# Integral equations and the Fredholm alternative / theory

The Fredholm alternative states that either: $$0 = \lambda \phi(x) - \int_a^b K(x,y) \phi(y) dy$$ has a non-trivial solution, or: $$f(x) = \lambda \phi(x) - \int_a^b K(x,y) \phi(y) dy$$ always has a unique solution for any f(x)

A sufficient condition is for the kernel K to be square-integrable, but depending on sources there is some confusion whether $$\lambda$$ must be a non-zero complex number. For example, Wikipedia says so but also refers to this page which doesn't...

(Note that $$\lambda=0$$ the integral equation is called Fredholm of the first kind, and $$\lambda \neq 0$$ is of second kind.)

Adding to the confusion, some authors prefer to have $$\lambda$$ scaling the integral rather than $$\phi(x)$$, in which case it makes sense to require it to be nonzero.

Could anyone clarify whether the Fredholm alternative is only true of second kind equations (i.e. $$\lambda \neq 0$$)? And if so, why...?

Thanks!

## 1 Answer

In abstract terms, what is important is that the operator $$\mathcal K\colon L^2(a, b)\to L^2(a, b)$$ defined by $$\mathcal K\phi(x)=\int_a^b K(x, y)\phi(y)\, dy,$$ is compact. Such operators are never surjective. Therefore, for the Fredholm equation of the first kind $$\mathcal K \phi = f$$ there always exists some $$f\in L^2$$ such that the equation has no solution.

CONCRETE EXAMPLE. Let $$K(x, y)=\begin{cases} 1, & x>y, \\ 0, & x\le y, \end{cases}$$ so that $$\newcommand{\Kcal}{\mathcal{K}} \Kcal\phi(x)=\int_{-1}^x\phi(y)\, dy$$. Here $$(a, b)=(-1,1)$$. Notice that, if $$\phi\in L^2(-1,1)$$, then $$\Kcal\phi$$ is a continuous function (actually, $$\phi\in L^1(-1,1)$$ would suffice). Therefore, if $$f$$ is not continuous, then the equation $$\Kcal \phi =f$$ has no solutions.

Let us now add a tiny $$\lambda \ne 0$$ and consider the equation $$\Kcal \phi + \lambda \phi = f.$$ The Fredholm theory guarantees that this equation has a unique solution for all $$f\in L^2(-1, 1)$$. Let us plug in a discontinuous function, such as $$f(x)=H(x)=\begin{cases} 1, & x>0, \\ 0, & x\le 0,\end{cases}$$ ($$H$$ stands for Heaviside), and see what happens. Taking a formal derivative, the equation becomes $$\phi + \lambda \phi' = \delta(x),$$ where $$\delta$$ is the Dirac distribution. Now, $$\phi(x)=e^{-x/\lambda} H(x)$$ does satisfy this equation, because, if we plug it into the left-hand side, we get $$e^{-\frac x \lambda} H(x) -\frac{\lambda}{\lambda} e^{-\frac x \lambda} H(x) +e^{-\frac x \lambda} \delta(x)= \delta(x),$$ where we used the identity $$e^{-\frac x \lambda} \delta(x)= e^{-0}\delta(x)=\delta(x)$$.

(WARNING! This is not 100% rigorous, as we should now verify that $$\phi$$ solves the integral equation).

Summarizing, the presence of that parameter $$\lambda\ne 0$$, no matter how tiny it is, enabled the surjectivity of the operator. If the parameter tends to $$0$$, however, the solution $$\phi$$ we just found ceases to have a meaning, compatibly with the fact that $$\Kcal$$ cannot be surjective.

A REFERENCE. You asked for a reference. The book of Eidelman, Milman and Tsolomitis is one, chapter 5. See pag.82 for a remark on the parameters $$\lambda$$ and $$\mu$$, and the relationship between the classical notation on integral equation and the more abstract notation with the compact operators.

• I will read your link. However I am surprised that if the kernel is "invertible" (no zero eigenvalue in the homogeneous equation of second kind) then, according to your answer, there would still be some target function f in L2 that would have no solution? What is so special about first kind equations that would break the analogy with linear systems of equations? – phaedo Oct 23 '18 at 12:51
• The key is in the link. A compact operator cannot be invertible, because if it were, then the unit ball of $L^2$ would be compact. And this is false. In finite dimension, this difficulty is nonexistent. – Giuseppe Negro Oct 23 '18 at 13:11
• Are you also saying that the operator $K - \lambda I, \lambda \neq 0$ used in second kind equations is surjective (and therefore NOT compact)? – phaedo Oct 23 '18 at 13:31
• Yes, it is not compact. Without loss of generality, let $\lambda=1$, and suppose for a contradiction that $K-I$ were compact. Let $f_n$ be a bounded sequence in $L^2$. Up to passing to a subsequence, $g_n:=Kf_n-f_n$ converges. Since $K$ is also compact, $Kf_n$ converges too, up to subsequences. Therefore, a subsequence of $f_n$ is the difference of the two convergent sequences $Kf_n$ and $g_n$, and so it converges. This is a contradiction in infinite dimension, because it would mean that every bounded sequence has a convergent subsequence. – Giuseppe Negro Oct 23 '18 at 14:28
• Yes, that's the point: with a restriction. There always is a subspace of $L^2$ such that, if $f$ belongs to this subspace, then there is solution to $K\phi=f$. But this subspace is a proper subset of $L^2$ if $K$ is a compact operator. – Giuseppe Negro Oct 23 '18 at 19:06