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 $\| f \|_{L^1(0,1)} = 1$.
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.