Compute the eigenvalues and eigenfunctions of $$ \varphi(x) - \lambda \int_0^1 e^{x+s} \varphi(s) ds = f(x) $$ Are there functions $f$ such that the inhomogenous equation has for every real $\lambda$ at least one solution?

Attempt at my solution: I have to solve $$ \int_0^1 e^{x+s} \varphi(s) ds = \frac{1}{\lambda} \varphi(x) $$ with $$ \int_0^1 e^{x+s} \varphi(s) ds = e^x \int_0^1 e^s \varphi(s) ds $$ we have the eigenvalue $1/\lambda = \int_0^1 e^s \varphi(s) ds$ and the eigenfunctions $$ \varphi(x) = C \cdot e^x $$ for some constant $C$. Is this correct? I just derived it with looking at the equation, is there some theory by which I get this result? And are that all eigenvalues/eigenfunctions? And how to solve the question for functions $f$ such that there always exists eigenfunctions for real $\lambda$?

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    $\begingroup$ Wait, what is exactly the operator of which you are searching the eigenvectors? Is it $\phi(x)\mapsto\phi(x)-\lambda\int_0^1e^{x+s}\phi(s)dx$? $\endgroup$ – Daniel Robert-Nicoud Jul 26 '14 at 23:40
  • $\begingroup$ I bet it is $$\phi(x)\to \int_{0}^{1}e^{x+s}\phi(s)\,ds$$. In this case, the only eigenfunctions are the constant functions and they all belong to the $\lambda=\frac{2}{e^2-1}$ eigenspace. $\endgroup$ – Jack D'Aurizio Jul 26 '14 at 23:46
  • $\begingroup$ Yes Jack D'Aurizio wrote the right operator. But why the constant functions? On the RHS we have $\int_0^1 e^{x+s} \varphi(s)ds = e^x \int_0^1 e^s\varphi(s) ds$ which is $e^x$ times a constant, so $\varphi(x)$ must also be $e^x$ times a constant, and not constant, or what is wrong with my reasoning? $\endgroup$ – StefanH Jul 26 '14 at 23:56
  • $\begingroup$ Nothing wrong, I mean "constant times $e^x$". $\endgroup$ – Jack D'Aurizio Jul 27 '14 at 1:42

Suppose $\int_{0}^{1}f(s)e^{s}\,ds=0$. Then your equation always has the solution $\varphi = f$ which holds for all $\lambda$. So this condition is sufficient. Actually, this condition is also necessary in order to have a solution for all real $\lambda$, but I'll leave that to you to see. (Hint: choose $\lambda = 1/\int_{0}^{1}e^{2x}\,dx$, and show that $h(x)=\varphi(x)-\lambda\int_{0}^{1}\varphi(s)e^{s+x}\,ds$ satisfies $\int_{0}^{1}h(x)e^{x}\,dx=0$.)


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