Existence of a solution to $f(x) = \int_0^1 k(x,y) f(y) dy$ Let $X = (0,1)\times (0,1)$ with the Lebesgue measure, and $k\colon X \to \mathbb{R}$ be a measurable non-negative function such that
$$ \int_0^1 k(x,y) dy = 1$$
for every $x \in (0,1)$.
My question is:

What (more) should we assume on $k$ to guarantee that there exists a unique (up to equivalence class) function $f \in L^1(0,1)$ such that
  $$f(x) = \int_0^1 k(y,x) f(y) dy$$
  for every $x \in (0,1)$ and $\| f \|_{L^1(0,1)} = 1$.

Additionally, if we can prove that such unique $f$ exists, when it is strictly positive a.e.?
I would be grateful for any reference.

Edit: I realized that it makes no sense to ask for a unique solution, since if there is one solution, then all its scalar multiple are also solutions. Hence, I added the requirement that $f \in L^1(0,1)$ and $\| f \|_{L^1(0,1)} = 1$.
 A: May I change the order of $x$ and $y$ to more natural $K(x,y)=k(y,x)$. So we have 
$$
K(x,y)\ge 0,\quad \int_0^1 K(x,y)\,dx=1.
$$
First some examples.
Example 1: Let $K(x,y)=\delta(x-y)$. Not a function, but it is easy to see what happens here. We have for all continuous $f$
$$
\int_0^1K(x,y)f(y)\,dy=f(x).
$$
No uniqueness.
Example 2: Let's modify the example and take a function now
$$
K(x,y)=\left\{
\begin{array}{l}
2,\quad \text{if}\ (x,y)\in(0,\frac12)^2,\\
2,\quad \text{if}\ (x,y)\in(\frac12,1)^2,\\
0,\quad \text{otherwise}.
\end{array}
\right.
$$
It is straightforward to see that the function
$$
f(x)=\left\{
\begin{array}{l}
C_1,\quad \text{if}\ x\in(0,\frac12),\\
C_2,\quad \text{if}\ x\in(\frac12,1)
\end{array}
\right.
$$
satisfies the equation
$$
f(x)=\int_0^1K(x,y)f(y)\,dy
$$
for all constants $C_1$ and $C_2$. Clearly, no uniqueness again. Take e.g. $C_1=2$, $C_2=0$ and then $C_1=0$, $C_2=2$ to get two non-negative nomalized solutions.
Example 3: Take now the finite dimensional analogue of the problem: a non-negative square matrix $A$ with the condition
$$
e^TA=e^T,\qquad e^T=\left[\matrix{1 &1&\ldots&1}\right]
$$
and look for the solution to $f=Af$ (existence, uniqueness etc). Since the matrix is non-negative we are in the field of Perron-Frobenius theorem that tells us that there exists an eigenvalue $\rho(A)$ and a non-negative eigenvector $f$ such that $\rho(A) f=Af$. Pre-multiplying by $e^T$ we get
$$
\rho(A)e^Tf=e^TAf=e^Tf\ne 0\quad\Rightarrow\quad \rho(A)=1.
$$
So we have existence of a non-negative solution. Uniqueness is not granted by non-negativity of $A$, we need $A$ to be irreducible, then we have the strictly positive vector $f$ and uniqueness.
Conclusion: There is a generalization of Perron-Frobenius theorem which is called Krein-Rutman theorem (see also here) that says approximately the same for a positive compact operator. So you need more conditions on your kernel function to get compactness and irreducibility of the integral operator.
