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:

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 $C$).