_{t_0}^t f(t,\,x(t)) dt,\,\, a\le t_0 \le t \le t_0 + \delta \le b. \tag{15}\] We designate the right side of this equality by \(A(x).\) \(A\) is an operator with transforms \(x=x(t)\)\(\in C^E [t_0 ,\, t_0 + \delta ]\) into an element of the same space. We have \[\begin{align} \lVert A(x) - A(y) \rVert_C &= \left\lVert \int_{t_0}^t [f(t,\,x(t)) - f(t,\,y(t))]dt \right\rVert_C \\[6pt] &\le \int_{t_0}^{t_0 + \delta} \lVert f(t,\,x(t)) - f(t,\,y(t)) \rVert dt. \end{align}\] It follows, according to (14), that \[\begin{align} \lVert A(x) - A(y) \rVert_C &\le M\int_{t_0}^{t_0 + \delta} \lVert x(t) - y(t) \rVert dt\\[6pt] &\le M \delta \max_{t_0 \le t \le t_0 + \delta} \lVert x(t) - y(t) \rVert \\[6pt] &=M \delta \lVert x(t) - y(t) \rVert_C . \tag{16} \end{align}\] If \(M\delta < 1,\) then the operator \(A\) determines, according to (16), a contracting mapping of the space \(C^E [a,\,b]\) into itself and consequently there exists exactly one solution of (15).
The equation (15) is equivalent to (13) for the initial value \(x(t_0 ) - x_0 .\) Therefore (13) has exactly one solution in the interval \([t_0 ,\, t_0 + \delta ].\)
In particular, this equation has for an arbitrary initial value \(x(a) = x_0\) exactly one solution \(x(t)\) on the interval \([a,\,a+\delta ].\) \(x(t)\) can be continued to the whole interval \([a,\,b].\) For, if \(a+\delta < b\) and \(x(a+\delta ) = x_1 ,\) then we construct by repetition of this process a solution on the interval \([a+\delta ,\, a+2 \delta ]\) with the initial value \(x_1 ,\) etc.
Example 3. If \(E\) is the \(n\)-dimensional space, then we obtain the well-known existence theorem for a system of \(n\) differential equations.
Example 4. If \(E\) is one of the spaces \(l_p ,\) \(c,\) \(m,\) etc., then we obtain the existence theorem for a solution of the corresponding classes of infinite systems of differential equations.
Example 5. We consider the integro-differential equation \[\frac{dy}{dt} = f(t,\,y) + \int_a^b K(t,\,s) y(s) ds. \tag{17}\] Let \(f(t,\,y)\) satisfy the Lipschitz condition (14) and the kernel \(K(t,\,s)\) be bounded, i.e., \(\lvert K(t,\,s) \rvert < M_1 .\)
If the right side (17) is designated by \(F(t,\,y)\) we obtain the inequality \[\begin{align} \lvert F(t,\,y) - F(t,\,z) \rvert &\le M \lvert y-z \rvert + M_1 (b-a) \lvert y-z \rvert \\[6pt] &= (M+M_1 (b-a)) \lvert y-z \rvert . \end{align}\] \(F(t,\,y)\) satisfies the Lipschitz condition. Consequently, (17) has a unique solution for an arbitrary initial value \(y=y_0 .\)
Reference
- L. A. Liusternik ad V. J. Sobolev, 『Elements of Functional Analysis』 (Translated by Anthony E. Labarre), Frederick Ungar Publishing Company(New York), 173-178.