Square Integrable Functions, Norms, and the Trial Method

This is Part II of the Integral Equations series. If you haven’t read Part I on definitions and types, start there.

Here we cover the function spaces that integral equation theory lives in: square integrable functions (\( \mathfrak{L}_2 \) spaces), inner products, norms, and the key inequalities. Then we solve an integral equation using the trial method, step by step.

This article is adapted from my book Elements of Integral Equations, which covers the complete theory with solved examples, exercises, and proofs. If you’re studying integral equations for exams or research, the book gives you everything in one place.

Square Integrable Functions (\( \mathfrak{L}_2 \) Functions)

A function \( y(x) \) is said to be square integrable (or \( \mathfrak{L}_2 \)) on the interval \( (a,b) \) if

$$ \int_a^b |y(x)|^2 \, dx < \infty $$

or equivalently,

$$ \int_a^b y(x) \bar{y}(x) \, dx < \infty $$

Such a function is also called a regular function. The space of all square integrable functions on \( (a,b) \) is denoted \( L^2(a,b) \) and forms a Hilbert space, the natural setting for integral equation theory.

The kernel \( K(x,t) \), a function of two variables, is an \( \mathfrak{L}_2 \)-function if at least one of the following holds:

  • \( \int_{x=a}^b \int_{t=a}^b |K(x,t)|^2 \, dx \, dt < \infty \)
  • \( \int_{t=a}^b |K(x,t)|^2 \, dx < \infty \)
  • \( \int_{x=a}^b |K(x,t)|^2 \, dt < \infty \)

Inner Product of \( \mathfrak{L}_2 \) Functions

The inner product (or scalar product) \( (\phi, \psi) \) of two complex \( \mathfrak{L}_2 \) functions \( \phi \) and \( \psi \) of a real variable \( x \) on \( [a, b] \) is defined as

$$ (\phi, \psi) = \int_a^b \phi(x) \bar{\psi}(x) \, dx $$

where \( \bar{\psi}(x) \) is the complex conjugate of \( \psi(x) \).

When \( (\phi, \psi) = 0 \), the functions \( \phi \) and \( \psi \) are called orthogonal to each other. Orthogonality is central to the spectral theory of integral equations, where eigenfunctions of symmetric kernels form orthogonal systems.

Norm of a Function

The norm of a complex function \( y(x) \) on \( [a,b] \) is

$$ \| y(x) \| = \sqrt{\int_a^b |y(x)|^2 \, dx} $$

The norm measures the “size” of a function in \( L^2 \). A function with \( \|y\| = 0 \) is the zero function (almost everywhere). Two fundamental inequalities govern how norms interact:

Schwarz’s Inequality: \( |(\phi, \psi)| \le \|\phi\| \cdot \|\psi\| \)

Triangle Inequality (Minkowski): \( \| \phi + \psi \| \le \|\phi\| + \|\psi\| \)

These inequalities are not just theoretical niceties. They’re the tools you need to prove convergence of iterative methods for solving integral equations, which we’ll cover in later parts of this series.

Solution of Integral Equations by the Trial Method

The trial method is the simplest approach to solving an integral equation: you’re given (or guess) a candidate solution and verify it satisfies the equation. It’s not a general-purpose solver, but it builds the intuition you need before tackling iterative and resolvent kernel methods.

Worked Example: Volterra Equation of the Second Kind

Problem. Show that \( y(x) = (1+x^2)^{-3/2} \) is a solution of

$$ y(x) = \frac{1}{1+x^2} – \int_0^x \frac{t}{1+x^2} \, y(t) \, dt \quad \ldots (1) $$

This is a Volterra equation of the second kind with lower limit \( a = 0 \) and upper limit the variable \( x \).

Solution. We have \( y(x) = (1+x^2)^{-3/2} \), so \( y(t) = (1+t^2)^{-3/2} \) by replacing \( x \) with \( t \).

Substituting into the RHS of (1):

$$ \text{RHS} = \frac{1}{1+x^2} – \int_0^x \frac{t}{1+x^2} (1+t^2)^{-3/2} \, dt $$

Since \( \frac{1}{1+x^2} \) is independent of \( t \), it factors out:

$$ = \frac{1}{1+x^2} – \frac{1}{1+x^2} \int_0^x \frac{t}{(1+t^2)^{3/2}} \, dt $$

Evaluating the integral (substituting \( u = 1 + t^2 \)):

$$ \int_0^x \frac{t}{(1+t^2)^{3/2}} \, dt = 1 – \frac{1}{\sqrt{1+x^2}} $$

Substituting back:

$$ \text{RHS} = \frac{1}{1+x^2} + \frac{1}{1+x^2}\left(\frac{1}{\sqrt{1+x^2}} – 1\right) = \frac{1}{(1+x^2)^{3/2}} = y(x) $$

Hence \( y(x) = (1+x^2)^{-3/2} \) is a solution of (1). \( \square \)

Continue the Series

This article is Part II. The complete integral equations series:

For the complete theory with all solved examples, exercises, iterative methods, resolvent kernels, and Fredholm theory, get Elements of Integral Equations. I wrote this book to cover everything a student needs for exams and research in one well-organized reference.

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