I am a physicist and I am aware that this forum is for professional mathematical questions, but please be not too hard on my notation.

I encountered the following integral equation for functions $f:[0,\infty) \rightarrow [0,\infty)$ that is

$$f(x) = \frac{C}{(1+x)} \int_{0}^{\infty} \frac{f(y)}{1+x+y} \ dy$$

and I would like to show that there is at most one solution up to a constant factor (frankly I do not care so much about existence, but the uniqueness is important to me).

Now I also found a pedestrian method to verify this by considering the map

$$(Tf)(x) = \frac{C}{(1+x)} \int_{0}^{\infty} \frac{f(y)}{1+x+y} \ dy$$

Then, one can show using the $L^1$ norm that

$$\left\lVert Tf \right\rVert_{L^1} \le C \int_0^{\infty} \frac{1}{(1+x)^2} \ dx \left\lVert f \right\rVert_{L^1}= C\left\lVert f \right\rVert_{L^1}.$$

Now, if $C<1$ I'm all good, as the Banach fixed point theorem gives existence and uniqueness. But what if $C\ge 1.$

Does the uniqueness really fail or is there a trick to extend the uniqueness also into the regime $C \ge 1$?

I would like to motivate my question by the fact that also when proving the Picard-Lindelöf theorem, one can remove the restriction to short time-intervals by using an equivalent norm

see here

So perhaps an equivalent norm does the trick in this case as well?


If you see a different space where uniqueness works, please also let me know as $L^1$ was just a natural choice given the structure of the equation.

  • $\begingroup$ Just a remark (although you noted that you do not really care about existence): the inequality $\|Tf\|_1 \le \|f\|_1$ implies that the operator $T$ has spectral radius at most $C$, so if $C < 1$, then there is no non-zero solution at all. [Of course, this is also a consequence of the Banach fixed point theorem: the theorem tells you that there is exactly one solution (not merely a solution up to scalar multiples); since zero is obviously a solution, there cannot be a non-zero solution.] $\endgroup$ Jul 22, 2018 at 8:00
  • $\begingroup$ By the way, for $C = 1$ the space of solutions is indeed either $\{0\}$ or it is one-dimensional; this follows from infinite-dimensional Perron-Frobenius theory. However, I'm not sure right now about the case $C > 1$. $\endgroup$ Jul 22, 2018 at 8:04
  • $\begingroup$ @JochenGlueck do you have a reference for the infinite-dimensional Frobenius theorem you have in mind? For which class of operators does it apply? $\endgroup$ Jul 22, 2018 at 16:53
  • $\begingroup$ $f$ seems to be infinitely differentiable and totally monotone in sense of this (en.wikipedia.org/wiki/…) definition, so it has a lot of good qualities. Not sure if it helps. $\endgroup$
    – Kaban-5
    Jul 22, 2018 at 17:21
  • 1
    $\begingroup$ @AndreaTauber: Sure; please see my answer below which contains a reference. $\endgroup$ Jul 22, 2018 at 21:49

3 Answers 3


The answer to the question is as follows:

Theorem 1. For every $C > 0$ there is, up to scalar multiples, only one function $0 \le f \in L^1 := L^1((0,\infty))$ such that $Tf = f$ (where $T$ is the $C$-dependent operator from the question).

For the proof we need a few preparations.

First we quote the following result (which is actually true for all Banach lattices with order continuous norm, but we restrict ourselves to $L^p$-spaces here); in what follows, all inequalites are to be understood almost everywhere.

Theorem 2. Let $(\Omega,\mu)$ be an arbitrary measure space and let $p \in [1,\infty)$. Let $0 \le u \in L^p := L^p(\Omega,\mu)$ and suppose that $S$ is a bounded linear operator on $L^p$ such that for each $f \in L^p$ there exists a number $c \ge 0$ for which we have $\lvert Sf\rvert \le c u$. Then:

  • If $p \in (1,\infty)$ the operator $T$ is compact.

  • If $p=1$ the operator $T^2$ is compact.

Theorem 2 is a consequence of standard results about Dunford-Pettis operators on Banach lattices which can for instance be found in [P. Meyer-Nieberg: Banach Lattices (1991)]; however, I don't know an explicit reference for Theorem 2 itself in a classical monograph. An explicit reference for Theorem 2 in a recent article (including a proof and detailed references) is [D. Daners, J. Glück: The Role of Domination and Smoothing Conditions in the Theory of Eventually Positive Semigroups, Theorem 2.3] (where the result is formulated in the more general setting of Banach lattices).

As a consequence of the general Theorem 2 we obtain the following result for the operator $T$ in our concrete situation:

Proposition 3. The operator $T^2$ is compact from $L^1$ to $L^1$. In particular, the spectrum of $T$ is at most countable, and all spectral valued of $T$, except for possibly the number $0$, are eigenvalues of finite algebraic multiplicity and poles of the resolvent of $T$.

Proof. For every $f \in L^1$ we have $\lvert Tf\vert \le T\lvert f \vert \le C \lVert f\rVert_{L^1} u$, where $u \in L^1$ is the function given by $u(x) = 1/(1+x)^2$. Thus, the assertion follows from Theorem 2.

Remark. Mateusz Kwaśnicki noted in his answer that $T$ is compact (even a Hilbert-Schmidt operator) from $L^2$ to $L^2$, but Proposition 3 above deals with the operator on $L^1$.

Now we are going to use infinite dimensional Perron-Frobenius theory. We first recall a few notions for readers not familiar with this theory:

  • Let $(\Omega,\mu)$ be a $\sigma$-finite measure space, let $p \in [1,\infty)$; we set $L^p := L^p(\Omega,\mu)$.

  • We call a function $f \in L^p$ positive and denote this by $f \ge 0$ if $f(\omega) \ge 0$ for almost all $\omega \in \Omega$.

  • Let $S$ be a bounded linear operator on $L^p$.

  • The operator $S$ is called positive if $Sf \ge 0$ for all $f \ge 0$.

  • The operator $S$ is called reducible if there exists a measurable set $M \subseteq \Omega$ such that both $M$ and $\Omega \setminus M$ have non-zero measure and such that $Sf$ vanishes almost everywhere on $M$ whenever $f$ vanishes almost everywhere on $M$.

  • The operator $S$ is called irreducible if it is not reducible.

Remarks. (a) All the above notions can also be defined in the more general context of Banach lattices, and the subsequent theorem is also true on Banach lattices; however, I think this post might be more accessible if we restrict our attention to $L^p$-spaces).

(b) In particular, one can also define reducible and irreducible operators on $L^p$ if $(\Omega,\mu)$ is not $\sigma$-finite (since $L^p$ is still a Banach lattice in this case); however, the above way to define those notions is not appropriate in this setting; instead one has to use more abstract notions from Banach lattice theory to get a reasonable theory of irreducible operators on $L^p$-spaces over non-$\sigma$-finite measure spaces.

Now we quote the following result which belongs to what is usually called infinite-dimensional Perron--Frobenis theory:

Theorem 4. Let $(\Omega,\mu)$ be a $\sigma$-finite measure space and let $S$ be a positive and irreducible bounded linear operator on $L^p := L^p(\Omega,\mu)$ of spectral radius $r(S) = 1$. If some power of $S$ is compact, then the following assertions hold:

(a) The fixed space $\ker(1-S)$ is one-dimensional.

(b) The number $1$ is the only eigenvalue of $S$ with a positive eigenvector.

This result can, for instance, be found in [H. H. Schaefer: Banach Lattices and Positive Operators (1974), Theorem V.5.2 and part (ii) of this theorem's corollary].

Now we can prove Theorem 1:

Proof of Theorem 1. First we note that the operator $T$ is irreducible since we have $Tf(x) > 0$ for almost all $x \in [0,\infty)$ whenever $0 \not= f \ge 0$.

Let $r(T) \in [0,\infty)$ denote the spectral radius of $T$. We distinguish between three case:

First case: $r(T) < 1$. In this case we obviously have $\ker(1 - T) = \{0\}$.

Second case: $r(T) = 1$. In this case the space $\ker(1-T)$ is one-dimensional according to Theorem 4(a).

Third case: $r(T) > 1$. By applying Theorem 4(b) to the operator $S := T/r(T)$ we can see that $r(T)$ is the only eigenvalue of $T$ with a positive eigenvector. Hence, $\ker(1-T)$ does not contain any non-zero functions $f$ which fulfil $f \ge 0$. This completes the proof.

Remark. Actually, I'm not sure whether the spectral radius of $T$ is really non-trivial or whether we have $r(T) = 0$ (which is true for some $C$ if and only if it is true for all $C$).

  • $\begingroup$ this is really very nice and a beautiful outline of that theory. I am impressed (and was not aware at all of this) $\endgroup$ Jul 22, 2018 at 22:17
  • $\begingroup$ @AndreaTauber: Thank you. It seems that, simultaneously to my answer, Mateusz Kwaśnicki has updated his answer and proved the same result with an argument which also belongs to the realm of Perron-Frobenius theory, but is a bit more concrete. $\endgroup$ Jul 22, 2018 at 22:23
  • $\begingroup$ can I asked you a brief question? In the theorem [H. H. Schaefer: Banach Lattices and Positive Operators (1974), Theorem V.5.2] you mentioned, there is the implication in the proof of the first part that if all (non-zero) elements in $F_{+}$ are quasi-interior points then the underlying vector lattice is simple. I am not too familiar with the definitions, but do you think you could explain to me briefly how that follows? $\endgroup$ Jul 23, 2018 at 16:09
  • $\begingroup$ @AndreaTauber: I'm not sure how Schaefer does it (I don't have his book at hand right now), but here is an argument which I found in an article of Lotz (who was a student of Schaefer): Let $x, y \in F$. The positive vectors $(x-y)^+$ and $(x-y)^-$ are disjoint (i.e. they have infimum zero), so they cannot both be quasi-interior points of $F_+$. By assumption on $F_+$ it follows that at least one of the vectors is $0$, so $x \le y$ or $y \le x$. Hence, the vector lattice $F$ is totally ordered. [to be continued] $\endgroup$ Jul 23, 2018 at 19:49
  • $\begingroup$ @AndreaTauber: [continuation] I'm not sure what Schaefer means by "simple", but it's not difficult to show that a totally ordered Archimedean vector latttice is one-dimensional; thus $F$ is one-dimensional, and that's what you need for the proof of the theorem. (Please feel free to ask if you have further questions). $\endgroup$ Jul 23, 2018 at 19:53

This is an extended comment rather than an answer. EDIT: scroll to the bottom for the actual answer.

The kernel $$ K(x, y) = \frac{1}{1 + x} \, \frac{1}{1 + x + y} $$ satisfies $$ \int_0^\infty \int_0^\infty (K(x, y))^2 dx dy < \infty $$ (the integral evaluates to $\tfrac{1}{2}$), and so the operator $$ K f(x) = \int_0^\infty K(x, y) f(y) dy $$ is a Hilbert–Schmidt operator on $L^2((0, \infty))$. In particular, $K$ is a compact operator, the spectrum of $K$ is purely discrete, and the equation $f = C K f$ has a non-zero solution for at least countably many $C$.

If $f \in L^1((0, \infty))$, then $K f \in L^1((0, \infty)) \cap L^\infty((0, \infty))$, and hence $K f \in L^2((0, \infty))$. Therefore, any $L^1((0, \infty))$ solution of $f = K f$ is automatically an $L^2((0, \infty))$ solution. Therefore, your equation has a non-zero solution for at most countably many $C$. Additionally, for every $C \ne 0$ the space of solutions is finitely dimensional.

Your question is thus equivalent to non-degeneracy of the eigenvalues of $K$. I do not know any general method to tackle this kind of problems, and I believe it is often rather difficult. But then, I am not an expert in the area.

EDIT: I just noticed that $f$ is required to take non-negative values only. Then everything is much simpler: due to Perron–Frobenius theorem, there is only one non-negative eigenfunction (up to multiplication by a constant, of course). The first reference suggested by Google is this one, but I am sure there are better ones.

However, due to special character of the kernel of $K$, there is no need to refer to general theory, one can simply repeat the relevant parts of the proof of Perron–Frobenius theorem for self-adjoint operators (even though $K$ is not self-adjoint).

Suppose $f, g \geqslant 0$ are non-zero solutions: $f = C K f$, $g = D K f$ for some constants $C, D > 0$. Since the kernel of $K$ is strictly positive, we have $f, g > 0$ on $[0, \infty)$. Note that $$ \frac{1}{D} \int_0^\infty (1 + x) f(x) g(x) dy = \int_0^\infty \int_0^\infty \frac{f(x) g(y)}{1 + x + y} \, dx dy = \frac{1}{C} \int_0^\infty (1 + y) f(y) g(y) dy . $$ (That is, $T$ is self-adjoint on a weighted space $L^2((0, \infty), (1 + x) dx)$). If $C \ne D$, we necessarily have $\int_0^\infty (1 + x) f(x) g(x) dx = 0$, which is impossible. Therefore, $C = D$. It remains to prove that $f / g$ is constant.

First of all, by monotone convergence, $$\lim_{x \to \infty} (x^2 f(x)) = C \lim_{x \to \infty} \int_0^\infty \frac{x^2 f(y)}{(1 + x) (1 + x + y)} \, dy = C \|f\|_1$$ exists and it is positive. Similarly, $\lim_{x \to \infty} (x^2 g(x)) = C \|g\|_1 > 0$. Furthermore, by dominated convergence, $f$ and $g$ are continuous on $[0, \infty)$. It follows that $$ A := \inf \left\{\frac{f(x)}{g(x)} : x \geqslant 0\right\} > 0 . $$ Consider $h = f - A g$. Clearly, $h \geqslant 0$ and $h = C K h$, that is, $h$ is the solution of our problem. If $h$ is non-zero, then $h > 0$ on $[0, \infty)$, and $\lim_{x \to \infty} (x^2 h(x)) = \|h\|_1 > 0$, which implies that $$ B := \inf \left\{\frac{h(x)}{g(x)} : x \geqslant 0\right\} > 0 . $$ But this implies hat $f(x) / g(x) = (A g(x) + h(x)) / g(x) \geqslant A + B$, contradicting the definition of $A$. Therefore, $h$ is identically zero, and so $f = A g$, as desired.


Also not an answer, but there is only one solution ($f = 0$) if $C > 2$. First thing to notice is that $f$ is continuous function, because $f(x)$ is product of continuous function $\frac{C}{1 + x}$ and $\int\limits_0^{+\infty} \frac{f(y)}{1 + x + y} ~dy$, which is continuous function of $x$ (easy to check by definition).

Now, $f(x) = \frac{C}{1 + x} \int\limits_0^{+\infty} \frac{f(y)}{1 + x + y} ~dy \geqslant \frac{C}{1 + x} \int\limits_0^1 \frac{f(y)}{1 + x + y} ~dy \geqslant \frac{C}{1 + x} \int\limits_0^1 \frac{f(y)}{2 + x} ~dy \geqslant \frac{C\delta}{(1 + x)(2 + x)} \geqslant \frac{C\delta/2}{(1 + x)^2}$ due to non-negativity of $f$, where $\delta = \int\limits_0^1 f(y) ~dy > 0$, as long as $f \not = 0$ (because $f \not = 0$ implies $f(0) > 0$)

Now suppose that we know that $f(x) \geqslant \frac{k}{(1 + x)^2}$ holds for all $x$ for some $k > 0$. Then $f(x) = \frac{C}{1 + x} \int\limits_0^{+\infty} \frac{f(y)}{1 + x + y} \geqslant \frac{C}{1 + x} \int\limits_0^{+\infty} \frac{k}{(1 + y)^2} \frac{1}{1 + x + y} ~dy \geqslant \frac{C}{1 + x} \int\limits_0^{+\infty} \frac{k}{(1 + y)^2} \frac{1}{1 + x} \frac{1}{1 + y} ~dy = \mbox{$\frac{Ck}{(1 + x)^2} \int\limits_0^{+\infty} \frac{1}{(1 + y)^3} ~dy$} = \frac{Ck}{2 (1 + x)^2}$.

So $f \not = 0$ implies $f(x) > \frac{k}{(1 + x)^2}$, where $k = C\delta/2 > 0$, which implies that $f(x) > \frac{Ck/2}{(1 + x)^2}$, which implies that $f(x) > \frac{(C/2)^2 k}{(1 + x)^2}$, e t. c., which can't happen if $C > 2$.


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