Does there exists three continuous functions $K : [0,1]^2 \mapsto \mathbb{R}_+^*$, $f : [0,1] \mapsto \mathbb{R}_+^*$, such that for all $x \in [0,1]$, $g(x) = \displaystyle{\int_0^1} K(x,y)f(y)dy$, $f(x)=\displaystyle{\int_0^1} K(x,y)g(y)dy$, and $f \neq g$ ?
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If you only assume $K$ to be a real-valued function, it is easy to find counter-examples. I think that with the above hypothesis, the answer is no, but what I have tried (combine both equations and iterate for instance) was not enough to conlude
The answer is indeed no. Since $f(x),g(x)$ are positive and continuous for $x \in [0,1]$, the function $f(x)/g(x)$ is also positive and continuous for $x \in [0,1]$. Let $m$ be the greatest lower bound for $f(x)/g(x), x \in [0,1]$. A continuous function on a compact set achieves its minimum, so there is $x_0 \in [0,1]$ such that $m = f(x_0)/g(x_0)$. Further, since $m$ is a lower bound, we have $$\frac{f(x)}{g(x)} \ge m \,\,\,\,\,\,\, \implies \,\,\,\,\,\,\, f(x) -mg(x) \ge 0, \,\,\,\,\, x \in [0,1].$$ Assume that $f(x)-mg(x)$ is not identically zero on $[0,1]$. Then there is $y_0 \in [0,1]$ such that $f(y_0) - mg(y_0) > 0$. By continuity, there must be $\varepsilon > 0$ such that $f(y) - m g(y) >0 $ for all $y\in [y_0 - \varepsilon, y_0 + \varepsilon]$.
Let $T: C[0,1] \to C[0,1]$ be an operator defined by $$[Th](x) = \int^1_0 h(y) K(x,y) dy$$ for $h \in C[0,1]$. Then $T$ is a linear operator with $[Tf](x) = g(x), [Tg](x) = f(x)$. Also by positivity of $K(x,y)$, we have $$[T(f-mg)](x) =\int^1_0 (f(y) - mg(y)) K(x,y) dy \ge \int^{y_0+\varepsilon}_{y_0-\varepsilon} (f(y) - mg(y)) K(x,y) dy > 0$$ for all $x \in [0,1].$ But $[T(f-mg)](x) = g(x) - mf(x).$ So $g(x) - mf(x) >0, \, x\in [0,1].$ Applying the operator again, we see $$[T(g-mf)](x) = \int^1_0 (g(y) - mf(y)) K(x,y) dy > 0 $$ for all $x \in [0,1]$. But $[T(g-mf)](x) = f(x)-mg(x).$ So $f(x) - mg(x) > 0, x\in [0,1]$. However, this would imply that $f(x)/g(x) > m$ for all $x \in [0,1]$. This contradicts the existence of $x_0$ as defined above. Thus we conclude that our assumption was wrong and $f(x)-mg(x) = 0$ for all $x \in [0,1]$. Then $$f(x) = mg(x) \,\,\, \implies \,\,\, [Tf](x) = m[Tg](x) \,\,\, \implies \,\,\, g(x) = mf(x), \,\,\, x\in [0,1].$$ Adding the first and last equation from the above line yields $$(f(x) + g(x)) = m(f(x) + g(x)), \,\,\, x\in [0,1].$$ We can divide by the positive quantity $f(x)+g(x)$ to get $m=1$. Then $f(x) = mg(x) \implies f(x) = g(x), x\in[0,1].$