We have the integral Equation: $$y(x)=f(x)+λ\int_{a}^{b} K(x,t)y(t) dt ,x\in [a,b],K\in L^{2}([a,b]\times[a,b]),f\in L^{2}([a,b])$$ Prove that $R(x,t;\lambda)$ is a one-to-one function in $\lambda$.
I have tried to solve this, but I got stuck. I tried to prove that with the definition of bijective function: $$R(x,t;\lambda_{1})=R(x,t;\lambda_{2}) \Rightarrow \sum_{m=1}^{+\infty} \lambda_{1}K_{m}(x,t) =\sum_{m=1}^{+\infty} \lambda_{2}K_{m}(x,t) \Rightarrow \sum_{m=1}^{+\infty} (\lambda_{1}-\lambda_{2})K_{m}(x,t)=0 $$ (from uniform continuity)
At this moment I believe that the Cauchy-Scwartz theorem helps to finish the proof.