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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!

Giuseppe Negro
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phaedo
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1 Answers1

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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.

Giuseppe Negro
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  • 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
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    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
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    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
  • What about Volterra equations of the first kind? If $K(x,y) = 0$ for y>x and $K(x,x)\neq 0$, differentiation shows that they are equivalent to second-kind equations which always have a unique solution. This would seem to contradict your answer? – phaedo Oct 23 '18 at 16:16
  • Additionally, it would seem that a unique solution exists to first-kind equations with injective kernel and the restriction that $ f \in \mathcal K[ L^2(a,B) ] $... – phaedo Oct 23 '18 at 17:30
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    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
  • How about the operator K - λI - not necessarily surjective, even if λ = !0. If K is compact, then K - λI is too, hence not surjective. The Fredholm alternative for second kind integral guarantee solution for all f in L2(a,b), but doesn't guarantee that the solution is in the range of K - λI. The range of K - λI willbe a proper subspace of L2(a,b). The important point is that operator K - λI is still well-defined and bounded, and its range is a closed subspace of L2(a,b). Allowing Hilbert spaces and bounded linear operators to study the properties of the solution to the equation Kϕ + λϕ = f. – Anon Imus Mar 08 '23 at 01:06
  • @AnonImus: You wrote "If $K$ is compact, then $K-\lambda I$ is too". NO. That's exactly the point of my answer: I wrote "no matter how tiny $\lambda$ is, $K-\lambda I$ is surjective", hence it is not compact. – Giuseppe Negro Mar 08 '23 at 10:31