Lectures on Integral Equations

By Harold Widom

This concise and classic volume presents the main results of integral equation theory as consequences of the theory of operators on Banach and Hilbert spaces. In addition, it offers a brief account of Fredholm's original approach. The self-contained treatment requires only some familiarity with elementary real variable theory, including the elements of Lebesgue integration, and is suitable for advanced undergraduates and graduate students of mathematics.
Other material discusses applications to second order linear differential equations, and a final chapter uses Fourier integral techniques to investigate certain singular integral equations of interest for physical applications as well as for their own sake. A helpful index concludes the text.

• Language: English
• Publisher: Dover Publications
• Release date: Nov 28, 2016
• ISBN: 9780486817828

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INTRODUCTION

The aim of this course is to investigate the equation

In (1) we have:

With these data, we look for a function f satisfying (1).

We rewrite (1) in operator notation

where Tf(·)· = ∫X K(·, y) f(y)dy. The theory of linear transformations on vector spaces lends much power and elegance to the study of (1) and (2); hence we restrict ourselves to g in Lp(X) (1 ≤ p ≤ ∞) and seek an f in Lp(X) satisfying (1) and (2). Implicit in this restriction is the use only of kernels K(x, y) for which

is in Lp when f is. A sufficient condition on K is contained in the following.

PROPOSITION 1.Let 1 ≤ p ≤ ∞, and . Define

and assume

Then the integral operator T with kernel K maps Lp into Lp, and moreover ||Tf||p ≤ |||K|||p || f ||p.

Proof.We consider only p ≠ 1, ∞; the cases p = 1, ∞ are easier and are left as exercises. We have by Hölder’s inequality:

Thus Tf (x) = ∫ K(x, y) f(y)dy exists almost everywhere and

Remark: If p = 2, then q = 2 and the condition (3) becomes

The condition on K of Proposition 1 is far from necessary. Before giving a simple example of this, we recall the fact from elementary measure theory that

Granting this, we show that if K(x, y) = k(x − y), with

then K is the kernel of a linear transformation on Lp(−∞, ∞).

By Fubini’s theorem,

Thus Tf(x) = ∫k(x − y) f(y)dy exists, a.e., and ||Tf||p ≤ ||k||1||p, so K is the kernel of a transformation T from Lp to Lp. But unless k = 0, a.e., T does not satisfy the condition of Proposition 1:

PRODUCTS OF OPERATORS: Let T and T′ be integral operators on Lp with kernels K and K′. The operator (TT′) also maps Lp to Lp, where (TT′) f = T(T′ f). If the kernels K and K′ are sufficiently well-behaved to justify an interchange of the order of integration, we may derive a formula for K″, the kernel of TT′, in terms of K and K′. For example, if both K and K′ satisfy the hypothesis of Proposition 1, the following manipulations are justified by Fubini’s theorem.

i.e., TT′ is the operator with kernel K″(x, y) = ∫ K(x, z) K′(z, y)dz.

If we take T′ = T, we see that T² has kernel

in general, the kernel of Tn is

Definition: The Kn(x, y) described above are called the iterated kernels of K(x, y).

In developing the theory of integral equations, we find that many of the techniques and results are also valid for a more general situation, that of linear transformations on complete normed vector spaces (Banach spaces). Conversely, studying linear transformations in this more abstract situation lends much unity to the theory of integral equations. We turn now to this subject.

CHAPTER I

BANACH SPACES

Definition.Let E be a vector space over the real or complex numbers. E is said to be normed if there is a real-valued function

such that

(a)||x|| ≥ 0; ||x|| = 0 if and only if x = 0;

(b)||αx|| = |α| ||x||;

(c)||x + y|| ≤ ||x|| + ||y||.

Remark.Let d(x, y) = ||x − y||. Then (E, d) is a metric space.

Definition.E is a Banach Space if E is also complete under the metric d.

Remark.The topology of E which we will use hereafter is that induced by d.

Let T: E → F; E, F normed spaces. We consider only linear T; i.e., those for which

Definition.T is bounded if there is an M < ∞ such that ||Tx|| ≤ M||x||, all x ∈ E. The least such M for which this is true is called the norm of T, ||T||. Thus, by definition,

THEOREM 1.Let T be linear. Then T is continuous T is bounded.

ProofLet xn → x. Then ||Txn − Tx|| ≤ M||xn − x|| → 0.

If no such M exists, we may find xn, ||xn|| = 1, such that ||Txn|| > n. Then the sequence yn = n−1/2 xn → 0, whereas ||Tyn|| → ∞.

Example.We have seen (Proposition 1 of the Introduction) that if T is the integral operator on Lp with kernel K(x, y), and if |||K|||P < ∞, then T is bounded, and ||T|| ≤ |||K||P.

Definition.L(E, F) is the set of bounded linear transformations from E to F. If T ∈ L(E, F) ≡ L(E), T is called an operator on E.

L(E, F) forms a normed space under pointwise operations. In addition we have the important

THEOREM 2.If F is a Banach space, so is L(E, F).

Proof.Let {Tn} be a Cauchy sequence in L(E, F). Then for each x, ||Tnx − Tmx|| ≤ ||Tn − Tm|| ||x|| → 0. Since F is complete, Tnx converges and we define Tx = lim Tnx. It is easy to see that T is a linear transformation from E to F. We shall show that T ∈ L(E, F) and Tn → T. We have

Now, given ε > 0, there is an n0 such that n, m > n0 implies

Then ||(Tn − Tm)x|| ≤ ½ε ||x|| and (letting n → ∞) ||(T − Tm)x|| ≤ ½ε||x||. Thus, n > n0 implies ||(Tn − T)x|| ≤