NOTE: Cross-posted on Mathematics Stack Exchange

I am trying to solve an equation of the form $$ (\mathbb{I} + K)\phi = f $$ where $(\mathbb{I} + K): L^2([0,1];\mathbb{R}) \rightarrow L^2([0,1];\mathbb{R}) $ is a Fredholm integral operator of the second kind, $\mathbb{I}$ is the identity in $L^2([0,1];\mathbb{R})$, and $f$ and $\phi$ are $L^2([0,1];\mathbb{R})$ functions. I am considering, in particular, that the equation I want to solve is $$ \phi(x)+\int_0^1k(x,y)\phi(y) dy = f(x) $$ for a.e. $x\in[0,1]$, with $k\in H^1([0,1]\times[0,1];\mathbb{R})$ (the regularity of the kernel is specific to the problem I am considering but I don't think it hurts).

My question is the following: assuming a solution exists and is unique (i.e. $-1\notin \sigma(K)$), is there any specific structure that we can a priori assume for the inverse operator w.l.o.g. ? For instance, does the solution satisfy an integral equation $$ \phi(x)= f(x) + \int_0^1 l(x,y)f(y) dy $$ for some $l\in H^1([0,1]\times[0,1];\mathbb{R})$ ?

I know that if K were a Volterra integral (with integration limits from $0$ to $x$), one would look for an inverse of the form identity + Volterra integral but I haven't been able to find a similar result spelled out clearly in the case of Fredholm integral equations unless the kernels have some additional structure. I tried to read the original paper by Fredholm "Sur une classe d'équations fonctionnelles" (from Acta Mathematica 27, 1903), yet I am unsure of the sense of convergence for the minors considered in the paper (and thus in what space I should look for my $l$).

Any simpler references to point out (with a more modern notation perhaps) ? I don't want to re-develop what is probably already out there and I want to make sure no counter-examples exist.

  • $\begingroup$ Customarily (and assuming that all the hypotheses for its existence hold), the solution of a Fredholm equation is represented as a Neumann' series: $$\varphi(x) = f(x) + \sum_{p=1}^\infty K^p f(x) $$ where $$Kf(x)=\int_G k(x,y) f(y) \mathrm{d} y$$ and $$K^{p+1} f(x) = Kf(x)=\int_G k(x,y) K^p f(y) \mathrm{d} y\quad p\in\Bbb N $$. $\endgroup$ Mar 3 at 12:00
  • $\begingroup$ @DanieleTampieri Thanks for the reply but, doesn't the representation of the solution using the Neumann series hold only if the operator $K$ has a spectral radius less than 1? As far as I can tell, having a spectral radius less than one is only a sufficient condition for the existence and uniqueness of solutions to the equation, but not necessary (as long as $-1$ does not belong to the spectrum of $K$ the equation should still be solvable). $\endgroup$
    – a1228d
    Mar 4 at 13:32
  • $\begingroup$ Is your K compact? In this case I think the best thing you could do is write k(x,y) as double Fourier series e.g., with terms lambda_j Re{exp_i_pi_j_x)]*exp[(i_pi_j_y)]} and then compute the inverse series for I + K. $\endgroup$
    – Derek
    Mar 7 at 10:18

1 Answer 1


Not really an answer, but some remarks:

  1. Even if the spectral radius of $K$ is less than $1$, there are counterexamples that the resolvent need not have the required form: This is related to the fact that the composition of two integral operators is not an integral operator, in general. A sufficient condition for the latter is that the operator is regular, that is, that the integral operator with kernel $|k|$ acts in $L^2$ as well. In case $k\in L_2([0,1]^2)$ the integral operator is regular (actually Hilbert-Schmidt). In that case, having spectral radius less than $1$ is certainly sufficient.

  2. AFAIK, the general case is still an open problem for regular integral operators. As you observed, the series of Fredholm determinants converges only under additional hypotheses. I think that it is known that it converges in trace norm if the operator is of trace class. I am not sure whether your regularity condition on the kernel is sufficient for the latter.


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