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Let us consider the following integral equation$$f(x) + \lambda \int_0^1 {K(s,x)f(s)ds = 0,{\text{ x}} \in {\text{(0}}{\text{,1)}}{\text{.}}} $$ I'm looking of the values of $\lambda$ so that the above equation has only $f=0$ as solution with a constant kernel. Suppose that $K(s,x)=K$, we obtain $$f(x) + \lambda K\int_0^1 {f(s)ds = 0,{\text{ x}} \in {\text{(0}}{\text{,1)}}{\text{.}}} $$ By taking the integral over $(0,1)$, we get $$(1 + \lambda K)\int_0^1 {f(s)ds = 0} $$. for all $f$. Now, if $\lambda$ is different of $-1/K$, then $$\int_0^1 {f(s)ds = 0} $$. I don't see how this can be helpful. Any suggestions?. Thank you.

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    $\begingroup$ The function has mean value 0 in integral sense over interval 0 to 1. It removes one degree of freedom. This means you have infinite set of solutions. Any function fulfilling the mean value equation $=0$ will do. $\endgroup$ – mathreadler Jan 4 at 15:48
  • $\begingroup$ So if $\lambda= - 1/K$ we have only one solution? $\endgroup$ – Gustave Jan 4 at 15:54
  • $\begingroup$ If the other factor is $0$ then it does not matter what $f$ is, since the product will always be $0$ so then all functions $f$ will satisfy it. $\endgroup$ – mathreadler Jan 4 at 16:01
  • $\begingroup$ Thanks. I understand, but what I can say about the uniqueness of the trivial solution with respect to $\lambda$? $\endgroup$ – Gustave Jan 4 at 16:09
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Your step of taking the integral is too crude, at least initially. When you have $$ f(x)+\lambda K\int_0^1f(s)\,ds=0, $$ you can write this as $$ f(x)=-\lambda K\int_0^1f(s)\,ds $$ to conclude that $f$ is constant. If $\lambda=0$, you get $f=0$. If $\lambda\ne0$ and $\lambda\ne-1/K$, your trick of integrating again gives you that $\int_0^1 f=0$, so $f=0$.

When $\lambda=-1/K$ the solution is not unique, as any constant $f$ will be a solution.

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